Prove that \(\mathbb{F}_p(x,y)/\mathbb{F}_p(x^p,y^p)\) is not a simple extension by explicitly exhibiting an infinite number of infinite subfields.
Let \(F = \mathbb{F}_p(x^p,y^p)\), and consider the extensions \(F(x^{pn+1}+y)\), where \(n \in \mathbb{Z}^+ \cup \{0\}\). They are degree \(p\) over \(F\) (due to the lone \(y\) term) and so are proper subfields of the degree \(p^2\) extension \(\mathbb{F}_p(x,y)\).
Suppose \(F(x^{pi+1}+y) = F(x^{p(i+k)+1}+y)\) where \(i \ge 0, k > 0\) (denote the field \(K\)). Then \(x^{p(i+k)+1}-x^{pi+1} = x^{pi+1}(x^{pk}-1) \in K\), and dividing by \(x^{pi}(x^{pk}-1)\) (an element of \(F\)) gives \(x \in K\). In particular, \(x^{pi+1} \in K\) so \(y \in K\), a clear contradiction. Hence the extensions \(F(x^{pn+1}+y)\) are all distinct, and there are infinitely many of them.
Friday, 19 June 2020
Wednesday, 10 June 2020
Galois group of \(\mathbb{Q}(\sqrt{2+\sqrt{2}})\) (D&F 14.2.14)
Show that \(\mathbb{Q}(\sqrt{2+\sqrt{2}})\) is a cyclic quartic field.
Let \(a_+ = \sqrt{2+\sqrt{2}}\) and \(a_- = \sqrt{2-\sqrt{2}}\). It can be seen that \(a_+\) is a root of the minimal polynomial \(p(x) = (x^2-2)^2-2 = x^4-4x^2+2\) (being irreducible by Eisenstein), and solving for all roots of \(p(x)\) gives \(\pm a_+, \pm a_-\). Also note that since \({a_+}^2-2 = \sqrt{2}\) and \({a_+}{a_-} = \sqrt{2}\ (*)\), it follows that \(a_- \in \mathbb{Q}(a_+)\), and so \(\mathbb{Q}(a_+)\) contains all roots of \(p(x)\). Since \(\mathbb{Q}(a_+)\) has degree 4 over \(\mathbb{Q}\), it is in fact the splitting field of \(p(x)\) and is thus Galois.
We need only find 4 automorphisms in \(\text{Gal}(\mathbb{Q}(a_+)/\mathbb{Q})\) and show that they form a cyclic group. Note that the automorphisms are fully determined by their action on \(a_+\) (due to the starred expression above). Let \(\varphi\) be such an automorphism; then \(\varphi(a_+)\varphi(a_-) = \varphi(a_+)^2-2\), so \(\varphi(a_-) = (\varphi(a_+)^2-2)/\varphi(a_+)\). If \(\varphi(a_+) = a_+\), then clearly \(\varphi(a_-) = a_-\). If \(\varphi(a_+) = a_-\), then \(\varphi(a_-) = -a_-\). If \(\varphi(a_+) = a_-\), then a manual calculation gives \(\varphi(a_-) = -a_+\). Lastly if \(\varphi(a_+) = -a_-\), then we similarly get \(\varphi(a_-) = a_+\).
Associate \(a_+, -a_+, a_-, -a_-\) with the numbers \(1,2,3,4\) in order. Then the automorphisms listed above correspond to the permutations 1, \((1\ 2)(3\ 4)\), \((1\ 3\ 2\ 4)\) and \((1\ 4\ 2\ 3)\). It can be checked that \((1\ 3\ 2\ 4)\) generates this set and thus the conclusion follows.
Galois group of \(x^4-14x^2+9 \in \mathbb{Q}[x]\) (D&F 14.2.12)
Determine the Galois group of the splitting field over \(\mathbb{Q}\) of \(x^4-14x^2+9\).
We can solve for the roots explicitly: they are \(\pm\sqrt{7 \pm \sqrt{40}}\). For convenience, let \(a_+ = \sqrt{7+\sqrt{40}}\) and \(a_- = \sqrt{7-\sqrt{40}}\), so that the roots are \(\pm a_+, \pm a_-\). Since \(x^4-14x^2+9\) is separable, the splitting field is Galois over \(\mathbb{Q}\), i.e. there are exactly 4 automorphisms fixing \(\mathbb{Q}\).
Observe that for any automorphism \(\varphi\), we have \(\varphi(a_+)\varphi(a_-) = \varphi(\sqrt{49-40}) = 3\). Hence the action of \(\varphi\) on \(a_+\) determines its action on \(a_-\), and hence the remaining roots. In particular, the automorphisms induced by mapping \(a_+\) to each of the 4 roots exhaust the Galois group of the splitting field. It can then be verified that the group of automorphisms is isomorphic to \(V_4\).
Simple extensions in splitting fields with Galois group \(S_4\) (D&F 14.2.11)
Suppose \(f(x) \in \mathbb{Z}[x]\) is an irreducible quartic whose splitting field has Galois group \(S_4\) over \(\mathbb{Q}\). Let \(\theta\) be a root of \(f(x)\) and set \(K = Q(\theta)\). Prove that \(K\) is an extension of \(\mathbb{Q}\) of degree 4 which has no proper subfields.
Let the splitting field be \(L\). We have \(\text{Gal}(L/K) \cong S_3\) as one root is fixed, but the other 3 roots can be freely permuted. If \(M \subset K\) is a proper subfield, it must have degree 2 and hence \(\text{Gal}(L/M) \cong A_4\), since \(A_4\) is the unique index 2 subgroup in \(S_4\). However, we show that this leads to a contradiction as \(A_4\) does not contain an isomorphic copy. Simply, note that there are 3 elements of order 2 in both \(S_3\) and \(A_4\), hence any embedding of \(S_3\) and \(A_4\) gives a one-to-one correspondence between these. But only the order 2 elements of \(A_4\), together with the identity, form a subgroup (namely \(Z_4\)), meaning there is no such embedding.
Let the splitting field be \(L\). We have \(\text{Gal}(L/K) \cong S_3\) as one root is fixed, but the other 3 roots can be freely permuted. If \(M \subset K\) is a proper subfield, it must have degree 2 and hence \(\text{Gal}(L/M) \cong A_4\), since \(A_4\) is the unique index 2 subgroup in \(S_4\). However, we show that this leads to a contradiction as \(A_4\) does not contain an isomorphic copy. Simply, note that there are 3 elements of order 2 in both \(S_3\) and \(A_4\), hence any embedding of \(S_3\) and \(A_4\) gives a one-to-one correspondence between these. But only the order 2 elements of \(A_4\), together with the identity, form a subgroup (namely \(Z_4\)), meaning there is no such embedding.
Wednesday, 3 June 2020
Automorphisms of a rational function field (D&F 14.1.8)
Prove that the automorphisms of the rational function field \(k(t)\) which fix \(k\) are precisely the fractional linear transformations determined by \(t \rightarrow \frac{at+b}{ct+d}\) for \(a,b,c,d \in k\), \(ad-bc \neq 0\).
It is easy to verify that fractional linear transformations are automorphisms of \(k(t)\) fixing \(k\). Conversely, let \(\varphi \in \text{Aut}(k(t)/k)\). Then \(k(\varphi(t)) = \varphi(k(t)) = k(t)\) (1st equality follows from the definition of a homomorphism, 2nd equality follows from surjectivity), so that \([k(t) : k(\varphi(t))] = 1\). If \(\varphi(t) = p(t)/q(t)\) for \(p(t),q(t) \in k[t]\), where \(p(t),q(t)\) are relatively prime and \(q(t) \neq 0\), then by 13.2.18 the LHS is equal to \(\max (\deg p, \deg q)\). Thus \(p(t) = at+b\) and \(q(t) = ct+d\) for \(a,b,c,d \in k\). Lastly, if \(ad-bc = 0\), then \(a/c = x = b/d\) for some \(x \in k\) and so \(p(t) = xq(t)\), contradicting the fact that they are relatively prime. Notice in particular that we cannot have \(a = c = 0\).
It is easy to verify that fractional linear transformations are automorphisms of \(k(t)\) fixing \(k\). Conversely, let \(\varphi \in \text{Aut}(k(t)/k)\). Then \(k(\varphi(t)) = \varphi(k(t)) = k(t)\) (1st equality follows from the definition of a homomorphism, 2nd equality follows from surjectivity), so that \([k(t) : k(\varphi(t))] = 1\). If \(\varphi(t) = p(t)/q(t)\) for \(p(t),q(t) \in k[t]\), where \(p(t),q(t)\) are relatively prime and \(q(t) \neq 0\), then by 13.2.18 the LHS is equal to \(\max (\deg p, \deg q)\). Thus \(p(t) = at+b\) and \(q(t) = ct+d\) for \(a,b,c,d \in k\). Lastly, if \(ad-bc = 0\), then \(a/c = x = b/d\) for some \(x \in k\) and so \(p(t) = xq(t)\), contradicting the fact that they are relatively prime. Notice in particular that we cannot have \(a = c = 0\).
Tuesday, 2 June 2020
Algebraic extensions: Degrees (part 2)
We start by mentioning the following result: If a field extension \(K/F\) has degree \(n\), then any \(\alpha \in K\) satisfies a polynomial with degree \(\le n\). Observe that if \(K = F(\alpha)\) then equality always holds, but this need not always be the case. For example, the extension \(\mathbb{Q}(\sqrt[4]{2})/\mathbb{Q}\) is of degree 4, but the element \(\sqrt{2}\) satisfies a degree 2 (< 4) polynomial.
The next major result is that \(F \subseteq K \subseteq L\) then \([L:F] = [L:K][K:F]\). Intuitively, a basis for the vector space of \(L\) over \(F\) is \(\alpha_i \beta_j\), where \(\alpha_i\) and \(\beta_j\) run through bases of \(L\) over \(K\) and \(K\) over \(F\) respectively (one can check this by manual expansion). A special case is where \(K = F(\alpha)\) and \(L = F(\alpha,\beta) = F(\alpha)(\beta)\). The degree \(d\) of \(\beta\) over \(F(\alpha)\) is \(\le\) its degree over \(F\), and an \(F\)-basis for \(L\) is \(\alpha^i \beta^j\), where \(1 \le i \le [K:F]\) and \(1 \le j \le d\).
This leads to the following result: \(K/F\) is finite iff it is generated by finitely many algebraic elements \(\alpha_i\), and its degree is \(\le \prod_{i} \text{deg}\ \alpha_i\). This is useful in proving that the set of algebraic elements of \(K/F\) form a subfield (in particular an algebraic extension of \(F\)). For instance, the set of algebraic elements over \(\mathbb{Q}\) in \(\mathbb{C}\) form an infinite extension of \(\mathbb{Q}\).
Sunday, 31 May 2020
Product of primitive roots of unity (D&F 13.6.1)
Suppose \(m\) and \(n\) are relatively prime positive integers. Let \(\zeta_m\) and \(\zeta_n\) be primitive \(m\)th and \(n\)th roots of unity. Prove that \(\zeta_m \zeta_n\) is a primitive \(mn\)th root of unity.
This is essentially proving that if \(a\) and \(b\) are integers such that \((a,m) = (b,n) = 1\) then \((an+bm,mn) = 1\) (if the conclusion holds then \((an+bm-kmn,mn) = 1\) for any integer \(k\)). Note that any prime divisor \(d\) of \(an+bm\) cannot appear in the prime factorization of \(m\) or \(n\). For instance if \(d \mid m\) then it follows that \(d \mid an\), and since \((m,n) = 1\) we have \(d \mid a\), a contradiction. Thus the prime factorization of \(an+bm\) is disjoint from that of \(mn\), and the conclusion follows.
This is essentially proving that if \(a\) and \(b\) are integers such that \((a,m) = (b,n) = 1\) then \((an+bm,mn) = 1\) (if the conclusion holds then \((an+bm-kmn,mn) = 1\) for any integer \(k\)). Note that any prime divisor \(d\) of \(an+bm\) cannot appear in the prime factorization of \(m\) or \(n\). For instance if \(d \mid m\) then it follows that \(d \mid an\), and since \((m,n) = 1\) we have \(d \mid a\), a contradiction. Thus the prime factorization of \(an+bm\) is disjoint from that of \(mn\), and the conclusion follows.
Friday, 29 May 2020
Composite and intersection of splitting fields (D&F 13.4.6)
Let \(K_1\) and \(K_2\) be finite extensions of \(F\) contained in the field \(K\), and assume both are splitting fields over \(F\).
(a) Prove that their composite \(K_1K_2\) is a splitting field over \(F\).
(b) Prove that \(K_1 \cap K_2\) is a splitting field over \(F\).
(a) \(K_1\) and \(K_2\) are splitting fields over sets of polynomials \(I\) and \(J\) respectively, whose roots make up sets \(A\) and \(B\). Note that \(K_1 = F(A)\) and \(K_2 = F(B)\). Let \(P\) be the set of all possible products of the polynomials; then \(K_1K_2\) contains \(A\) and \(B\). Furthermore, there cannot be a proper subfield \(F \subseteq K \subset K_1K_2\) with this property. Otherwise, \(K\) would also contain \(A\) and \(B\), and hence \(F(A)\) and \(F(B)\), contradicting the fact that \(K_1K_2\) is the smallest field containing \(F(A)\) and \(F(B)\).
(b) Let \(f(x)\) be an irreducible polynomial in \(F[x]\) with a root \(\alpha \in K_1 \cap K_2\). If \(f\) is linear, we are done. Otherwise, since \(f(x)\) splits completely in \(K_1[x]\) we can write \(f(x) = (x-\alpha)(x-\beta)g(x)\) for \(\beta \in K_1\). Since \(K_2(\alpha) = K_2 \cong K_2(\beta)\), it follows that \(\beta \in K_2\) as well. By induction we can prove similarly for irreducible factors of \(g(x)\), so that \(f(x)\) factors completely in \((K_1 \cap K_2)[x]\).
(a) Prove that their composite \(K_1K_2\) is a splitting field over \(F\).
(b) Prove that \(K_1 \cap K_2\) is a splitting field over \(F\).
(a) \(K_1\) and \(K_2\) are splitting fields over sets of polynomials \(I\) and \(J\) respectively, whose roots make up sets \(A\) and \(B\). Note that \(K_1 = F(A)\) and \(K_2 = F(B)\). Let \(P\) be the set of all possible products of the polynomials; then \(K_1K_2\) contains \(A\) and \(B\). Furthermore, there cannot be a proper subfield \(F \subseteq K \subset K_1K_2\) with this property. Otherwise, \(K\) would also contain \(A\) and \(B\), and hence \(F(A)\) and \(F(B)\), contradicting the fact that \(K_1K_2\) is the smallest field containing \(F(A)\) and \(F(B)\).
(b) Let \(f(x)\) be an irreducible polynomial in \(F[x]\) with a root \(\alpha \in K_1 \cap K_2\). If \(f\) is linear, we are done. Otherwise, since \(f(x)\) splits completely in \(K_1[x]\) we can write \(f(x) = (x-\alpha)(x-\beta)g(x)\) for \(\beta \in K_1\). Since \(K_2(\alpha) = K_2 \cong K_2(\beta)\), it follows that \(\beta \in K_2\) as well. By induction we can prove similarly for irreducible factors of \(g(x)\), so that \(f(x)\) factors completely in \((K_1 \cap K_2)[x]\).
Thursday, 28 May 2020
Elements of field extensions as matrices (D&F 13.2.19)
Let \(K\) be an extension of \(F\) of degree \(n\).
(a) For any \(\alpha \in K\) prove that \(\alpha\) acting by left multiplication on \(K\) is an \(F\)-linear transformation of \(K\).
(b) Prove that \(K\) is isomorphic to a subfield of the ring \(M_n(F)\), so \(M_n(F)\) contains an isomorphic copy of every extension of \(F\) of degree \(\le n\).
(a) Trivial.
(b) Each element \(\alpha \in K\) can be associated with a unique matrix \(M_\alpha \in M_n(F)\) as follows: let \(k_1, \cdots, k_n\) be a basis for \(K\) over \(F\). For each \(k_i\), we can write \(k_i \alpha = m_{i1}k_1 + \cdots m_{in}k_n\), where \(m_{ij} \in F\). Then let \((M_\alpha)_{ij} = m_{ij}\). Now we show that this is an isomorphism.
Let \(\alpha, \beta \in K\), \(m_{ij} = (M_\alpha)_{ij}\) and \(n_{ij} = (M_\beta)_{ij}\). Then \(k_i(\alpha+\beta) = \sum_{j} m_{ij}k_j + \sum_{j} n_{ij}k_j = \sum_{j} (m_{ij}+n_{ij})k_j = \sum_{j} (M_\alpha + M_\beta)_{ij}k_j\), so the map is additive.
Now for simplicity's sake assume that \(k_1 = 1\), so \(\beta = \sum_{j} n_{1j}k_j\) (hopefully it works). Thus \(k_i(\alpha\beta) = (\sum_{j} m_{ij}k_j)(\sum_{j} n_{1j}k_j) = \sum_{a,b} m_{ia}k_a(n_{1b}k_b) = \sum_{a,b} m_{ia}n_{ab}k_b\) \(= \sum_b (\sum_{a} m_{ia}n_{ab}) k_b = \sum_{b} (M_{\alpha}M_{\beta})_{ib}k_b\).
(a) For any \(\alpha \in K\) prove that \(\alpha\) acting by left multiplication on \(K\) is an \(F\)-linear transformation of \(K\).
(b) Prove that \(K\) is isomorphic to a subfield of the ring \(M_n(F)\), so \(M_n(F)\) contains an isomorphic copy of every extension of \(F\) of degree \(\le n\).
(a) Trivial.
(b) Each element \(\alpha \in K\) can be associated with a unique matrix \(M_\alpha \in M_n(F)\) as follows: let \(k_1, \cdots, k_n\) be a basis for \(K\) over \(F\). For each \(k_i\), we can write \(k_i \alpha = m_{i1}k_1 + \cdots m_{in}k_n\), where \(m_{ij} \in F\). Then let \((M_\alpha)_{ij} = m_{ij}\). Now we show that this is an isomorphism.
Let \(\alpha, \beta \in K\), \(m_{ij} = (M_\alpha)_{ij}\) and \(n_{ij} = (M_\beta)_{ij}\). Then \(k_i(\alpha+\beta) = \sum_{j} m_{ij}k_j + \sum_{j} n_{ij}k_j = \sum_{j} (m_{ij}+n_{ij})k_j = \sum_{j} (M_\alpha + M_\beta)_{ij}k_j\), so the map is additive.
Now for simplicity's sake assume that \(k_1 = 1\), so \(\beta = \sum_{j} n_{1j}k_j\) (hopefully it works). Thus \(k_i(\alpha\beta) = (\sum_{j} m_{ij}k_j)(\sum_{j} n_{1j}k_j) = \sum_{a,b} m_{ia}k_a(n_{1b}k_b) = \sum_{a,b} m_{ia}n_{ab}k_b\) \(= \sum_b (\sum_{a} m_{ia}n_{ab}) k_b = \sum_{b} (M_{\alpha}M_{\beta})_{ib}k_b\).
Field \(\subseteq\) ring \(\subseteq\) field (D&F 13.2.16)
Let \(K/F\) be an algebraic extension and let \(R\) be a ring contained in \(K\) and containing \(F\). Show that \(R\) is a subfield of \(K\) containing \(F\).
By virtue of \(R\) being a subring of \(K\), many properties are automatically satisfied, such as the ring axioms, commutativity and 'integral-domain-ness'. The only remaining property to check is closure under inverses.
Let \(\alpha\) be a nonzero element of \(R\). Since \(\alpha\) is algebraic, its degree over \(F\) is finite. Each element of \(F(\alpha)\) is the form \(a_0 + a_1\alpha + \cdots + a_{n-1}\alpha^{n-1}\), where \(n\) is the degree of \(\alpha\) over \(F\). Thus \(F(\alpha) \subseteq R\) and so \(R\) contains \(\alpha^{-1}\) since \(F(\alpha)\) is a field.
Irreducible factors of composite polynomials (D&F 13.2.17)
Let \(f(x) \in F[x]\) be an irreducible degree-\(n\) polynomial, and \(g(x) \in F[x]\). Prove that every irreducible factor of the composite polynomial \(f(g(x))\) has degree divisible by \(n\).
If \(\alpha\) be some root of \(f(x)\), then the field extension \(F(\alpha)/F\) has degree \(n\). If \(\beta\) is a root of \(f(g(x))\), then \(\beta\) must also be a root of the polynomial \(g(x)-\alpha \in F(\alpha)[x]\). Thus the degree of the extension \(F(\alpha,\beta)/F\) is divisible by \(n\).
Now observe the following chain: \(F \subseteq_a F(\beta) \subseteq_b F(\beta)(\alpha)\). Note that \(\beta\) is a root of some irreducible factor \(p(x)\) of \(f(g(x))\), thus \(a = \text{deg}\ p\). Conversely for every irreducible factor \(p(x)\) we can find a root \(\beta\) such that \([F(\beta) : F] = \text{deg}\ p\). Suppose the degree of some irreducible factor \(p(x)\) is not divisible by \(n\). Then \(b\) divides \(n\), since \(ab\) divides \(n\) (as shown in the previous paragraph). Furthermore, \(b\) is the degree of the minimal polynomial of \(\alpha\) in \(F(\beta)[x]\).
The above statement implies that if we can exhibit a degree \(n\) polynomial in \(F(\beta)[x]\) with \(\alpha\) as a root, then it must be a unit-multiple of the minimal polynomial, meaning it is irreducible. But \(f(x)\) is precisely such a polynomial. However, since \(g(\beta) = \alpha\) is a root of \(f(x)\), \(f(x)\) has a linear factor \((x-g(\beta))\), contradicting its irreducibility (it is critical that we are working in \(F(\beta)[x]\) and not just \(F[x]\)). Thus \(b\) cannot divide \(n\) and the conclusion follows.
Condition for a biquadratic extension (D&F 13.2.8)
Let \(F\) be a field of characteristic \(\neq 2\). Let \(D_1\) and \(D_2\) be elements of \(F\), neither of which is a square of \(F\). Prove that \(F(\sqrt{D_1}, \sqrt{D_2})\) is of degree 4 over \(F\) if \(D_1D_2\) is not a square in \(F\) and is of degree 2 over \(F\) otherwise (when \(F(\sqrt{D_1}, \sqrt{D_2})\) is of degree 4 over \(F\) it is called a biquadratic extension of \(F\)).
Remark: note that the condition \(\text{chr}(F) \neq 2\) is required, to exclude e.g. \(\mathbb{F}_2\), all of whose elements are squares.
Let \(F' = F(\sqrt{D_1D_2})\) and \(F'' = F(\sqrt{D_1},\sqrt{D_2})\). Consider the chain \(F \subseteq_{n} F' \subseteq_{m} F''\), where \(n\) and \(m\) denote the respective degrees. Note the second inclusion is valid since each element of \(F'\) can be written in the form \(a+b\sqrt{D_1D_2}\) for \(a,b \in F\). Furthermore, \(m > 1\) because \(\sqrt{D_1}\) cannot be expressed in that form (otherwise \(\sqrt{D_2}^{-1} \in F\) which is a contradiction).
\(D_1D_2\) not being a square in \(F\) means \(n = 2\), forcing \(nm = 4\) as \(nm \le 4\) and \(m > 1\). On the other hand, \(nm \neq 2\) forces \(nm = 4\) (as \(nm = 1\) is not a valid possibility), so \(n = 2\) and \(D_1D_2\) is not a square in \(F\).
Remark: note that the condition \(\text{chr}(F) \neq 2\) is required, to exclude e.g. \(\mathbb{F}_2\), all of whose elements are squares.
Let \(F' = F(\sqrt{D_1D_2})\) and \(F'' = F(\sqrt{D_1},\sqrt{D_2})\). Consider the chain \(F \subseteq_{n} F' \subseteq_{m} F''\), where \(n\) and \(m\) denote the respective degrees. Note the second inclusion is valid since each element of \(F'\) can be written in the form \(a+b\sqrt{D_1D_2}\) for \(a,b \in F\). Furthermore, \(m > 1\) because \(\sqrt{D_1}\) cannot be expressed in that form (otherwise \(\sqrt{D_2}^{-1} \in F\) which is a contradiction).
\(D_1D_2\) not being a square in \(F\) means \(n = 2\), forcing \(nm = 4\) as \(nm \le 4\) and \(m > 1\). On the other hand, \(nm \neq 2\) forces \(nm = 4\) (as \(nm = 1\) is not a valid possibility), so \(n = 2\) and \(D_1D_2\) is not a square in \(F\).
Wednesday, 27 May 2020
Algebraic extensions: The very basics (part 1)
Let \(F\) be a field. The degree of a field extension \(E/F\), denoted \([E : F]\), is the dimension of \(E\) as a vector space over \(F\). For example \([\mathbb{C} : \mathbb{R}] = 2\), where a basis is \(\{1,i\}\). Note that it is not necessarily finite, and finding an explicit basis is not always possible.
There is a particular type of field extension whose aforementioned properties are known. Consider an irreducible polynomial \(p(x) \in F[x]\). Then we can construct a field \(F'\) containing a root \(\alpha\), and an isomorphic copy of \(F\). In particular, we let \(F'\) be the quotient \(F[x]/(p(x))\), whose elements can be represented by polynomials of degree \(< \text{deg}
\ p\).
Note that the condition that \(p(x)\) is irreducible is required for this quotient to be a field. Furthermore, it forces the \(p(x)\) to be of minimal degree. (If irreducible polynomials \(a(x),b(x)\) have the same root \(\alpha\), but \(a(x)\) is of minimal degree while \(b(x)\) is not, we obtain a contradiction from dividing \(b(x)\) by \(a(x)\) and considering the remainder.)
If we impose the further condition that \(p(x)\) be monic, then we obtain the minimal polynomial \(m_{\alpha,F}(x)\) which is unique. In particular, \([F' : F] = \deg m_{\alpha,F}\) (where \(F\) is identified with its isomorphic copy). Also, the element \(\bar{x} \in F'\) is the desired root of \(p(x)\), and \(F'\) has basis \(\{1,\bar{x},\cdots,\bar{x}^{n-1}\}\), where \(n = \deg m_{\alpha,F}\). Furthermore, the image of \(F\) under the projection is precisely the isomorphic copy of \(F\).
As an example, the field \(F = \mathbb{R}[x]/(x^2+1)\) contains a root \(\alpha\) to the polynomial \(x^2+1\). The degree of \(F\) over \(\mathbb{R}\) is 2, and every element of \(F\) can be represented in the form \(a+b\alpha\), where \(a,b \in \mathbb{R}\). \(F\) contains a copy of \(\mathbb{R}\), namely the set of elements with \(b=0\). The rules of complex addition and multiplication can then be derived via usual addition and multiplication of polynomials, mod \(\alpha^2+1\).
If we know a priori some field extension \(E/F\) containing a root \(\alpha\) to \(p(x)\), then we use \(F(\alpha)\) to denote the smallest subfield containing \(F\) and \(\alpha\). One key isomorphism which arises is \(F(\alpha) \cong F[x]/(p(x))\), so it does not matter whether we construct a field containing a root, or simply take an existing one. Taking the abovementioned example, we have \(\mathbb{C} \cong \mathbb{R}(i) \cong \mathbb{R}[x]/(x^2+1)\).
Now, suppose there are 2 distinct roots \(\alpha,\beta\) of \(p(x) \in F[x]\) in some field \(F'\). Then \(F(\alpha) \cong F[x]/(p(x)) \cong F(\beta)\), i.e. the roots are algebraically indistinguishable. Furthermore, there is a canonical isomorphism \(F(\alpha) \rightarrow F(\beta)\) mapping \(a_0+a_1\alpha+\cdots+a_n\alpha^n\) to \(a_0+a_1\beta+\cdots+a_n\beta^n\). For example, the map \(a+bi \rightarrow a-bi\) is an isomorphism.
This isomorphism can be generalized to the case where the fields \(F\) and \(F'\) where \(\alpha\) and \(\beta\) are adjoined to are isomorphic. Let \(\varphi : F \rightarrow F'\) be an isomorphism. Then \(\varphi\) extends naturally to an isomorphism \(F[x] \rightarrow F'[x]\) and thus we have \(F[x]/(p(x)) \cong F'[x]/(p(x))\). So we can conclude that \(F(\alpha) \cong F'(\beta)\). This fact is useful later on in proving the uniqueness of splitting fields up to isomorphism.
There is a particular type of field extension whose aforementioned properties are known. Consider an irreducible polynomial \(p(x) \in F[x]\). Then we can construct a field \(F'\) containing a root \(\alpha\), and an isomorphic copy of \(F\). In particular, we let \(F'\) be the quotient \(F[x]/(p(x))\), whose elements can be represented by polynomials of degree \(< \text{deg}
\ p\).
Note that the condition that \(p(x)\) is irreducible is required for this quotient to be a field. Furthermore, it forces the \(p(x)\) to be of minimal degree. (If irreducible polynomials \(a(x),b(x)\) have the same root \(\alpha\), but \(a(x)\) is of minimal degree while \(b(x)\) is not, we obtain a contradiction from dividing \(b(x)\) by \(a(x)\) and considering the remainder.)
If we impose the further condition that \(p(x)\) be monic, then we obtain the minimal polynomial \(m_{\alpha,F}(x)\) which is unique. In particular, \([F' : F] = \deg m_{\alpha,F}\) (where \(F\) is identified with its isomorphic copy). Also, the element \(\bar{x} \in F'\) is the desired root of \(p(x)\), and \(F'\) has basis \(\{1,\bar{x},\cdots,\bar{x}^{n-1}\}\), where \(n = \deg m_{\alpha,F}\). Furthermore, the image of \(F\) under the projection is precisely the isomorphic copy of \(F\).
As an example, the field \(F = \mathbb{R}[x]/(x^2+1)\) contains a root \(\alpha\) to the polynomial \(x^2+1\). The degree of \(F\) over \(\mathbb{R}\) is 2, and every element of \(F\) can be represented in the form \(a+b\alpha\), where \(a,b \in \mathbb{R}\). \(F\) contains a copy of \(\mathbb{R}\), namely the set of elements with \(b=0\). The rules of complex addition and multiplication can then be derived via usual addition and multiplication of polynomials, mod \(\alpha^2+1\).
If we know a priori some field extension \(E/F\) containing a root \(\alpha\) to \(p(x)\), then we use \(F(\alpha)\) to denote the smallest subfield containing \(F\) and \(\alpha\). One key isomorphism which arises is \(F(\alpha) \cong F[x]/(p(x))\), so it does not matter whether we construct a field containing a root, or simply take an existing one. Taking the abovementioned example, we have \(\mathbb{C} \cong \mathbb{R}(i) \cong \mathbb{R}[x]/(x^2+1)\).
Now, suppose there are 2 distinct roots \(\alpha,\beta\) of \(p(x) \in F[x]\) in some field \(F'\). Then \(F(\alpha) \cong F[x]/(p(x)) \cong F(\beta)\), i.e. the roots are algebraically indistinguishable. Furthermore, there is a canonical isomorphism \(F(\alpha) \rightarrow F(\beta)\) mapping \(a_0+a_1\alpha+\cdots+a_n\alpha^n\) to \(a_0+a_1\beta+\cdots+a_n\beta^n\). For example, the map \(a+bi \rightarrow a-bi\) is an isomorphism.
This isomorphism can be generalized to the case where the fields \(F\) and \(F'\) where \(\alpha\) and \(\beta\) are adjoined to are isomorphic. Let \(\varphi : F \rightarrow F'\) be an isomorphism. Then \(\varphi\) extends naturally to an isomorphism \(F[x] \rightarrow F'[x]\) and thus we have \(F[x]/(p(x)) \cong F'[x]/(p(x))\). So we can conclude that \(F(\alpha) \cong F'(\beta)\). This fact is useful later on in proving the uniqueness of splitting fields up to isomorphism.
Tuesday, 26 May 2020
Index of a subgroup's normalizer (FGT 1A.10)
Let \(H \subseteq G\).
(a) Show that \(\left|N_G(H) : H\right|\) is equal to the number of left cosets of \(H\) in \(G\) that are invariant under left multiplication of \(H\).
(b) Suppose that \(\left|H\right|\) is a power of the prime \(p\) and that \(\left|G : H\right|\) is divisible by \(p\). Show that \(\left|N_G(H) : H\right|\) is divisible by \(p\).
(a) \(h(aH) = aH\) for all \(h \in H\) \(\iff\) \(a^{-1}ha \in H\) \(\iff\) \(a^{-1}\), hence \(a\) in \(N_G(H)\) \(\iff\) \(aH \subseteq N_G(H)\)
(b) Let \(H\) act on the left cosets of \(H\) by left multiplication. Then the orbit of the coset \(1H\) has size 1. But all orbit sizes must be of some \(p\)-power (by the condition imposed on \(\left|H\right|\)) and sum up to \(\left|G : H\right|\) which is itself a \(p\)-multiple, so the number of size-1 orbits must be divisible by \(p\). The elements in these orbits are precisely the left cosets invariant under left multiplication by \(H\), so by (a) the conclusion follows.
(a) Show that \(\left|N_G(H) : H\right|\) is equal to the number of left cosets of \(H\) in \(G\) that are invariant under left multiplication of \(H\).
(b) Suppose that \(\left|H\right|\) is a power of the prime \(p\) and that \(\left|G : H\right|\) is divisible by \(p\). Show that \(\left|N_G(H) : H\right|\) is divisible by \(p\).
(a) \(h(aH) = aH\) for all \(h \in H\) \(\iff\) \(a^{-1}ha \in H\) \(\iff\) \(a^{-1}\), hence \(a\) in \(N_G(H)\) \(\iff\) \(aH \subseteq N_G(H)\)
(b) Let \(H\) act on the left cosets of \(H\) by left multiplication. Then the orbit of the coset \(1H\) has size 1. But all orbit sizes must be of some \(p\)-power (by the condition imposed on \(\left|H\right|\)) and sum up to \(\left|G : H\right|\) which is itself a \(p\)-multiple, so the number of size-1 orbits must be divisible by \(p\). The elements in these orbits are precisely the left cosets invariant under left multiplication by \(H\), so by (a) the conclusion follows.
Friday, 22 May 2020
The ring \(\mathbb{Q}[x,y,z]/(xy-z^2)\) (D&F 9.1.12)
Let bars denote passage to \(\mathbb{Q}[x,y,z]/(xy-z^2)\). Prove that \(\bar{P} = (\bar{x}, \bar{z})\) is a prime ideal. Show that \(\bar{xy} \in \bar{P}^2\) but that no power of \(\bar{y}\) lies in \(\bar{P}\). (This shows \(\bar{P}\) is a prime ideal whose square is not a primary ideal.)
Let \(I = (xy-z^2)\). For any coset \(r+I\) (where \(r\) is any representative), define \(\varphi(r+I)\) as \(r\) but with all terms containing \(x\) and \(z\) removed. Note that this map is well defined as for any \(k \in \mathbb{Q}[x,y,z]\), we have \(\varphi(r+k(xy-z^2)) = \varphi(r)\). Furthermore, it can be seen that \(\varphi\) is a ring homomorphism to \(\mathbb{Q}[y]\), an integral domain. Thus \((x+I, z+I) = \ker \varphi\) is a prime ideal.
As for the second part, \(z^2+I\) = \((z^2+xy-z^2)+I\) = \(xy+I\). The latter statement is evident.
Let \(I = (xy-z^2)\). For any coset \(r+I\) (where \(r\) is any representative), define \(\varphi(r+I)\) as \(r\) but with all terms containing \(x\) and \(z\) removed. Note that this map is well defined as for any \(k \in \mathbb{Q}[x,y,z]\), we have \(\varphi(r+k(xy-z^2)) = \varphi(r)\). Furthermore, it can be seen that \(\varphi\) is a ring homomorphism to \(\mathbb{Q}[y]\), an integral domain. Thus \((x+I, z+I) = \ker \varphi\) is a prime ideal.
As for the second part, \(z^2+I\) = \((z^2+xy-z^2)+I\) = \(xy+I\). The latter statement is evident.
Monday, 18 May 2020
A maximal ideal wrt not containing a finitely generated ideal (D&F 7.4.35)
Let \(A = (a_1, \ldots, a_n)\) be a nonzero finitely generated ideal of \(R\). Prove that there is an ideal \(B\) which is maximal with respect to the property that it does not contain \(A\).
Let \(S\) be the set of ideals not containing \(A\). If \(C\) is a chain in \(S\), then define \(J = \cup_{I \in C} I\). Suppose that \(A \subseteq J\); then in particular each \(a_i\) is in some \(I_i \in C\). Thus the ideal \(I' = \cup_{1 \leq i \leq n} I_i\) contains each \(a_i\) and hence \(A\) itself. But this is a contradiction since \(I' \in C \subseteq S\). Thus \(J\) does not contain \(A\) and is an upper bound for \(C\). By Zorn's Lemma, there exists a maximal ideal \(B\) with respect to not containing \(A\).
Let \(S\) be the set of ideals not containing \(A\). If \(C\) is a chain in \(S\), then define \(J = \cup_{I \in C} I\). Suppose that \(A \subseteq J\); then in particular each \(a_i\) is in some \(I_i \in C\). Thus the ideal \(I' = \cup_{1 \leq i \leq n} I_i\) contains each \(a_i\) and hence \(A\) itself. But this is a contradiction since \(I' \in C \subseteq S\). Thus \(J\) does not contain \(A\) and is an upper bound for \(C\). By Zorn's Lemma, there exists a maximal ideal \(B\) with respect to not containing \(A\).
A minimal prime ideal (D&F 7.4.36)
Assume \(R\) is commutative. Prove that the set of prime ideals in \(R\) has a minimal element with respect to inclusion (possibly the zero ideal).
The prime ideals form a partially ordered set \(S\) by inclusion. If \(C\) is a chain in \(S\), then define \(P = \cap_{I \in C} I\). Suppose \(ab \in P\) and \(a,b \not\in P\). Then there is some \(I_a, I_b \in C\) such that \(a \not\in I_a\) and \(b \not\in I_b\). If \(I_a \subseteq I_b\), then \(b \not\in I_a\), contradicting the fact that \(I_a\) is a prime ideal (likewise if \(I_b \subseteq I_a\)). Thus \(P\) is a prime ideal and a lower bound for \(C\). By Zorn's Lemma, there exists a minimal prime ideal.
The prime ideals form a partially ordered set \(S\) by inclusion. If \(C\) is a chain in \(S\), then define \(P = \cap_{I \in C} I\). Suppose \(ab \in P\) and \(a,b \not\in P\). Then there is some \(I_a, I_b \in C\) such that \(a \not\in I_a\) and \(b \not\in I_b\). If \(I_a \subseteq I_b\), then \(b \not\in I_a\), contradicting the fact that \(I_a\) is a prime ideal (likewise if \(I_b \subseteq I_a\)). Thus \(P\) is a prime ideal and a lower bound for \(C\). By Zorn's Lemma, there exists a minimal prime ideal.
Saturday, 16 May 2020
Order of conductor \(f\) in \(\mathbb{Q}(\sqrt{D})\) (D&F 7.1.23)
Let \(\mathcal{O}\) be the quadratic ring of integers in \(\mathbb{Q}(\sqrt{D})\). For any positive integer \(f\) prove that the set \(\mathcal{O}_f = \{a + bf\omega \mid a,b \in \mathbb{Z}\}\) is a subring of \(\mathcal{O}\) containing the identity, and that \(\left[\mathcal{O} : \mathcal{O}_f\right] = f\). Prove conversely that a subring of \(\mathcal{O}\) containing the identity and having finite index \(f\) (as additive group) is equal to \(\mathcal{O}_f\).
\(\implies\): Evident.
\(\impliedby\):Lemma. Let \(G = \mathbb{Z} \times \mathbb{Z}\) and \(H \le G\). If \(\left|G : H\right|\) is finite, then \(H \cong a\mathbb{Z} \times b\mathbb{Z}\) for positive integers \(a,b\).
Proof. Suppose there are nonzero integers \(m,n\) such that \((m,0),(0,n) \in H\). We can take \(m,n\) to be the smallest such positive integer, which implies \(H \cong m\mathbb{Z} \times n\mathbb{Z}\). Now suppose \((m,0) \not\in H\) for all \(m \in \mathbb{Z}-\{0\}\). Then any two cosets \((i,0) + H\) and \((j,0) + H\) with \(i,j \in \mathbb{Z}-\{0\}\), \(i \neq j\) are distinct, since \((i,0)-(j,0) \not\in H\). This contradicts the fact that \(\left|G : H\right|\) is finite. The same argument can be applied if \((0,n) \not\in H\) for all \(n \in \mathbb{Z}-\{0\}\). \(\blacksquare\)
As an additive group, \(\mathcal{O}\) is isomorphic to \(\mathbb{Z} \times \mathbb{Z}\). Any subgroup \(H\) of index \(f\) is of the form \(m\mathbb{Z} \times n\mathbb{Z}\), where \(m,n\) are the smallest positive integers appearing in their respective components and \(mn = f\). However by assumption \(H\) contains the multiplicative identity and so \(m = 1\), forcing \(H = \mathbb{Z} \times f\mathbb{Z}\). As a subring, \(H\) is precisely \(\mathcal{O}_f\).
Friday, 15 May 2020
Divisors in commutative rings
There is a general notion of divisibility and in particular, that of a greatest common divisor (GCD) for any commutative ring \(R\):
1. \(a \mid b\) if \(b = ra\) for some \(r \in R\).
2. A GCD \(d\) of \(a\) and \(b\) must satisfy the following: (1) \(d \mid a\) and \(d \mid b\); (2) For any \(d'\) satisfying 1, \(d' \mid d\).
These statements can be expressed in terms of ideals:
1*. \(a \mid b\) if \((b) \subseteq (a)\).
2*. A GCD \(d\) of \(a\) and \(b\) must satisfy the following: (1) \((a,b) \subseteq (d)\); (2) For any \(d'\) satisfying 1, \((d) \subseteq (d')\).
(Note that \((a,b)\) is the smallest ideal containing \((a)\) and \((b)\), so \((a),(b) \subseteq (d)\) automatically implies \((a,b) \subseteq (d)\).)
Based on the definition, we see that \((d)\) must be the unique smallest principal ideal containing \((a,b)\). This rules out situations where there are at least 2 minimal principal ideals containing \((a,b)\). Also note that there can exist multiple GCDs, provided they generate the same principal ideal. To take an extreme example, every real number is a GCD of every pair of real numbers (this is true for fields in general).
Generally, we do not require \((a,b) = (d)\). For example, in the polynomial ring \(\mathbb{Z}[x]\), \((2,x)\) is a maximal non-principal ideal (because the associated quotient is a field). Thus 1 is the GCD of 2 and \(x\) even though \((2,x) \subset (1)\).
On the other hand, there is a hierarchy of rings satisfying \((a,b) = (d)\) for all \(a,b \in R\):
Euclidean domains \(\subset\) PIDs \(\subset\) Bezout domains \(\subset\) Commutative rings
Bezout domains are defined precisely to satisfy this condition. However, one must note that not all Bezout domains are principal ideal domains. For instance, they can have infinitely generated non-principal ideals.
PIDs satisfy this condition trivially; in fact all PIDs are Bezout domains. In particular, Euclidean domains enable one to algorithmically compute \((d)\) and express \(d\) as an \(R\)-linear combination of \(a\) and \(b\), via the Euclidean algorithm.
1. \(a \mid b\) if \(b = ra\) for some \(r \in R\).
2. A GCD \(d\) of \(a\) and \(b\) must satisfy the following: (1) \(d \mid a\) and \(d \mid b\); (2) For any \(d'\) satisfying 1, \(d' \mid d\).
These statements can be expressed in terms of ideals:
1*. \(a \mid b\) if \((b) \subseteq (a)\).
2*. A GCD \(d\) of \(a\) and \(b\) must satisfy the following: (1) \((a,b) \subseteq (d)\); (2) For any \(d'\) satisfying 1, \((d) \subseteq (d')\).
(Note that \((a,b)\) is the smallest ideal containing \((a)\) and \((b)\), so \((a),(b) \subseteq (d)\) automatically implies \((a,b) \subseteq (d)\).)
Based on the definition, we see that \((d)\) must be the unique smallest principal ideal containing \((a,b)\). This rules out situations where there are at least 2 minimal principal ideals containing \((a,b)\). Also note that there can exist multiple GCDs, provided they generate the same principal ideal. To take an extreme example, every real number is a GCD of every pair of real numbers (this is true for fields in general).
Generally, we do not require \((a,b) = (d)\). For example, in the polynomial ring \(\mathbb{Z}[x]\), \((2,x)\) is a maximal non-principal ideal (because the associated quotient is a field). Thus 1 is the GCD of 2 and \(x\) even though \((2,x) \subset (1)\).
On the other hand, there is a hierarchy of rings satisfying \((a,b) = (d)\) for all \(a,b \in R\):
Euclidean domains \(\subset\) PIDs \(\subset\) Bezout domains \(\subset\) Commutative rings
Bezout domains are defined precisely to satisfy this condition. However, one must note that not all Bezout domains are principal ideal domains. For instance, they can have infinitely generated non-principal ideals.
PIDs satisfy this condition trivially; in fact all PIDs are Bezout domains. In particular, Euclidean domains enable one to algorithmically compute \((d)\) and express \(d\) as an \(R\)-linear combination of \(a\) and \(b\), via the Euclidean algorithm.
Tuesday, 12 May 2020
Group with order \(pm\), with \(p > m\) and \(p\) prime (FGT 1A.9)
Suppose \(\left|G\right| = pm\), where \(p > m\) and \(p\) is prime. Show that \(G\) has a unique subgroup of order \(p\).
Assume there are \(k\) distinct order \(p\) subgroups. Since they have trivial intersection, the number of order \(p\) elements is \(k(p-1)\). By 1A.8, we have \(k(p-1) \equiv -1\) mod \(p\), or \(k(p-1) = lp-1\) for some \(l \in \mathbb{Z}^+\).
The sequences \(\{k(p-1)\}\) and \(\{lp-1\}\) form an arithmetic progression with step size \(p-1\) and \(p\), and clearly they meet at \(p-1\) when \(k = l = 1\). Since \(p\) and \(p-1\) are relative prime, the sequences next meet at \(1(p-1)+p(p-1) = (p+1)(p-1)\), so the next smallest solution has \(k = p+1\). But then there are a total of \((p+1)(p-1)+1 = p^2 > pm\) distinct elements, a contradiction.
Assume there are \(k\) distinct order \(p\) subgroups. Since they have trivial intersection, the number of order \(p\) elements is \(k(p-1)\). By 1A.8, we have \(k(p-1) \equiv -1\) mod \(p\), or \(k(p-1) = lp-1\) for some \(l \in \mathbb{Z}^+\).
The sequences \(\{k(p-1)\}\) and \(\{lp-1\}\) form an arithmetic progression with step size \(p-1\) and \(p\), and clearly they meet at \(p-1\) when \(k = l = 1\). Since \(p\) and \(p-1\) are relative prime, the sequences next meet at \(1(p-1)+p(p-1) = (p+1)(p-1)\), so the next smallest solution has \(k = p+1\). But then there are a total of \((p+1)(p-1)+1 = p^2 > pm\) distinct elements, a contradiction.
McKay's proof of Cauchy's theorem (FGT 1A.8)
Let \(G\) be a finite group, let \(n > 0\) be an integer. Let \(A\) be the set of \(n\)-tuples \((x_1, \ldots, x_n)\) of \(G\) such that \(x_1x_2 \ldots x_n = 1\).
(a) Show that \(\mathbb{Z}/n\mathbb{Z}\) acts on \(A\) according to the formula \(\overline{k} \cdot (x_1, \ldots, x_n) = (x_{1+k}, \ldots, x_{n+k})\), where indices are taken mod \(n\).
(b) Suppose \(n = p\) is a prime dividing \(\left|G\right|\). Show that \(p\) divides the number of size 1 orbits, and deduce that the number of elements of order \(p\) in \(G\) is congruent to -1 mod \(p\).
(a) Notice that if \(x_1 \ldots x_n = (x_1 \ldots x_k)(x_{k+1} \ldots x_n) = 1\) then \(x_{1+k} \ldots x_{n+k} = (x_{1+k} \ldots x_n)(x_{n+1} \ldots x_{n+k}) = (x_1 \ldots x_k)(x_{k+1} \ldots x_n) = 1\).
(b) The size of any orbit must divide \(\left|\mathbb{Z}/p\mathbb{Z}\right| = p\), hence the only possibilities are 1 and \(p\). Let the number of size 1 and size \(p\) orbits be \(n_1\) and \(n_p\). The number of possible \(p\)-tuples is easily checked to be \(\left|G\right|^{p-1}\), so we have \(n_1 + pn_p = \left|G\right|^{p-1}\). Since \(p \mid pn_p\) and \(p \mid \left|G\right|^{p-1}\), it follows that \(p \mid n_1\).
If \(\{(x_1, \ldots, x_n)\}\) is a size 1 orbit, then we have \(x_1 = \ldots = x_n\) and \({x_1}^p = 1\). Thus \(x_1\) is either 1 or an order \(p\) element. Eliminating the former possibility yields the congruence relation.
(a) Show that \(\mathbb{Z}/n\mathbb{Z}\) acts on \(A\) according to the formula \(\overline{k} \cdot (x_1, \ldots, x_n) = (x_{1+k}, \ldots, x_{n+k})\), where indices are taken mod \(n\).
(b) Suppose \(n = p\) is a prime dividing \(\left|G\right|\). Show that \(p\) divides the number of size 1 orbits, and deduce that the number of elements of order \(p\) in \(G\) is congruent to -1 mod \(p\).
(a) Notice that if \(x_1 \ldots x_n = (x_1 \ldots x_k)(x_{k+1} \ldots x_n) = 1\) then \(x_{1+k} \ldots x_{n+k} = (x_{1+k} \ldots x_n)(x_{n+1} \ldots x_{n+k}) = (x_1 \ldots x_k)(x_{k+1} \ldots x_n) = 1\).
(b) The size of any orbit must divide \(\left|\mathbb{Z}/p\mathbb{Z}\right| = p\), hence the only possibilities are 1 and \(p\). Let the number of size 1 and size \(p\) orbits be \(n_1\) and \(n_p\). The number of possible \(p\)-tuples is easily checked to be \(\left|G\right|^{p-1}\), so we have \(n_1 + pn_p = \left|G\right|^{p-1}\). Since \(p \mid pn_p\) and \(p \mid \left|G\right|^{p-1}\), it follows that \(p \mid n_1\).
If \(\{(x_1, \ldots, x_n)\}\) is a size 1 orbit, then we have \(x_1 = \ldots = x_n\) and \({x_1}^p = 1\). Thus \(x_1\) is either 1 or an order \(p\) element. Eliminating the former possibility yields the congruence relation.
Monday, 11 May 2020
Burnside's lemma (FGT 1A.6)
Let \(\chi\) be the permutation character associated with the action of a group \(G\) on a set \(A\). Prove that \(\sum_{g \in G} \chi(g) = \sum_{a \in A} \left|G_a\right| = n\left|G\right|\), where \(n\) is the number of orbits.
The action of \(G\) on \(A\) can be represented by a table where the rows are elements of \(G\) and the columns are elements of \(A\). For example, here is the action of \(D_6\) on the three vertices of a triangle:
If we highlight all cells \((g,a)\) where \(g \cdot a = a\), then we get the following table:
Notice that the sum \(\sum_{g \in G} \chi(g)\) corresponds to be a horizontal summation of the highlighted squares, while \(\sum_{a \in A} \left|G_a\right|\) corresponds to a vertical summation.
Lastly, \(\sum_{a \in A} \left|G_a\right| = \sum_{\mathcal{O}} \sum_{a \in \mathcal{O}} \left|G_a\right| = \sum_{\mathcal{O}} \sum_{a \in \mathcal{O}} \frac{\left|G\right|}{\left|\mathcal{O}\right|} = n\left|G\right|\).
The action of \(G\) on \(A\) can be represented by a table where the rows are elements of \(G\) and the columns are elements of \(A\). For example, here is the action of \(D_6\) on the three vertices of a triangle:
If we highlight all cells \((g,a)\) where \(g \cdot a = a\), then we get the following table:
Notice that the sum \(\sum_{g \in G} \chi(g)\) corresponds to be a horizontal summation of the highlighted squares, while \(\sum_{a \in A} \left|G_a\right|\) corresponds to a vertical summation.
Lastly, \(\sum_{a \in A} \left|G_a\right| = \sum_{\mathcal{O}} \sum_{a \in \mathcal{O}} \left|G_a\right| = \sum_{\mathcal{O}} \sum_{a \in \mathcal{O}} \frac{\left|G\right|}{\left|\mathcal{O}\right|} = n\left|G\right|\).
Friday, 8 May 2020
Ring projection from \(\mathbb{Z}/m\mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z}\)
Let \(n,m \in \mathbb{Z}^+\) with \(n \mid m\). Prove that the natural surjective ring projection \(\mathbb{Z}/m\mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z}\) is also surjective on the units: \((\mathbb{Z}/m\mathbb{Z})^\times \rightarrow (\mathbb{Z}/n\mathbb{Z})^\times\).
Let \(\varphi\) be the ring projection. Suppose that \(m = p^a\) and \(n = p^b\) where \(p\) is a prime and \(a,b \in \mathbb{Z}^+ \cup \{0\}\). Clearly if \((x,n) = 1\), then \((x,m) = 1\) as well. Thus \(\varphi\) is surjective on the units.
Now suppose \(m,n\) are any positive integers; let \(m = {p_1}^{a_1} \ldots {p_n}^{a_n}\) and \(n = {p_1}^{b_1} \ldots {p_n}^{b_n}\) be their prime factorization (where \(b_i\) is set to 0 where needed). By the Chinese Remainder Theorem we have \((\mathbb{Z}/m\mathbb{Z})^\times \cong (\mathbb{Z}/{p_1}^{a_1}\mathbb{Z})^\times \times \ldots \times (\mathbb{Z}/{p_n}^{a_n}\mathbb{Z})^\times\) (likewise for \((\mathbb{Z}/n\mathbb{Z})^\times\)). Since the natural projections corresponding to each factor is surjective on the units as shown in the previous paragraph, \(\varphi\) is surjective.
Let \(\varphi\) be the ring projection. Suppose that \(m = p^a\) and \(n = p^b\) where \(p\) is a prime and \(a,b \in \mathbb{Z}^+ \cup \{0\}\). Clearly if \((x,n) = 1\), then \((x,m) = 1\) as well. Thus \(\varphi\) is surjective on the units.
Now suppose \(m,n\) are any positive integers; let \(m = {p_1}^{a_1} \ldots {p_n}^{a_n}\) and \(n = {p_1}^{b_1} \ldots {p_n}^{b_n}\) be their prime factorization (where \(b_i\) is set to 0 where needed). By the Chinese Remainder Theorem we have \((\mathbb{Z}/m\mathbb{Z})^\times \cong (\mathbb{Z}/{p_1}^{a_1}\mathbb{Z})^\times \times \ldots \times (\mathbb{Z}/{p_n}^{a_n}\mathbb{Z})^\times\) (likewise for \((\mathbb{Z}/n\mathbb{Z})^\times\)). Since the natural projections corresponding to each factor is surjective on the units as shown in the previous paragraph, \(\varphi\) is surjective.
Wednesday, 29 April 2020
Nilpotent ring elements (D&F 7.1.13)
(a) Show that if \(n = a^k b\) for some integers \(a\) and \(b\) then \(\overline{ab}\) is a nilpotent element of \(\mathbb{Z}/n\mathbb{Z}\).
(b) If \(a \in \mathbb{Z}\), show that \(\overline{a}\) is nilpotent if and only if every prime divisor of \(n\) is also a divisor of \(a\). In particular, determine the nilpotent elements of \(\mathbb{Z}/72\mathbb{Z}\) explicitly.
(c) Let \(R\) be the ring of functions from a nonempty set \(X\) to a field \(F\). Prove that \(R\) contains no nonzero nilpotent elements.
(a) \(\overline{ab}^k = \overline{a^k b} \cdot \overline{b^{k-1}} = \overline{0} \cdot \overline{b^{k-1}} = \overline{0}\)
(b) Suppose \(\overline{a}\) is nilpotent; then \(a^m = kn\) for integers \(k\) and \(m\). We then have \(p \mid n \implies p \mid kn \implies p \mid a^m \implies p \mid a\). The last implication holds because \(p\) is prime (the reader can convince himself of this fact).
Suppose the converse is true; let \(p_1, \ldots, p_k\) be the prime divisors of \(n\) (which are also prime divisors of \(a\)), and \(p_{k+1}, \ldots, p_n\) be the prime divisors of \(a\) that do not divide \(n\). Then we have \(n = \prod_{i=1}^k {p_i}^{a_i}\) and \(a = \prod_{j=1}^n {p_j}^{b_j}\) for integers \(a_i\), \(b_j\). Notice that \(a^m = \prod_{j=1}^n {p_j}^{mb_j}\), and in fact we can choose \(m\) such that \(mb_i \ge a_i\) for all \(1 \le i \le k\). This implies \(n \mid a\) (we implicitly assumed that \(a\) is positive, but the proof clearly holds for \(a < 0\) as well with slight modifications).
The nilpotent elements of \(\mathbb{Z}/72\mathbb{Z}\) are precisely the elements \(\overline{6n}\) where \(n \in \mathbb{Z}\).
(c) A field \(F\) contains no zero divisors, so it follows that all nonzero elements are not nilpotent.
(b) If \(a \in \mathbb{Z}\), show that \(\overline{a}\) is nilpotent if and only if every prime divisor of \(n\) is also a divisor of \(a\). In particular, determine the nilpotent elements of \(\mathbb{Z}/72\mathbb{Z}\) explicitly.
(c) Let \(R\) be the ring of functions from a nonempty set \(X\) to a field \(F\). Prove that \(R\) contains no nonzero nilpotent elements.
(a) \(\overline{ab}^k = \overline{a^k b} \cdot \overline{b^{k-1}} = \overline{0} \cdot \overline{b^{k-1}} = \overline{0}\)
(b) Suppose \(\overline{a}\) is nilpotent; then \(a^m = kn\) for integers \(k\) and \(m\). We then have \(p \mid n \implies p \mid kn \implies p \mid a^m \implies p \mid a\). The last implication holds because \(p\) is prime (the reader can convince himself of this fact).
Suppose the converse is true; let \(p_1, \ldots, p_k\) be the prime divisors of \(n\) (which are also prime divisors of \(a\)), and \(p_{k+1}, \ldots, p_n\) be the prime divisors of \(a\) that do not divide \(n\). Then we have \(n = \prod_{i=1}^k {p_i}^{a_i}\) and \(a = \prod_{j=1}^n {p_j}^{b_j}\) for integers \(a_i\), \(b_j\). Notice that \(a^m = \prod_{j=1}^n {p_j}^{mb_j}\), and in fact we can choose \(m\) such that \(mb_i \ge a_i\) for all \(1 \le i \le k\). This implies \(n \mid a\) (we implicitly assumed that \(a\) is positive, but the proof clearly holds for \(a < 0\) as well with slight modifications).
The nilpotent elements of \(\mathbb{Z}/72\mathbb{Z}\) are precisely the elements \(\overline{6n}\) where \(n \in \mathbb{Z}\).
(c) A field \(F\) contains no zero divisors, so it follows that all nonzero elements are not nilpotent.
Thursday, 23 April 2020
Groups of order 75 (D&F 5.5.8)
Construct a non-abelian group of order 75. Classify all groups of order 75.
Let \(H = Z_5 \times Z_5\) and \(K = Z_3 = \langle x \rangle\). It is known that \(\text{Aut}(H)\) has order \(5(5-1)^2(5+1) = 2^5 \cdot 3 \cdot 5\), so it has a Sylow 3-subgroup \(H_3\ = \langle y \rangle\). Hence we can define a homomorphism \(\varphi : K \rightarrow \text{Aut}(H)\) by \(\varphi(x) = y\), inducing a semidirect product \(G = H \rtimes_{\varphi} K\) of order 75. Note that \(G\) is the unique (nontrivial) semidirect product of \(H\) and \(K\) by 5.5.6 since all Sylow 3-subgroups of \(\text{Aut}(H)\) are conjugate. Furthermore \(G\) is non-abelian, since \(\varphi\) is non-trivial meaning \(K \not\trianglelefteq G\).
Let \(G\) be an order \(75 = 3 \cdot 5^2\) group, and \(G_3\) and \(G_5\) be a Sylow 3- and 5-subgroup. \(n_3\) can either be 1 and 25. If \(n_3 = 25\), then the Sylow 3-subgroups contain \(2 \cdot 25 = 50\) nonidentity elements, forcing any Sylow 5-subgroup to be normal. If \(G_5 \cong Z_{25}\), then \(\text{Aut}(G_5) \cong Z_{20}\), so any homomorphism from \(G_3\) to \(\text{Aut}(G_5)\) must be trivial. Hence we can only construct the abelian group \(Z_{25} \times Z_3 \cong Z_{75}\). If \(G_5 \cong Z_5 \times Z_5\), then we effectively get the unique non-abelian group in the previous paragraph.
Suppose \(n_3 = 1\). Since \(\text{Aut}(G_3) \cong Z_2\), any homomorphism from \(G_5\) to \(\text{Aut}(G_3)\) must be trivial. Thus we only get one new abelian group \(Z_3 \times Z_5 \times Z_5\) (\(Z_3 \times Z_{25}\) was constructed in the previous paragraph).
Let \(H = Z_5 \times Z_5\) and \(K = Z_3 = \langle x \rangle\). It is known that \(\text{Aut}(H)\) has order \(5(5-1)^2(5+1) = 2^5 \cdot 3 \cdot 5\), so it has a Sylow 3-subgroup \(H_3\ = \langle y \rangle\). Hence we can define a homomorphism \(\varphi : K \rightarrow \text{Aut}(H)\) by \(\varphi(x) = y\), inducing a semidirect product \(G = H \rtimes_{\varphi} K\) of order 75. Note that \(G\) is the unique (nontrivial) semidirect product of \(H\) and \(K\) by 5.5.6 since all Sylow 3-subgroups of \(\text{Aut}(H)\) are conjugate. Furthermore \(G\) is non-abelian, since \(\varphi\) is non-trivial meaning \(K \not\trianglelefteq G\).
Let \(G\) be an order \(75 = 3 \cdot 5^2\) group, and \(G_3\) and \(G_5\) be a Sylow 3- and 5-subgroup. \(n_3\) can either be 1 and 25. If \(n_3 = 25\), then the Sylow 3-subgroups contain \(2 \cdot 25 = 50\) nonidentity elements, forcing any Sylow 5-subgroup to be normal. If \(G_5 \cong Z_{25}\), then \(\text{Aut}(G_5) \cong Z_{20}\), so any homomorphism from \(G_3\) to \(\text{Aut}(G_5)\) must be trivial. Hence we can only construct the abelian group \(Z_{25} \times Z_3 \cong Z_{75}\). If \(G_5 \cong Z_5 \times Z_5\), then we effectively get the unique non-abelian group in the previous paragraph.
Suppose \(n_3 = 1\). Since \(\text{Aut}(G_3) \cong Z_2\), any homomorphism from \(G_5\) to \(\text{Aut}(G_3)\) must be trivial. Thus we only get one new abelian group \(Z_3 \times Z_5 \times Z_5\) (\(Z_3 \times Z_{25}\) was constructed in the previous paragraph).
Sunday, 19 April 2020
The subgroup \(H^x K^y\), where \(H,K \le G\) and \(x,y \in G\) (FGT 1A.4)
Suppose that \(G = HK\), where \(H\) and \(K\) are subgroups. Show that also \(G = H^x K^y\) for all elements \(x,y \in G\). Deduce that if \(G = HH^x\) for a subgroup \(H\) and an element \(x \in G\), then \(H = G\).
We first show that \(HK = HxK\). Firstly we can write \(x = h_0 k_0\) for \(h_0 \in H\) and \(k_0 \in H\). Clearly for any \(hk \in HK\) there is some \(h_1 \in H\) and \(k_1 \in K\) such that \(hk = h_1 h_0 k_0 k_1 = h_1 x k_1 \in HxK\), and the opposite is true as well. Furthermore we know that \(H^x\) is a subgroup with order \(\left|H\right|\), so \(\left|H^xK\right| = \frac{\left|H^x\right|\left|K\right|}{\left|H^x \cap K\right|} = \frac{\left|H\right|\left|K\right|}{\left|H^x \cap K\right|} = \left|HxK\right| = \left|G\right|\) and so \(G = H^xK\). We can do likewise with \(K^y\) to obtain the conclusion.
If \(G = HH^x\), then \(G = H(H^x)^{x^{-1}} = H\).
We first show that \(HK = HxK\). Firstly we can write \(x = h_0 k_0\) for \(h_0 \in H\) and \(k_0 \in H\). Clearly for any \(hk \in HK\) there is some \(h_1 \in H\) and \(k_1 \in K\) such that \(hk = h_1 h_0 k_0 k_1 = h_1 x k_1 \in HxK\), and the opposite is true as well. Furthermore we know that \(H^x\) is a subgroup with order \(\left|H\right|\), so \(\left|H^xK\right| = \frac{\left|H^x\right|\left|K\right|}{\left|H^x \cap K\right|} = \frac{\left|H\right|\left|K\right|}{\left|H^x \cap K\right|} = \left|HxK\right| = \left|G\right|\) and so \(G = H^xK\). We can do likewise with \(K^y\) to obtain the conclusion.
If \(G = HH^x\), then \(G = H(H^x)^{x^{-1}} = H\).
Saturday, 18 April 2020
The \(p\)th power map in order \(p^3\) groups, where \(p\) is an odd prime (D&F 5.4.9)
Prove that if \(p\) is an odd prime and \(P\) is a group of order \(p^3\) then the \(p\)th power map is a homomorphism of \(P\) into \(Z(P)\). If \(P\) is not cyclic, show that the kernel of the \(p\)th power map has order \(p^2\) or \(p^3\). Is the squaring map a homomorphism in non-abelian groups of order 8? Where is the oddness of \(p\) needed in the above proof?
If \(P\) is abelian, then \(Z(P)\) has order \(p^3\), and the first statement immediately follows. Otherwise, then \(Z(P)\) can only have order \(p\), since \(Z(P)\) is nontrivial and \(G/Z(P)\) cannot have order \(p\). So the order \(p^2\) group \(G/Z(P)\) is abelian, implying \(P' \le Z(P)\). But \(P'\) is non-trivial since \(P\) is non-abelian, forcing equality. In particular, every element of \(P'\) has order \(p\). Now notice that if \(p\) is odd then \(\frac{p-1}{2}\) is an integer. Since \([y,x]^{\frac{p-1}{2}} \in P'\), we have \([y,x]^{\frac{p(p-1)}{2}} = ([y,x]^{\frac{p-1}{2}})^p = 1 \in Z(P)\). Thus \((xy)^p = x^p y^p\) (by 5.4.8), proving the map is a homomorphism. (Note that this conclusion fails if \(p = 2\), since \((xy)^2 = x^2 y^2 [y,x]\).)
The next step is to show that for non-abelian groups \(P\), \(x^p \in Z(P)\) for all \(x\). This is trivial if \(x\) has order \(p\), so suppose it has order \(p^2\). Then the subgroup \(\langle x \rangle\) is normal in \(P\) (having index \(p\)). If \(Z(P) \nleq \langle x \rangle\), then their intersection is trivial; this combined with the fact that both groups are normal implies \(P \cong Z(P) \times \langle x \rangle\), contradicting the fact that \(P\) is non-abelian. So \(Z(P)\) must be the unique subgroup \(\langle x^p \rangle\), and the conclusion follows.
If \(P\) is non-abelian, then the the image is a subgroup of \(Z(P)\) having order \(1\) or \(p\). If \(P\) is abelian and non-cyclic, then it is isomorphic to \(Z_{p^2} \times Z_p\) or \(E_3\). Letting \(Z_{p^2} = \langle x \rangle\) in the former case, its image is \(\langle x^p \rangle \times 1\) which has order \(p\). In the latter case, the image is trivial. Applying the First Isomorphism Theorem to all these cases, we can see that the kernel has order \(p^2\) or \(p^3\).
If \(P\) is abelian, then \(Z(P)\) has order \(p^3\), and the first statement immediately follows. Otherwise, then \(Z(P)\) can only have order \(p\), since \(Z(P)\) is nontrivial and \(G/Z(P)\) cannot have order \(p\). So the order \(p^2\) group \(G/Z(P)\) is abelian, implying \(P' \le Z(P)\). But \(P'\) is non-trivial since \(P\) is non-abelian, forcing equality. In particular, every element of \(P'\) has order \(p\). Now notice that if \(p\) is odd then \(\frac{p-1}{2}\) is an integer. Since \([y,x]^{\frac{p-1}{2}} \in P'\), we have \([y,x]^{\frac{p(p-1)}{2}} = ([y,x]^{\frac{p-1}{2}})^p = 1 \in Z(P)\). Thus \((xy)^p = x^p y^p\) (by 5.4.8), proving the map is a homomorphism. (Note that this conclusion fails if \(p = 2\), since \((xy)^2 = x^2 y^2 [y,x]\).)
The next step is to show that for non-abelian groups \(P\), \(x^p \in Z(P)\) for all \(x\). This is trivial if \(x\) has order \(p\), so suppose it has order \(p^2\). Then the subgroup \(\langle x \rangle\) is normal in \(P\) (having index \(p\)). If \(Z(P) \nleq \langle x \rangle\), then their intersection is trivial; this combined with the fact that both groups are normal implies \(P \cong Z(P) \times \langle x \rangle\), contradicting the fact that \(P\) is non-abelian. So \(Z(P)\) must be the unique subgroup \(\langle x^p \rangle\), and the conclusion follows.
If \(P\) is non-abelian, then the the image is a subgroup of \(Z(P)\) having order \(1\) or \(p\). If \(P\) is abelian and non-cyclic, then it is isomorphic to \(Z_{p^2} \times Z_p\) or \(E_3\). Letting \(Z_{p^2} = \langle x \rangle\) in the former case, its image is \(\langle x^p \rangle \times 1\) which has order \(p\). In the latter case, the image is trivial. Applying the First Isomorphism Theorem to all these cases, we can see that the kernel has order \(p^2\) or \(p^3\).
Sunday, 12 April 2020
Orders of elements in finite abelian groups (D&F 5.2.5)
Let \(G\) be a finite abelian group of type \((n_1,n_2,\ldots,n_t)\). Prove that \(G\) contains an element of order \(m\) if and only if \(m \mid n_1\). Deduce that \(G\) is of exponent \(n_1\).
Let \(x = (x_1,x_2,\ldots,x_t) \in G\) have order \(m\). By defining \(a_i\) to be the \(t\)-tuple with \(x_i\) in its \(i\) component and 0 elsewhere, we can write \(x\) as a product of \(a_i\)'s. For all \(i\), we have that \(\left|a_i\right| \mid n_i \mid n_1\). Thus \(m\), which is precisely the LCM of all \(\left|a_i\right|\)'s, divides \(n_1\).
Conversely, if \(m \mid n_1\), then by virtue of being cyclic we can find an element \(a \in Z_{n_1}\) with order \(m\). Then the element \((a,1,\ldots,1)\) clearly has order \(m\) in \(G\). The statement that \(G\) has exponent \(n_1\) follows immediately from the fact that \(G\) has an order \(n_1\) element (namely \((b,1,\ldots,1)\) where \(Z_{n_1} = \langle b \rangle\)), and that orders of any element must divide \(n_1\).
Let \(x = (x_1,x_2,\ldots,x_t) \in G\) have order \(m\). By defining \(a_i\) to be the \(t\)-tuple with \(x_i\) in its \(i\) component and 0 elsewhere, we can write \(x\) as a product of \(a_i\)'s. For all \(i\), we have that \(\left|a_i\right| \mid n_i \mid n_1\). Thus \(m\), which is precisely the LCM of all \(\left|a_i\right|\)'s, divides \(n_1\).
Conversely, if \(m \mid n_1\), then by virtue of being cyclic we can find an element \(a \in Z_{n_1}\) with order \(m\). Then the element \((a,1,\ldots,1)\) clearly has order \(m\) in \(G\). The statement that \(G\) has exponent \(n_1\) follows immediately from the fact that \(G\) has an order \(n_1\) element (namely \((b,1,\ldots,1)\) where \(Z_{n_1} = \langle b \rangle\)), and that orders of any element must divide \(n_1\).
Friday, 10 April 2020
Subgroups of \(Q_8 \times E_{2^n}\) (D&F 5.1.6)
Show that all subgroups of \(Q_8 \times E_{2^n}\) are normal.
(Warning: potentially ugly/confusing working!)
The proof is by induction on \(n\). We first note 2 key properties of \(Q_8\): all subgroups are normal, and cosets of subgroups are conjugation-invariant (this can be verified manually). Let \(G = Q_8 \times Z_2\), and suppose \(H \le G\). We shall consider the restriction of the projection \(\pi_{Z_2} : G \rightarrow Z_2\) to \(H\): its image is a subgroup of \(Z_2\) and its kernel \(K\) is the set \(\{(q,1) \mid q \in Q\}\), where \(Q \le Q_8\). Using the First Isomorphism Theorem, we know \(K\) has index 1 or 2. If the former, then \(H = K \cong Q\), and for any \((g_1,g_2) \in G\) and \((h_1,h_2) \in H\) we have \((h_1,h_2)^{(g_1,g_2)} = (g_1h_1g_1^{-1},h_2) \in H\) (since \(Q \trianglelefteq Q_8\)).
Suppose the latter is true instead. Then \(H\) is partitioned into 2 cosets \(K\) and \(aK\), where \(a = (a_1,a_2)\) is any element with \(a_2 \neq 1\). It is easy to check that \(K \triangleleft G\), so it suffices to prove that \(aK\) is conjugation-invariant. If \((g_1,g_2) \in G\) and \((a_1b,a_2) = a(b,1) \in aK\), then \((a_1b,a_2)^{(g_1,g_2)} = (g_1(a_1b)g_1^{-1},a_2) \in aK\) (since \(a_1Q\) is conjugation-invariant). Thus \(H\) is normal. Furthermore, cosets of \(H\) are conjugation-invariant: let \(h = (h_1,h_2) \in H\), \(b = (b_1,b_2) \in G\) and \(g = (g_1,g_2) \in G\). Then \((bh)^g = (g_1(b_1h_1)g_1^{-1}, b_2h_2)\). But \(b_1h_1\) is in either \(b_1Q\) and \((b_1a_1)Q\) which are conjugation-invariant. Thus \((bh)^g \in bH\).
In conclusion, we have established the two properties mentioned at the beginning, but for \(Q_8 \times Z_2\). We can repeat the same argument for \(n = 2\) by replacing \(Q_8\) with \(Q_8 \times Z_2\), and so on.
(Warning: potentially ugly/confusing working!)
The proof is by induction on \(n\). We first note 2 key properties of \(Q_8\): all subgroups are normal, and cosets of subgroups are conjugation-invariant (this can be verified manually). Let \(G = Q_8 \times Z_2\), and suppose \(H \le G\). We shall consider the restriction of the projection \(\pi_{Z_2} : G \rightarrow Z_2\) to \(H\): its image is a subgroup of \(Z_2\) and its kernel \(K\) is the set \(\{(q,1) \mid q \in Q\}\), where \(Q \le Q_8\). Using the First Isomorphism Theorem, we know \(K\) has index 1 or 2. If the former, then \(H = K \cong Q\), and for any \((g_1,g_2) \in G\) and \((h_1,h_2) \in H\) we have \((h_1,h_2)^{(g_1,g_2)} = (g_1h_1g_1^{-1},h_2) \in H\) (since \(Q \trianglelefteq Q_8\)).
Suppose the latter is true instead. Then \(H\) is partitioned into 2 cosets \(K\) and \(aK\), where \(a = (a_1,a_2)\) is any element with \(a_2 \neq 1\). It is easy to check that \(K \triangleleft G\), so it suffices to prove that \(aK\) is conjugation-invariant. If \((g_1,g_2) \in G\) and \((a_1b,a_2) = a(b,1) \in aK\), then \((a_1b,a_2)^{(g_1,g_2)} = (g_1(a_1b)g_1^{-1},a_2) \in aK\) (since \(a_1Q\) is conjugation-invariant). Thus \(H\) is normal. Furthermore, cosets of \(H\) are conjugation-invariant: let \(h = (h_1,h_2) \in H\), \(b = (b_1,b_2) \in G\) and \(g = (g_1,g_2) \in G\). Then \((bh)^g = (g_1(b_1h_1)g_1^{-1}, b_2h_2)\). But \(b_1h_1\) is in either \(b_1Q\) and \((b_1a_1)Q\) which are conjugation-invariant. Thus \((bh)^g \in bH\).
In conclusion, we have established the two properties mentioned at the beginning, but for \(Q_8 \times Z_2\). We can repeat the same argument for \(n = 2\) by replacing \(Q_8\) with \(Q_8 \times Z_2\), and so on.
Thursday, 9 April 2020
Sylow \(p\)-subgroups of direct products (D&F 5.1.4)
Let \(A, B\) be finite groups and let \(p\) be a prime. Prove that any Sylow \(p\)-subgroup of \(A \times B\) is of the form \(P \times Q\), where \(P \in Syl_p(A)\) and \(Q \in Syl_p(B)\). Prove that \(n_p(A \times B) = n_p(A)n_p(B)\). Generalize both of these results to a direct product of any finite number of finite groups.
Let \(\left|A\right| = p^\alpha m\) and \(\left|B\right| = p^\beta n\), where \(p \nmid m\) and \(p \nmid n\). Suppose \(C \in Syl_p(A \times B)\). Consider now the restriction of the projection homomorphism \(\pi_A : A \times B \rightarrow A\) to \(C\) (which we shall still denote as \(\pi_A\), and whose kernel and image we shall denote as \(K\) and \(I\)). \(I\) is a subgroup of \(A\) and \(K\) can be identified with a subgroup \(K'\) of \(B\), i.e. \(\left|I\right| \mid p^\alpha m\) and \(\left|K'\right| \mid p^\beta n\). This in addition with the fact that \(\left|C\right| = p^{\alpha+\beta}\) and \(C/K \cong I\) force both \(I \in Syl_p(A)\) and \(K' \in Syl_p(B)\). In particular, if \(a \in I\) and \(b \in K'\), then it is easily seen that \((a,1), (1,b) \in C\), so \((a,b) \in C\). Thus \(I \times K' \subseteq C\), and equality holds since their order is equal.
Clearly \(n_p(A \times B)\) is the number of ways to form direct products of Sylow \(p\)-subgroups of \(A\) and \(B\), which is simply \(n_p(A)n_p(B)\).
The last statement is trivial.
Let \(\left|A\right| = p^\alpha m\) and \(\left|B\right| = p^\beta n\), where \(p \nmid m\) and \(p \nmid n\). Suppose \(C \in Syl_p(A \times B)\). Consider now the restriction of the projection homomorphism \(\pi_A : A \times B \rightarrow A\) to \(C\) (which we shall still denote as \(\pi_A\), and whose kernel and image we shall denote as \(K\) and \(I\)). \(I\) is a subgroup of \(A\) and \(K\) can be identified with a subgroup \(K'\) of \(B\), i.e. \(\left|I\right| \mid p^\alpha m\) and \(\left|K'\right| \mid p^\beta n\). This in addition with the fact that \(\left|C\right| = p^{\alpha+\beta}\) and \(C/K \cong I\) force both \(I \in Syl_p(A)\) and \(K' \in Syl_p(B)\). In particular, if \(a \in I\) and \(b \in K'\), then it is easily seen that \((a,1), (1,b) \in C\), so \((a,b) \in C\). Thus \(I \times K' \subseteq C\), and equality holds since their order is equal.
Clearly \(n_p(A \times B)\) is the number of ways to form direct products of Sylow \(p\)-subgroups of \(A\) and \(B\), which is simply \(n_p(A)n_p(B)\).
The last statement is trivial.
Subgroups of index \(p\), \(p\) being the smallest prime dividing \(\left|G\right|\) (FGT 1A.1)
Let \(H\) be a subgroup of prime index \(p\) in the finite group \(G\), and suppose that no prime smaller than \(p\) divides \(\left|G\right|\). Prove that \(H \triangleleft G\).
Let \(\left|G\right| = pn\). The action of \(G\) on left cosets of \(H\) induces a homomorphism \(\varphi : G \rightarrow S_p\), whose kernel we shall denote \(K\). Since \(K \le H\), we have \(1 \le \left|K\right| \le n\), so that \(p \le \left|G/K\right| \le \left|G\right|\). Since \(p\) is the smallest prime dividing \(\left|G\right|\), it follows that \(\left|G/K\right|\) is a product of primes \(\ge p\). But \(\left|S_p\right| = p!\) is a product of primes \(\le p\), where the prime \(p\) appears just once. It can be seen that \(\left|G/K\right|\) divides \(p!\) only when \(n = 1\), in which case \(H = 1\) is surely normal. If it does not, we have a contradiction, since \(G/K\) is isomorphic to a subgroup of \(S_p\). This completes the proof.
Let \(\left|G\right| = pn\). The action of \(G\) on left cosets of \(H\) induces a homomorphism \(\varphi : G \rightarrow S_p\), whose kernel we shall denote \(K\). Since \(K \le H\), we have \(1 \le \left|K\right| \le n\), so that \(p \le \left|G/K\right| \le \left|G\right|\). Since \(p\) is the smallest prime dividing \(\left|G\right|\), it follows that \(\left|G/K\right|\) is a product of primes \(\ge p\). But \(\left|S_p\right| = p!\) is a product of primes \(\le p\), where the prime \(p\) appears just once. It can be seen that \(\left|G/K\right|\) divides \(p!\) only when \(n = 1\), in which case \(H = 1\) is surely normal. If it does not, we have a contradiction, since \(G/K\) is isomorphic to a subgroup of \(S_p\). This completes the proof.
Chain of simple subgroups (D&F 4.6.5)
Prove that if there exists a chain of subgroups \(G_1 \le G_2 \le \ldots \le G\) such that \(G = \cup_{i=1}^\infty G_i\) and each \(G_i\) is simple then \(G\) is simple.
Suppose \(H \trianglelefteq G\) and \(H \neq 1,G\); so \(H \cap G_i \trianglelefteq G_i\) for all \(i\). We want to show that there is some \(i\) such that \(H \cap G_i\) is a proper nontrivial subgroup of \(G_i\), which contradicts the fact that \(G_i\) is simple. What follows is simply accounting for all the possibilities:
1. \(H \cap G_i = G_i\) for all \(i\)
In this case, we have \(G_i \le H\) for all \(i\) and so \(G = \cup_{i=1}^\infty G_i \le H\), a contradiction.
2. \(H \cap G_i = 1\) for all \(i\)
It follows that \(H \cap G = H \cap (\cup_{i=1}^\infty G_i) = 1\), yet another contradiction.
3. \(H \cap G_i = 1\) and \(H \cap G_j = G_j\) for some \(i < j\)
The second statement implies that \(G_i \le G_j \le H\), contradicting the first statement.
4. \(H \cap G_i = 1\) and \(H \cap G_j = G_j\) for some \(i > j\)
The first statement implies that \(H \cap G_j \le H \cap G_i = 1\), contradicting the second statement.
Suppose \(H \trianglelefteq G\) and \(H \neq 1,G\); so \(H \cap G_i \trianglelefteq G_i\) for all \(i\). We want to show that there is some \(i\) such that \(H \cap G_i\) is a proper nontrivial subgroup of \(G_i\), which contradicts the fact that \(G_i\) is simple. What follows is simply accounting for all the possibilities:
1. \(H \cap G_i = G_i\) for all \(i\)
In this case, we have \(G_i \le H\) for all \(i\) and so \(G = \cup_{i=1}^\infty G_i \le H\), a contradiction.
2. \(H \cap G_i = 1\) for all \(i\)
It follows that \(H \cap G = H \cap (\cup_{i=1}^\infty G_i) = 1\), yet another contradiction.
3. \(H \cap G_i = 1\) and \(H \cap G_j = G_j\) for some \(i < j\)
The second statement implies that \(G_i \le G_j \le H\), contradicting the first statement.
4. \(H \cap G_i = 1\) and \(H \cap G_j = G_j\) for some \(i > j\)
The first statement implies that \(H \cap G_j \le H \cap G_i = 1\), contradicting the second statement.
Wednesday, 8 April 2020
Non-abelian simple groups of order \(<\) 100 (D&F 4.5.29)
If \(G\) is a non-abelian simple group of order \(<\) 100, prove that \(G \cong A_5\).
We proceed by systematically filtering out all groups of order \(\neq 60\):
1. \(p\)-groups
These have a nontrivial center which is normal in \(G\), thus they are not simple.
2. Groups with order \(pq\)
These groups are guaranteed to have a normal Sylow \(q\)-subgroup.
3. Groups with order \(p^2 q\)
These groups have a normal Sylow subgroup for either \(p\) or \(q\).
4. Groups with order \(pqr\)
It has been proven (in 4.5.16) that these groups have a normal Sylow subgroup for either \(p\), \(q\) or \(r\).
5. Groups with \(n_p = 1\) for some prime p
Unique Sylow subgroups are normal.
6. Groups where the no. of distinct elements in Sylow subgroups (if not normal) exceeds its order
Let \(\left|G\right| = p^a m\), where \(p \nmid m\). We derive a lower bound for the number of distinct elements in all Sylow \(p\)-subgroups by assuming the largest possible intersection, which has size \(p^{a-1}\). These elements are only counted once, leaving \(p^{a-1}(p-1)\) unique elements which are counted \(k\) times, where \(k\) is the second smallest possible value of \(n_p\) after 1. Excluding the identity, the total number of distinct elements in Sylow \(p\)-subgroups is then \((p^{a-1}-1) + k p^{a-1}(p-1)\). If the sum over all \(p\) plus one exceeds \(\left|G\right|\), then some Sylow \(p\)-subgroup must be normal.
7. Groups where \(\left|G\right| \nmid m!\), where \(m\) is the index of some Sylow subgroup
Let \(P \in Syl_p(G)\). \(G\)'s action of left multiplication on the left cosets of \(P\) induces a permutation representation \(\varphi : G \rightarrow S_m\). If \(G\) is simple, then \(\ker \varphi\) is forced to be 1, since it is a normal proper subgroup. Hence \(G\) is isomorphic to a subgroup of \(S_m\). But this cannot be since \(\left|G\right| \nmid m!\), thus \(G\) is not simple.
8. Groups where \(\left|G\right| \nmid k!\), for all possible Sylow numbers \(k\) of a prime \(p\)
\(G\) acts by conjugation on \(Syl_p(G)\), inducing a permutation representation \(\varphi : G \rightarrow S_k\), where \(k\) is a possible Sylow number (only satisfying the congruence condition). Let \(P \in Syl_p(G)\). If \(G\) is simple, then the statements \(\ker \varphi = \cap_{i} N_G(g_i P g_i^{-1}) \le N_G(P) < G\) and \(\ker \varphi \trianglelefteq G\) force \(\ker \varphi = 1\). The argument continues as per 7.
9. Analysis of order \(90 = 2 \cdot 3^2 \cdot 5\) groups
The possible Sylow numbers for \(n_2\), \(n_3\) and \(n_5\) are \(\{1,3,5,9,15,45\}\), \(\{1,10\}\) and \(\{1,6\}\) respectively. Note that \(n_2 < 5\), because otherwise the number of distinct elements in all Sylow subgroups would exceed 90 (see 6). But 90 does not divide \(1!\) and \(3!\), so \(G\) is not simple.
These 9 steps are sufficient to filter out everything except order 60 groups. For completeness, here is the code used to perform all the calculations (note that it does not filter 90 as it requires special analysis):
We proceed by systematically filtering out all groups of order \(\neq 60\):
1. \(p\)-groups
These have a nontrivial center which is normal in \(G\), thus they are not simple.
2. Groups with order \(pq\)
These groups are guaranteed to have a normal Sylow \(q\)-subgroup.
3. Groups with order \(p^2 q\)
These groups have a normal Sylow subgroup for either \(p\) or \(q\).
4. Groups with order \(pqr\)
It has been proven (in 4.5.16) that these groups have a normal Sylow subgroup for either \(p\), \(q\) or \(r\).
5. Groups with \(n_p = 1\) for some prime p
Unique Sylow subgroups are normal.
6. Groups where the no. of distinct elements in Sylow subgroups (if not normal) exceeds its order
Let \(\left|G\right| = p^a m\), where \(p \nmid m\). We derive a lower bound for the number of distinct elements in all Sylow \(p\)-subgroups by assuming the largest possible intersection, which has size \(p^{a-1}\). These elements are only counted once, leaving \(p^{a-1}(p-1)\) unique elements which are counted \(k\) times, where \(k\) is the second smallest possible value of \(n_p\) after 1. Excluding the identity, the total number of distinct elements in Sylow \(p\)-subgroups is then \((p^{a-1}-1) + k p^{a-1}(p-1)\). If the sum over all \(p\) plus one exceeds \(\left|G\right|\), then some Sylow \(p\)-subgroup must be normal.
7. Groups where \(\left|G\right| \nmid m!\), where \(m\) is the index of some Sylow subgroup
Let \(P \in Syl_p(G)\). \(G\)'s action of left multiplication on the left cosets of \(P\) induces a permutation representation \(\varphi : G \rightarrow S_m\). If \(G\) is simple, then \(\ker \varphi\) is forced to be 1, since it is a normal proper subgroup. Hence \(G\) is isomorphic to a subgroup of \(S_m\). But this cannot be since \(\left|G\right| \nmid m!\), thus \(G\) is not simple.
8. Groups where \(\left|G\right| \nmid k!\), for all possible Sylow numbers \(k\) of a prime \(p\)
\(G\) acts by conjugation on \(Syl_p(G)\), inducing a permutation representation \(\varphi : G \rightarrow S_k\), where \(k\) is a possible Sylow number (only satisfying the congruence condition). Let \(P \in Syl_p(G)\). If \(G\) is simple, then the statements \(\ker \varphi = \cap_{i} N_G(g_i P g_i^{-1}) \le N_G(P) < G\) and \(\ker \varphi \trianglelefteq G\) force \(\ker \varphi = 1\). The argument continues as per 7.
9. Analysis of order \(90 = 2 \cdot 3^2 \cdot 5\) groups
The possible Sylow numbers for \(n_2\), \(n_3\) and \(n_5\) are \(\{1,3,5,9,15,45\}\), \(\{1,10\}\) and \(\{1,6\}\) respectively. Note that \(n_2 < 5\), because otherwise the number of distinct elements in all Sylow subgroups would exceed 90 (see 6). But 90 does not divide \(1!\) and \(3!\), so \(G\) is not simple.
These 9 steps are sufficient to filter out everything except order 60 groups. For completeness, here is the code used to perform all the calculations (note that it does not filter 90 as it requires special analysis):
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 | # Primes less than 100 primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97] def factorial(n): k = 1 for i in range(2, n+1): k *= i return k def prime_factorize(n): factors = dict() for p in primes: _n = n a = 0 while True: if _n % p != 0: break a += 1 factors.update({p: a}) _n /= p return factors def sylow_nums(n, factors): nums = dict() for p, a in factors.items(): i = 1 m = n / p**a possible = [] while True: if i > m: break if m % i == 0: possible.append(i) i += p nums.update({p: possible}) return nums for i in range(2, 100): factors = prime_factorize(i) # If this is false, then any group G of order i is guaranteed not # to be a non-abelian simple group. wanted = True # G is a p-group if len(factors) == 1: wanted = False # |G| has the form pq or (p**2)q if len(factors) == 2: if set(factors.values()) == {1} or \ set(factors.values()) == {1,2}: wanted = False # |G| has the form pqr if len(factors) == 3 and set(factors.values()) == {1}: wanted = False nums = sylow_nums(i, factors) if [1] in nums.values(): # G has a unique Sylow subgroup wanted = False else: # The number of distinct elements in Sylow subgroups (if they # are nonnormal) is greater than |G| j = 1 # Include the identity for p, ns in nums.items(): # Max no of 'shared' elements across all Sylow p-subgroups # (i.e. largest possible intersection) shared = p**(factors[p]-1) # No of elements in each Sylow p-subgroup not shared unique = p**factors[p] - shared j += (shared-1) + unique*ns[1] if j > i: wanted = False # |G| does not divide m!, where m is the index of some Sylow subgroup for p, a in factors.items(): if factorial(i // p**a) % i != 0: wanted = False # |G| does not divide k!, for all possible Sylow numbers k of some prime p for p, ns in nums.items(): for n in ns: if factorial(n) % i == 0: break else: wanted = False break if wanted: print(i) |
Monday, 6 April 2020
Sylow subgroups of groups with order \(pqr\), \(p < q < r\) (D&F 4.5.16)
Let \(\left|G\right| = pqr\), where \(p,q,r\) are primes with \(p < q < r\). Prove that \(G\) has a normal Sylow subgroup for either \(p\), \(q\) or \(r\).
Suppose \(G\) has no normal Sylow subgroups. Then \(n_p, n_q, n_r > 1\). This combined with the fact that \(n_p \mid qr\), \(n_q \mid pr\) and \(n_r \mid pq\) implies the possibilities for \(n_p\), \(n_q\) and \(n_r\) are \(\{q,r,qr\}\), \(\{p,r,pr\}\) and \(\{p,q,pq\}\) respectively. Furthermore, using the fact that \(p < q < r\) and \(n_p = 1+kp\) (likewise for \(n_q\) and \(n_r\)), we can narrow the possibilities for \(n_q\) and \(n_r\) to \(\{r,pr\}\) and \(\{pq\}\) respectively.
Now we shall choose the smallest possibilities for \(n_p\), \(n_q\) and \(n_r\) (that is, \(q\), \(r\) and \(pq\) respectively). Clearly all the Sylow subgroups have trivial intersection, so the total number of distinct nonidentity elements is \(q(p-1)+r(q-1)+pq(r-1) = pqr+qr-q-r > pqr\), which is a contradiction. Since these were the smallest possibilities, all other possibilities lead to the same contradiction. This completes the proof.
Suppose \(G\) has no normal Sylow subgroups. Then \(n_p, n_q, n_r > 1\). This combined with the fact that \(n_p \mid qr\), \(n_q \mid pr\) and \(n_r \mid pq\) implies the possibilities for \(n_p\), \(n_q\) and \(n_r\) are \(\{q,r,qr\}\), \(\{p,r,pr\}\) and \(\{p,q,pq\}\) respectively. Furthermore, using the fact that \(p < q < r\) and \(n_p = 1+kp\) (likewise for \(n_q\) and \(n_r\)), we can narrow the possibilities for \(n_q\) and \(n_r\) to \(\{r,pr\}\) and \(\{pq\}\) respectively.
Now we shall choose the smallest possibilities for \(n_p\), \(n_q\) and \(n_r\) (that is, \(q\), \(r\) and \(pq\) respectively). Clearly all the Sylow subgroups have trivial intersection, so the total number of distinct nonidentity elements is \(q(p-1)+r(q-1)+pq(r-1) = pqr+qr-q-r > pqr\), which is a contradiction. Since these were the smallest possibilities, all other possibilities lead to the same contradiction. This completes the proof.
Sylow \(p\)-subgroups of order-56 groups (D&F 4.5.13)
Prove that a group of order 56 has a normal Sylow \(p\)-subgroup for some prime \(p\) dividing its order.
The primes diving 56 are 2 and 7, and a quick check shows that the possibilities for \(n_2\) and \(n_7\) are \(1,7\) and \(1,8\) respectively. Suppose that \(n_2 = 7\) and \(n_7 = 8\). All nonidentity elements in a Sylow 7-subgroup have order 7, and the intersection of 2 distinct Sylow 7-subgroups is trivial. Thus there are \(8 \cdot (7-1) = 48\) distinct elements of order 7. Since elements of a Sylow 2-subgroup cannot have order 7, they must be contained in the remaining 8 elements. But a Sylow 2-subgroup already has 8 elements, so there cannot possibly be 7 of them, since they would all be equal. We can thus conclude that either \(n_2\) or \(n_7\) (or both) is 1.
The primes diving 56 are 2 and 7, and a quick check shows that the possibilities for \(n_2\) and \(n_7\) are \(1,7\) and \(1,8\) respectively. Suppose that \(n_2 = 7\) and \(n_7 = 8\). All nonidentity elements in a Sylow 7-subgroup have order 7, and the intersection of 2 distinct Sylow 7-subgroups is trivial. Thus there are \(8 \cdot (7-1) = 48\) distinct elements of order 7. Since elements of a Sylow 2-subgroup cannot have order 7, they must be contained in the remaining 8 elements. But a Sylow 2-subgroup already has 8 elements, so there cannot possibly be 7 of them, since they would all be equal. We can thus conclude that either \(n_2\) or \(n_7\) (or both) is 1.
Sylow 2-subgroups of \(D_{2n}\) (D&F 4.5.12)
Let \(2n = 2^a k\) where \(k\) is odd. Prove that the number of Sylow 2-subgroups of \(D_{2n}\) is \(k\).
One approach is to prove that \(N_{D_{2n}}(P) = P\) for any \(P \in Syl_2(D_{2n})\) and calculate \(n_2\) directly. However I have tried for hours to work out this approach to no avail. So here is an alternative approach:
There are \(k\) Sylow 2-subgroups of the form \(\langle sr^i, r^k \rangle\), where \(0 \le i \le k-1\), so \(n_2 \ge k\). Furthermore, \(n_2 \mid k\) so \(n_2 \le k\).
One approach is to prove that \(N_{D_{2n}}(P) = P\) for any \(P \in Syl_2(D_{2n})\) and calculate \(n_2\) directly. However I have tried for hours to work out this approach to no avail. So here is an alternative approach:
There are \(k\) Sylow 2-subgroups of the form \(\langle sr^i, r^k \rangle\), where \(0 \le i \le k-1\), so \(n_2 \ge k\). Furthermore, \(n_2 \mid k\) so \(n_2 \le k\).
Sylow \(p\)-subgroups of \(D_{2n}\) (D&F 4.5.5)
Show that a Sylow \(p\)-subgroup of \(D_{2n}\) is cyclic and normal for every odd prime \(p\).
Let \(2n = p^\alpha m\), where \(p \nmid m\). Note that \(m\) is even since \(2n\) is even and \(p^\alpha\) is odd. Thus we can consider the cyclic subgroup \(\langle r^{m/2} \rangle\), which has order \(p^\alpha\) and is hence a Sylow \(p\)-subgroup. Furthermore, \(\langle r^{m/2} \rangle \trianglelefteq D_{2n}\) so it is unique.
Let \(2n = p^\alpha m\), where \(p \nmid m\). Note that \(m\) is even since \(2n\) is even and \(p^\alpha\) is odd. Thus we can consider the cyclic subgroup \(\langle r^{m/2} \rangle\), which has order \(p^\alpha\) and is hence a Sylow \(p\)-subgroup. Furthermore, \(\langle r^{m/2} \rangle \trianglelefteq D_{2n}\) so it is unique.
Tuesday, 31 March 2020
Groups of order 3825 (D&F 4.4.12)
Let \(G\) be a group of order 3825. Prove that if \(H\) is a normal subgroup of order 17 in \(G\) then \(H \le Z(G)\).
Since \(H\) has prime order, \(\text{Aut}(H)\) is isomorphic to \((\mathbb{Z}/17\mathbb{Z})^\times\) which has order 16. Since \(H \triangleleft G\), \(G/C_G(H)\) is isomorphic to a subgroup of \(\text{Aut}(G)\), meaning its order is either 1, 2, 4, 8 or 16. But only 1 divides \(|G| = 3825\), so it follows that \(C_G(H) = G\) and \(H \le Z(G)\).
Since \(H\) has prime order, \(\text{Aut}(H)\) is isomorphic to \((\mathbb{Z}/17\mathbb{Z})^\times\) which has order 16. Since \(H \triangleleft G\), \(G/C_G(H)\) is isomorphic to a subgroup of \(\text{Aut}(G)\), meaning its order is either 1, 2, 4, 8 or 16. But only 1 divides \(|G| = 3825\), so it follows that \(C_G(H) = G\) and \(H \le Z(G)\).
Conjugation on normal subgroup by quotient space (D&F 4.4.10)
Let \(G\) be a group and \(A\) be an abelian normal subgroup of \(G\). Show that \(\bar{G}\) acts on conjugation on \(A\) by \(\bar{g} \cdot a = gag^{-1}\), where \(g\) is any representative of the coset \(\bar{g}\). Give an explicit example to show that this action is not well defined if \(A\) is non-abelian.
The proof of the first part is rather straightforward and hence omitted.
Consider the normal subgroup \(A_5\) in \(S_5\); clearly \(A_5\) is not abelian since \((1\ 2\ 3\ 4\ 5)(1\ 2\ 4\ 3\ 5) = (1\ 3)(2\ 5)\) while \((1\ 2\ 4\ 3\ 5)(1\ 2\ 3\ 4\ 5) = (1\ 4)(2\ 5)\). Let \(g = (1\ 2)(3\ 4)\), \(h = (1\ 2)(3\ 5)\) and \(a = (1\ 2)(3\ 4)\). All of them are in \(A_5\), i.e. \(\bar{g} = \bar{h}\). However we have \(\bar{g} \cdot a = (1\ 2)(3\ 4)\) and \(\bar{h} \cdot a = (1\ 2)(4\ 5)\) which are not equal, so the action is not well-defined.
The proof of the first part is rather straightforward and hence omitted.
Consider the normal subgroup \(A_5\) in \(S_5\); clearly \(A_5\) is not abelian since \((1\ 2\ 3\ 4\ 5)(1\ 2\ 4\ 3\ 5) = (1\ 3)(2\ 5)\) while \((1\ 2\ 4\ 3\ 5)(1\ 2\ 3\ 4\ 5) = (1\ 4)(2\ 5)\). Let \(g = (1\ 2)(3\ 4)\), \(h = (1\ 2)(3\ 5)\) and \(a = (1\ 2)(3\ 4)\). All of them are in \(A_5\), i.e. \(\bar{g} = \bar{h}\). However we have \(\bar{g} \cdot a = (1\ 2)(3\ 4)\) and \(\bar{h} \cdot a = (1\ 2)(4\ 5)\) which are not equal, so the action is not well-defined.
Monday, 30 March 2020
Non-abelian groups of order pq, p and q prime (D&F 4.3.27)
Let \(p\) and \(q\) be primes with \(p < q\). Prove that a non-abelian group \(G\) of order \(pq\) has a nonnormal subgroup of index \(q\), so that there exists an injective homomorphism into \(S_q\). Deduce that \(G\) is isomorphic to a subgroup of \(N_{S_q}(\left<(1\ 2\ \ldots\ q)\right>)\).
Note that \(Z(G) \neq G\) since \(G\) is non-abelian, and also \(Z(G)\) is neither \(p\) nor \(q\) since otherwise \(G/Z(G)\) would be cyclic. This forces \(Z(G) = 1\). Let \(g_1, g_2, \ldots, g_r\) be representatives of the conjugacy classes of \(G\) not in \(Z(G)\). For each \(g_i\), the subgroup \(G_i = \left<g_i\right>\) cannot have order \(pq\) since that means \(G\) is cyclic. Hence \(\left|G_i\right|\) is either \(p\) or \(q\). If \(\left|G_i\right| = p\), then \(G_i \le C_G(G_i) < G\), so \(\left|G : C_G(G_i)\right| = q\). Likewise if \(\left|G_i\right| = q\) then \(\left|G : C_G(G_i)\right| = p\). Hence the class equation of \(G\) is \(1 + mp + nq\), for \(m,n \in \mathbb{Z^+}\). It is clear that the value of any sum including the term \(1\) is either \(1\) or some \(k > p\). Thus any order-\(p\) subgroup cannot be expressed as a union of conjugacy classes, and so is nonnormal in \(G\).
By Cauchy's Theorem, there exists an order-\(p\) subgroup \(H\) and an order-\(q\) element \(x\). Since the only normal subgroup of \(G\) contained in \(H\) is the trivial subgroup, \(\ker \pi_H\) (the permutation representation of the action of \(G\) on \(G/H\)) is trivial, and so \(\pi_H\) is an injective homomorphism from \(G\) to \(S_{G/H} \cong S_q\). Now, \(G/H\) is cyclic since it has prime order. Since \(\left<x\right> \cap H = 1\) (which is up to the reader to justify because I don't feel like doing it), we can write \(G/H = \left<xH\right>\). In particular, \(\pi_H(x) = (H\ xH\ x^2H\ \ldots\ x^{q-1}H)\). Furthermore \(\left<x\right> \trianglelefteq G\) since it has index \(p\) (the smallest prime dividing \(G\)). This implies \(\left<\pi_H(x)\right> = \pi_H(\left<x\right>) \trianglelefteq \text{im}\ \pi_H\). By the First Isomorphism Theorem we can conclude that \(G \cong \text{im}\ \pi_H \le N_{S_{G/H}}(\left<\pi_H(x)\right>) = N_{S_{G/H}}(\left<(H\ xH\ x^2H\ \ldots\ x^{q-1}H)\right>)\).
(ok to be honest I looked up the solution for that last part)
Note that \(Z(G) \neq G\) since \(G\) is non-abelian, and also \(Z(G)\) is neither \(p\) nor \(q\) since otherwise \(G/Z(G)\) would be cyclic. This forces \(Z(G) = 1\). Let \(g_1, g_2, \ldots, g_r\) be representatives of the conjugacy classes of \(G\) not in \(Z(G)\). For each \(g_i\), the subgroup \(G_i = \left<g_i\right>\) cannot have order \(pq\) since that means \(G\) is cyclic. Hence \(\left|G_i\right|\) is either \(p\) or \(q\). If \(\left|G_i\right| = p\), then \(G_i \le C_G(G_i) < G\), so \(\left|G : C_G(G_i)\right| = q\). Likewise if \(\left|G_i\right| = q\) then \(\left|G : C_G(G_i)\right| = p\). Hence the class equation of \(G\) is \(1 + mp + nq\), for \(m,n \in \mathbb{Z^+}\). It is clear that the value of any sum including the term \(1\) is either \(1\) or some \(k > p\). Thus any order-\(p\) subgroup cannot be expressed as a union of conjugacy classes, and so is nonnormal in \(G\).
By Cauchy's Theorem, there exists an order-\(p\) subgroup \(H\) and an order-\(q\) element \(x\). Since the only normal subgroup of \(G\) contained in \(H\) is the trivial subgroup, \(\ker \pi_H\) (the permutation representation of the action of \(G\) on \(G/H\)) is trivial, and so \(\pi_H\) is an injective homomorphism from \(G\) to \(S_{G/H} \cong S_q\). Now, \(G/H\) is cyclic since it has prime order. Since \(\left<x\right> \cap H = 1\) (which is up to the reader to justify because I don't feel like doing it), we can write \(G/H = \left<xH\right>\). In particular, \(\pi_H(x) = (H\ xH\ x^2H\ \ldots\ x^{q-1}H)\). Furthermore \(\left<x\right> \trianglelefteq G\) since it has index \(p\) (the smallest prime dividing \(G\)). This implies \(\left<\pi_H(x)\right> = \pi_H(\left<x\right>) \trianglelefteq \text{im}\ \pi_H\). By the First Isomorphism Theorem we can conclude that \(G \cong \text{im}\ \pi_H \le N_{S_{G/H}}(\left<\pi_H(x)\right>) = N_{S_{G/H}}(\left<(H\ xH\ x^2H\ \ldots\ x^{q-1}H)\right>)\).
(ok to be honest I looked up the solution for that last part)
Saturday, 28 March 2020
Proving the Fundamental Theorem of Arithmetic via Jordan–Hölder
Let \(n \in \mathbb{Z}\). By Jordan–Hölder, the cyclic group \(Z_n\) has a composition series \(1 = G_0 \trianglelefteq G_1 \trianglelefteq \ldots \trianglelefteq G_s = Z_n\), where \(G_{i+1}/G_i\) is simple. It is known that quotient groups and subgroups of cyclic groups are simple. Thus all \(G_i\), and hence all \(G_{i+1}/G_i\) are cyclic. In particular, \(G_{i+1}/G_i\) must have prime order \(p_i\) due to its simplicity. So we can conclude that \(n = \left|Z_n\right| = \left|G_s\right| = \left|G_s/G_{s-1}\right|\left|G_{s-1}/G_{s-2}\right| \ldots \left|G_1/G_0\right| = p_{s-1}p_{s-2} \ldots p_0\) is a product of primes. Furthermore, the composition series is unique up to isomorphism of quotient groups and rearrangement, so the list of \(p_i\)'s is unique. Thus \(n\) has a unique prime factorization.
Friday, 27 March 2020
Conjugates in \(GL_2(\mathbb{C})\) (D&F 4.3.25)
Let \(G = GL_2(\mathbb{C})\) and \(H = \{\begin{bmatrix}a&b\\0&c\\\end{bmatrix} \mid a,b,c \in \mathbb{C}, ac \neq 0\}\). Prove that every element of \(G\) is conjugate to some element of \(H\) and deduce that \(G\) is the union of conjugates of \(H\).
Let \(g = \begin{bmatrix}i&j\\k&l\\\end{bmatrix} \in G\), and \(\beta = \{e_1,e_2\}\) be the standard basis of \(\mathbb{C}^2\). By the Fundamental Theorem of Algebra, the characteristic polynomial of \(G\) has some root \(\lambda \in \mathbb{C}\), which is a complex eigenvalue corresponding to the eigenvector \(v\). Clearly \(v\) is linearly independent from either \(e_1\) or \(e_2\) (or both); we shall assume it is linearly independent from \(e_1\) (the argument for \(e_2\) is similar). Then \(\gamma = \{v,e_1\}\) is another basis for \(\mathbb{C}^2\); now let \(h = [L_g]_\gamma = \begin{bmatrix}\lambda&i'\\0&j'\\\end{bmatrix}\). We know that \(\lambda \neq 0\) since \(h\) is invertible, and also \(j' \neq 0\) because otherwise \(g((\lambda/i')e_1) = (\lambda/i')i'v = \lambda v = gv\) which shows that \(e_1\) and \(v\) are not linearly independent. Thus \(h \in H\), and \(h\) conjugated by the change-of-basis matrix \([I]^\beta_\gamma\) is \(g\).
Let \(g = \begin{bmatrix}i&j\\k&l\\\end{bmatrix} \in G\), and \(\beta = \{e_1,e_2\}\) be the standard basis of \(\mathbb{C}^2\). By the Fundamental Theorem of Algebra, the characteristic polynomial of \(G\) has some root \(\lambda \in \mathbb{C}\), which is a complex eigenvalue corresponding to the eigenvector \(v\). Clearly \(v\) is linearly independent from either \(e_1\) or \(e_2\) (or both); we shall assume it is linearly independent from \(e_1\) (the argument for \(e_2\) is similar). Then \(\gamma = \{v,e_1\}\) is another basis for \(\mathbb{C}^2\); now let \(h = [L_g]_\gamma = \begin{bmatrix}\lambda&i'\\0&j'\\\end{bmatrix}\). We know that \(\lambda \neq 0\) since \(h\) is invertible, and also \(j' \neq 0\) because otherwise \(g((\lambda/i')e_1) = (\lambda/i')i'v = \lambda v = gv\) which shows that \(e_1\) and \(v\) are not linearly independent. Thus \(h \in H\), and \(h\) conjugated by the change-of-basis matrix \([I]^\beta_\gamma\) is \(g\).
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