(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.
Wednesday, 29 April 2020
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.
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