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 \(H \le Z(G)\).

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.

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)

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.

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 \(g\).

Maximal subgroups and conjugates (D&F 4.3.23)

Prove that if \(M\) is a maximal subgroup of \(G\) then either \(N_G(M) = M\) or \(N_G(M) = G\). Deduce that if \(M\) is a maximal subgroup of \(G\) that is not normal in \(G\) then the number of nonidentity elements in \(G\) that are contained in conjugates of \(M\) is at most \((\left|M\right|-1)\left|G : M\right|\).

The first statement is clearly true since \(M \le N_G(M) \le G\). As for the second part, we shall first prove the following lemma: if \(g_1M = g_2M\), then \(g_1Mg_1^{-1} = g_2Mg_2^{-1}\).

Let \(m \in M\). Since \(g_2^{-1}g_1\) and \(g_1^{-1}g_2\) are both in \(M\), it follows that for some \(m' \in M\), \((g_2^{-1}g_1)m(g_1^{-1}g_2) = m'\), or \(g_1mg_1^{-1} = g_2m'g_2^{-1}\). So \(g_1Mg_1^{-1} \subseteq g_2Mg_2^{-1}\). By symmetry, \(g_2Mg_2^{-1} \subseteq g_1Mg_1^{-1}\) as well, giving us the desired result.

The lemma tells us that there are only \(\left|G : M\right|\) distinct conjugates of \(M\). Furthermore, the element \(g1g^{-1} = 1 \in gMg^{-1}\) is not counted, so at most \(\left|M\right|-1\) nonidentity elements are contained in conjugates of \(M\). The total is as its maximum when all conjugates are distinct; then the total is precisely \((\left|M\right|-1)\left|G : M\right|\).

Note that if \(M\) is a normal subgroup of \(G\), then the maximum total is instead \(\left|M\right|-1\), since \(gMg^{-1} = M\).

Conjugacy classes of group products (D&F 4.3.3)

Find all conjugacy classes and their sizes in the following groups:
(a) \(Z_2 \times S_3\)
(b) \(S_3 \times S_3\)
(c) \(Z_3 \times A_4\)

First, let \(K_G\) be the set of all conjugacy classes of \(G\), and denote elements of cyclic groups as \(a^k\) where \(k \in \mathbb{Z}\). Second, observe that \((a^i)^{(a^j)} = a^j a^i a^{-j} = a^i\).

(a) Let \((a^i,x),(a^j,y) \in Z_2 \times S_3\). As noted above, \((a^i,x)^{(a^j,y)} = (a^i,x^y)\). Thus we can identify \((a^i,x)\) with \(x\) and \((a^j,y)\) with \(y\) and conclude that the set of all conjugacy classes of the set \(\{(a^i,b) \mid b \in S_3\}\) (denoted \(K_i\)) is precisely that of \(S_3\), but with each individual permutation \(b\) replaced by \((a^i,b)\). Thus \(K_{Z_2 \times S_3} = K_0 \cup K_1\). (It would be too troublesome to list it out explicitly.)

(b) Suppose \((a,b)\) and \((c,d)\) are conjugate in \(S_3 \times S_3\). Then \((c,d) = (a,b)^{(x,y)} = (a^x,b^y)\) for some \((x,y) \in S_3 \times S_3\), so \(a\) and \(c\) as well as \(b\) and \(d\) are conjugate, i.e. have the same cycle type. In other words, \(K_{S_3 \times S_3}\) consists of the Cartesian products of all ordered pairs containing sets in \(K_{S_3}\).

(c) As in (a), let \(K_i\) be the set of conjugacy classes of the set \(\{(a^i,b) \mid b \in A_4\}\). Then \(K_{Z_3 \times A_4} = K_0 \cup K_1 \cup K_2\).

Conjugacy classes of \(A_4\) (D&F 4.3.2(c))

Find all conjugacy classes and their sizes in \(A_4\).

\(A_4\) is precisely the set of permutations with cycle type \((1,1,1,1)\), \((1,3)\) or \((2,2)\). We shall handle these case by case.

\((1,1,1,1)\): corresponds to the identity, the conjugacy class is simply \(\{1\}\).

\((1,3)\): there are \((4 \cdot 3 \cdot 2)/3 = 8\) possible 3-cycles. We will pick a representative \(x = (1\ 2\ 3)\). Note that \(C_{A_4}(x) = \left<x\right>\), since the only permutation fixing 1, 2 and 3 is the identity. The centraliser has order 3, so the conjugacy class containing \(x\) has \(12/3 = 4\) elements, meaning there are \(8/4 = 2\) conjugacy classes. Observe that for any 3-cycle \((a\ b\ c)\), there is some \(\tau \in S_4\) such that \(x^\tau = (\tau(1)\ \tau(2)\ \tau(3)) = (a\ b\ c)\). However, if we restrict \(\tau\) to be in \(A_4\) then \(x^\tau\) can only be in the conjugacy class of \(x\). We can infer that the 3-cycles in the other conjugacy class are conjugates of \(x\) by an odd permutation, which only swap 2 elements. Using this fact, we can list the other conjugacy class: \(\{(1\ 3\ 2), (4\ 2\ 3), (1\ 4\ 3), (1\ 2\ 4)\}\) (the first element swaps 2 and 3, while the rest swap 1, 2, and 3 with 4 respectively). Thus the conjugacy class of \(x\) consists of the remaining 3-cycles: \(\{(1\ 2\ 3), (4\ 3\ 2), (1\ 3\ 4), (1\ 4\ 2)\}\).

\((2,2)\): there are only 3 such permutations, and it can be noted that \(((1\ 2)(3\ 4))^{(2\ 3\ 4)} = (1\ 3)(2\ 4)\) and \(((1\ 2)(3\ 4))^{(2\ 4\ 3)} = (1\ 4)(2\ 3)\). Thus \(\{(1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\ 4)(2\ 3)\}\) is a conjugacy class.

Classification of order 6 groups (D&F 4.2.10)

Prove that every non-abelian group of order 6 has a nonnormal subgroup of order 2. Use this to classify groups of order 6.

Let \(G\) be a non-abelian order 6 group. By Cauchy's Theorem, there are 2 subgroups \(H = \{1, h\}\) and \(K = \{1, k, k^2\}\). Note that \(h \neq k,k^2\) since \(h\) has order 2 but \(k\) and \(k^2\) do not. Also, \(K \trianglelefteq G\) as it has index 2. In particular, \(hkh^{-1} \in K\). However, \(k \neq 1\) so \(hk \neq h\) and thus \(hkh^{-1} \neq 1\). Furthermore \(hkh^{-1} \neq k\). To see why, observe that there are 2 cosets \(1K\) and \(hK\) in \(G\). This means that every element in \(G\) can be written in terms of \(h\) and \(k\), so \(G = \left<h,k\right>\). But \(hkh^{-1} = k\) implies that \(h\) and \(k\) commute, which means that all elements of \(G\) commute, contradicting the fact that \(G\) is non-abelian. So the only remaining possibility is \(hkh^{-1} = k^2\), or \(khk = h\). But \(khk^{-1} = khk^2 = hk \not\in H\), and so we conclude that \(H\) is nonnormal.

It is easily verified that if \(hkh^{-1} = k\), then \(G = \left<hk\right>\) and so \(G \cong Z_6\). On the other hand suppose \(hkh^{-1} = k^2\). Then \(\ker \pi_H \neq H\) as \(H\) is not normal in \(G\). So \(\ker \pi_H = 1\), i.e. \(\pi_H\) is injective. Since \(\pi_H(G) \le S_3\) (because there are 3 cosets of \(H\)) and \(|\pi_H(G)| = 6 = |S_3|\), it follows that \(\pi_H\) is an isomorphism. Thus \(G \cong S_3\).

Group actions: Permutation representations

A group action of a group \(G\) on a set \(A\) is a function \(\varphi : G \times A \to A\) satisfying the following properties (here \(\varphi(g,a)\) will be denoted by \(g \cdot a\) and so we shall dispense with \(\varphi\) altogether):

1. \(g \cdot (h \cdot a) = (gh) \cdot a\) for all \(g,h \in G\), \(a \in A\)
2. \(1 \cdot a = a\) for all \(a \in A\)

A fundamental result is the existence of a bijection between group actions and homomorphisms \(\phi\) from \(G\) to \(S_A\).

Let \(g \in G\) be arbitrary. The function \(\sigma_g\) is defined such that \(\sigma_g(a) = g \cdot a\) for all \(a \in A\). We can verify that \(\sigma_g\) is a permutation of \(A\): since \((\sigma_g\sigma_{g^{-1}})(a) = \sigma_g(\sigma_{g^{-1}}(a)) = g \cdot (g^{-1} \cdot a) = (gg^{-1}) \cdot a = a\) for all \(a \in A\), \(\sigma_g\sigma_{g^{-1}}\) is precisely the identity permutation. Thus \(\sigma_g\) has an inverse (namely \(\sigma_{g^{-1}}\)), and so it is a bijection in \(A\). Now, we can define the function \(\phi : g \to \sigma_g\) mapping \(G\) to \(S_A\), and show that it is indeed a homomorphism: \(\phi(gh)(a) = \sigma_{gh}(a) = (gh) \cdot a = g \cdot (h \cdot a) = (\sigma_g\sigma_h)(a) = (\phi(g)\phi(h))(a)\) for all \(a \in A\).

So far we have established the existence of a homomorphism \(\phi : G \to S_A\) for any group action. What remains is to prove the existence of a group action  given a homomorphism \(\phi\). This is in fact rather trivial: simply define \(g \cdot a = \phi(g)(a)\); it is easy to check that the properties of group actions are satisfied. In light of this result, we can say that a group action affords or induces the associated permutation representation \(\phi\).

We can proceed by defining the left regular representation of \(G\) to be the permutation representation afforded by the action of the group \(G\) on the set \(G\) by left multiplication (meaning that \(g \cdot h = gh\)). Similarly we can define right regular representations (where \(g \cdot h = hg\)). Given the intimate relation between operations on group elements and permutations, the following result seems quite natural:

(Cayley's Theorem) Every group of order \(n\) is isomorphic to a subgroup of \(S_n\).

Subgroups and quotient groups of a solvable group are solvable (D&F 3.4.5)

Prove that subgroups and quotient groups of a solvable group are solvable.

Subgroups: Let \(G\) be a solvable group and \(H\) be any subgroup. By definition, there is a chain of subgroups \(1 = G_0 \trianglelefteq G_1 \trianglelefteq \cdots \trianglelefteq G_s = G\) such that \(G_{i+1}/G_i\) is abelian for all \(0 \le i \le s-1\). Using this notation, let \(H_s = H\) and \(H_i = H_{i+1} \cap G_i\) for \(0 \le i \le s-1\) (note that \(H_0 = G_0 = 1\)). For convenience purposes, we shall also define \(H_{s+1} = G\). Since \(H_{i+1} = H_{i+2} \cap G_{i+1} \le G_{i+1} \le N_G(G_i)\), the Diamond Isomorphism Theorem states that \(H_i = H_{i+1} \cap G_i \trianglelefteq H_{i+1}\). Furthermore, \(G_i \le H_{i+1}G_i \le G_{i+1}\) (the latter inequality following from the fact that \(H_{i+1} = H_{i+2} \cap G_{i+1} \subseteq G_{i+1}\) and \(G_i \subseteq G_{i+1}\)), so \(H_{i+1}/H_i = H_{i+1}/(H_{i+1} \cap G_i) \cong H_{i+1}G_i/G_i\) is a subgroup of \(G_{i+1}/G_i\) (by the Lattice Isomorphism Theorem) and is thus abelian. Hence \(H\) is solvable.

Quotient groups: Let \(N\) be a normal subgroup of \(G\). Using the previously defined notation, we can construct a chain of subgroups \(N = NG_0 \le NG_1 \le \cdots \le NG_s = G\). Since \(G_{i+1}\) normalises \(N\) and \(G_i\), it also normalises \(NG_i\). Thus \(G_{i+1}(NG_i)\) is a subgroup and \(NG_i \trianglelefteq G_{i+1}(NG_i) = N(G_iG_{i+1}) = NG_{i+1}\). By an argument involving commutators (which I admittedly still don't understand) we can also show that \(NG_{i+1}/NG_i\) is abelian. Using the lattice isomorphism theorem, we can construct a chain \(1 = N/N = NG_0/N \trianglelefteq NG_1/N \trianglelefteq \cdots \trianglelefteq NG_s/N = G/N\). Furthermore, \((NG_{i+1}/N)/(NG_i/N) \cong (NG_{i+1}/NG_i)\) is abelian, and we can conclude that \(G/N\) is solvable.

Cosets, orbits and Lagrange's Theorem

The standard proof of Lagrange's Theorem for a subgroup \(H \le G\) involves establishing a bijection \(h \leftrightarrow gh\) between \(H\) and each coset of \(H\) in \(G\), and concluding that since \(\left|H\right| = \left|gH\right|\) and the cosets partition \(G\), \(\left|G\right|\) divides \(\left|H\right|\).

There is an equivalent formulation using orbits. Let \(H\) act on \(G\) by right multiplication. Then the orbit of any \(g \in G\) under \(H\) is defined to be \(\{gh \mid h \in H\}\), which is just the left coset \(gH\). By proving that each orbit has \(\left|H\right|\) elements and the orbits partition \(G\), we arrive at essentially the same proof.

Doubly transitive groups (D&F 4.1.8)

A transitive permutation group \(G\) on a set \(A\) is doubly transitive if for all \(a \in A\) the subgroup \(G_a\) is transitive on \(A-\{a\}\).

(a) Prove that \(S_n\) is doubly transitive on \(A = \{1,2,\cdots,n\}\) for all \(n \ge 2\).

Let \(G = S_n\). Clearly \(G\) is transitive on \(A\). Furthermore, for any \(i \in A\), \(G_i \cong S_{n-1}\) is also transitive on \(A-\{i\}\).

(b) Prove that a doubly transitive group is primitive. Deduce that \(D_8\) is not doubly transitive in its action on the 4 vertices of a square.

Let \(a \in A\), let \(B\) be any proper subset of \(A\) containing \(a\), and let \(x \in A-B\), \(y \in B-\{a\}\). By double-transitivity, there is some \(\sigma \in G_a\) such that \(\sigma(x) = y\). Since both \(a = \sigma^{-1}(a)\) and \(x\) are in \(\sigma^{-1}(B)\) but \(x \not\in B\) while \(a \in B\), it follows that \(\sigma^{-1}(B)\) is neither equal nor disjoint from \(B\). So \(B\) cannot be a block. Since every block must contain some \(a \in A\), we can conclude that the only blocks are the singletons and \(A\).

Transitive permutation groups and blocks (D&F 4.1.7)

Let \(G\) be a transitive permutation group on the finite set \(A\). A block is a nonempty subset \(B\) of \(A\) such that for all \(\sigma \in G\) either \(\sigma(B) = B\) or \(\sigma(B) \cap B = \varnothing\).

(a) Prove that if \(B\) is a block containing \(a \in A\), then \(G_B\) is a subgroup of \(G\) containing \(G_a\).

Trivial.

(b) Show that if \(B\) is a block, then the set of distinct images of \(B\) under the elements of \(G\) \(\{\sigma_1(B), \sigma_2(B), \ldots, \sigma_n(B)\}\) is a partition of \(A\).

First we show that \(A = \bigcup_{\sigma \in G} \sigma(B)\). Fix a point \(b \in B\) and let \(a \in A\) be arbitrary. Since \(G\) is transitive, there is a permutation \(\sigma\) such that \(a = \sigma(b) \in \sigma(B)\). Thus \(A \subseteq\bigcup_{\sigma \in G} \sigma(B)\); the other direction is obvious.

Next we show that for distinct permutations \(\sigma\) and \(\tau\), either \(\sigma(B) = \tau(B)\) or \(\sigma(B) \cap \tau(B) = \varnothing\). Suppose that neither is true. Then there is some \(a \in A\) which is in both \(\sigma(B)\) and \(\tau(B)\), and also some \(b \in A\) which is in \(\sigma(B)\) but not \(\tau(B)\). So \(\sigma^{-1}(a), \sigma^{-1}(b), \tau^{-1}(a) \in B\) but \(\tau^{-1}(b) \not\in B\).

Now consider the permutation \(\tau^{-1}\sigma \in G\), which maps \(\sigma^{-1}(a) \in B\) to \(\tau^{-1}(a) \in B\) and \(\sigma^{-1}(b) \in B\) to \(\tau^{-1}(b) \not\in B\). This means that \(B\) cannot be a block, because \(\tau^{-1}\sigma(B)\) is neither equal to \(B\) nor disjoint from \(B\). Hence \(\sigma(B)\) and \(\tau(B)\) are either equal or disjoint, and the conclusion follows.

(c) A transitive group \(G\) on a set \(A\) is primitive if the only blocks in \(A\) are the singletons and \(A\) itself. Show that \(S_4\) is primitive on \(A = \{1,2,3,4\}\) and that \(D_8\) is not primitive as a permutation group on the four vertices of a square.

\(S_4\): Let \(B\) be a block and \(n = \left|B\right|\). Clearly the distinct images of \(B\) under the elements of \(G\) are precisely the \(n\)-element subsets of \(A\). These only form a partition when \(n = 1\) or \(\left|A\right|\).

\(D_8\): Consider the subset \(B = \{1,3\}\), where the standard labelling of vertices applies. It can be verified that all permutations in \(D_8\) send \(B\) to either itself or \(\{2,4\}\), a disjoint set. Thus \(B\) is a block and so \(D_8\) is not primitive.

(d) Prove that the transitive group \(G\) is primitive on \(A\) if and only if for each \(a \in A\), \(G_a\) is a maximal subgroup of \(G\).

\(\Leftarrow\): Let \(B\) be a block containing \(a\). Then either \(G_B = G_a\) or \(G_B = G\). If the former case is true, then for all \(\sigma \in G\), \(\sigma(a) \in B\) if and only if \(\sigma(a) \in \{a\}\). Since \(G\) is transitive, for all \(b \in B\) there is some \(\sigma_b\) mapping \(a\) to \(b\), so \(B = \{a\}\). If the latter case is true, then for all \(\sigma \in G\) and \(b \in B\), \(\sigma(b) \in B\). By transitivity of \(G\), we also have that for all \(a \in A\), there is some \(\sigma_a\) such that \(a = \sigma_a(b) \in B\). So \(B = A\).

\(\Rightarrow\): We show that for any subgroup \(H\) of \(G\) containing \(G_a\), the set \(Ha = \{h \cdot a \mid h \in H\}\) is a block containing \(a\), and \(G_{Ha} = H\). For all \(h' \in H\) and \(h \cdot a \in Ha\), \(h' \cdot (h \cdot a) = (h'h) \cdot a \in Ha\). Furthermore, \(h'\) maps injectively (and hence surjectively) from \(Ha\) to itself, i.e. \(\sigma_{h'}(Ha) = Ha\).  On the other hand, for all \(g \in G-H\), \(g \cdot (h \cdot a) = (gh) \cdot a \not\in Ha\), otherwise \(g\) would be in \(H\). In other words \(\sigma_{g}(Ha) \cap Ha = \varnothing\). Furthermore, \(x \in G_{Ha}\) iff \(x \cdot (h \cdot a) = (xh) \cdot a \in Ha\) for all \(h \cdot a\in Ha\), iff \(x \in H\).

Based on the above result and the fact that \(G\) is primitive, it is evident that each subgroup \(H\) containing \(G_a\) is either \(G_a\) or \(G_A = G\).