Identifying \((1,2,3,4,5,6,7,8)\) with \((1,i,-1,-i,j,k,-j,-k)\) we see that \(i\) acts as \((1\ 2\ 3\ 4)(5\ 6\ 7\ 8)\) while \(j\) acts as \((1\ 5\ 3 \ 7)(2\ 8\ 4\ 6)\). Thus \(Q_8\) is isomorphic to the subgroup \[\langle (1\ 2\ 3\ 4)(5\ 6\ 7\ 8), (1\ 5\ 3 \ 7)(2\ 8\ 4\ 6)\rangle\] in the symmetric group \(S_8\).

Recall that the conjugacy class of an element is determined by its cycle type and the order of an element is the least common multiple of the lengths of the cycles in its cycle type. Thus, we need to list all partitions of \(8\) and \(12\) with parts \(4\), \(2\), and \(1\), where \(4\) occurs at least once. For \(8\), in multiplicative partition notation we have \(4^2\), \(4^1 2^2\), \(4^1 2^1 1^2\), and \(4^1 1^4\). For \(12\), we have \(4^3\), \(4^2 2^2\), \(4^2 2^1 1^2\), \(4^2 1^4\), \(4^1 2^4\), \(4^1 2^3 1^2\), \(4^1 2^2 1^4\), \(4^1 2^1 1^6\), and \(4^1 1^8\).

Thus, for \(S_8\) we have representatives

• \((1\ 2\ 3\ 4)(5\ 6\ 7\ 8)\)
• \((1\ 2\ 3\ 4)(5\ 6)(7\ 8)\)
• \((1\ 2\ 3\ 4)(5\ 6)\)
• \((1\ 2\ 3\ 4)\)

For \(S_{12}\) we have representatives

• \((1\ 2\ 3\ 4)(5\ 6\ 7\ 8)(9\ 10\ 11\ 12)\)
• \((1\ 2\ 3\ 4)(5\ 6\ 7\ 8)(9\ 10)(11\ 12)\)
• \((1\ 2\ 3\ 4)(5\ 6\ 7\ 8)(9\ 10)\)
• \((1\ 2\ 3\ 4)(5\ 6\ 7\ 8)\)
• \((1\ 2\ 3\ 4)(5\ 6)(7\ 8)(9\ 10)(11\ 12)\)
• \((1\ 2\ 3\ 4)(5\ 6)(7\ 8)(9\ 10)\)
• \((1\ 2\ 3\ 4)(5\ 6)(7\ 8)\)
• \((1\ 2\ 3\ 4)(5\ 6)\)
• \((1\ 2\ 3\ 4)\)

The following solution assumes \(\mathcal{K}\) is finite, since I think D+F’s intended solution uses this.

By construction, \(\mathcal{K}\) is the orbit of \(x\) under the conjugation action by \(G\). The stabilizer of \(x\) in \(G\) is \(C_G(x)\) so \(|\mathcal{K}|=[G:C_G(x)]\).

Let \(\mathcal{L}_x\) be the orbit of \(x\) under the conjugation action by \(H\). The stabilizer of \(x\) in \(H\) is \(C_H(x)\) so \(|\mathcal{L}_x|=[H:C_H(x)]\). Recall that if \(y = gxg^{-1}\) then \(C_H(y)=gC_H(x)g^{-1}\). Thus \(|\mathcal{L_x}|=|\mathcal{L_y}|\) for all \(y \in \mathcal{K}\). Thus, the \(k\) conjugacy classes in \(H\) have equal size \([ H : C_H(x)]\).

Since \(H\) is normal, the second isomorphism theorem tells us that \(HC_G(x)\) is a subgroup of \(G\); moreover, we have \[HC_G(x)/H \cong C_G(x)/(H \cap C_G(x)).\] We do not need the subgroup structure; we only care about the indices. Observe that \[[HC_G(x):H][H:H \cap C_G(x)]\] and \[[HC_G(x):C_G(x)][C_G(x):H \cap C_G(x)]\] are both equal to the same number \([HC_G(x):H \cap C_G(x)]\). Thus, assuming \(\mathcal{K}\) is finite, the equality \[[HC_G(x):H]=[C_G(x):H \cap C_G(x)]\] implies \[[HC_G(x):C_G(x)]=[H:H \cap C_G(x)].\] Since \(C_H(x)=H \cap C_G(x)\), we have \([HC_G(x):C_G(x)]=[H:C_H(x)]\). Thus \[\begin{align*} &|\mathcal{K}|\\ =&[G:C_G(x)]\\ =&[G:HC_G(x)][HC_G(x):C_G(x)]\\ =&[G:HC_G(x)]|\mathcal{L}_x| \end{align*}\] and we conclude that \(k=[G:HC_G(x)]\).

For the particular case of the symmetric group, we are in the situation of the above with \(H=A_n\) and \(G=S_n\). Since \(A_n\) has index \(2\) in \(S_n\), we must have \(k\) dividing \(2\).

We can compute the number of elements of order \(p\) in \(S_p\) as follows. Let \((a_1\ a_2\ \cdots\ a_p)\) be an element in \(S_p\) of order \(p\). Without loss of generality, we may assume \(a_1=1\). There are \(p-1\) choices for \(a_2\), then \(p-2\) for \(a_3\) and so on. Thus we conclude there are \((p-1)!\) elements of order \(p\) in \(S_p\).

Let \(x \in P\) be an element of order \(p\). Now \(C_{S_p}(x)\) is the stabilizer at \(x\) of the action of \(S_p\) on itself by conjugation. The orbit of \(x\) contains all elements of order \(p\). Thus, by orbit-stabilizer, we have \([S_p : C_{S_p}(x)]=(p-1)!\). Thus \(|C_{S_p}(x)|=p\). Note \(|C_{S_p}(P)|=|C_{S_p}(x)|\) since \(x\) is a generator.

A subgroup of order \(p\) has exactly \(p-1\) elements of order \(p\) and no non-trivial subgroups. Thus, two subgroups of order \(p\) cannot have non-trivial intersection. Thus there are \((p-1)!/(p-1)=(p-2)!\) subgroups of order \(p\) in \(S_p\). Now \(N_{S_p}(P)\) is the stabilizer at \(P\) of the action of \(S_p\) by conjugation on the set of subgroups of \(S_p\). The orbit is the set of subgroups of order \(p\). Thus, by orbit-stabilizer, we have \([S_p : N_{S_p}(P)]=(p-2)!\). Since \(|S_p|=p!\), we conclude that \(|N_{S_p}(P)|=p(p-1)\).

Recall that the cyclic group \(\mathbb{Z}/p\mathbb{Z}\) of order \(p\) has automorphism group \((\mathbb{Z}/p\mathbb{Z})^\times\) of order \(\varphi(p)=p-1\). Since \(N_{S_p}(P)\) acts on \(P\) by group automorphisms, we have a homomorphism \(N_{S_p}(P) \to \operatorname{Aut}(P)\). The kernel is precisely \(C_{S_p}(P)\). Thus the first isomorphism theorem give us an injective map \(N_{S_p}(P)/C_{S_p}(P) \to \operatorname{Aut}(P)\) which must be an isomorphism since they are sets of the same cardinality.

Conjugation induces a homomorphism \(\alpha: G \to \operatorname{Aut}(H)\). The image of \(\alpha\) must have order dividing \(|\operatorname{Aut}(H)|=\varphi(17)=16\). Note that \(3825=3^2\times 5^2\times 17\) so the image is trivial. Thus \(H \le Z(G)\).

By the Sylow theorems, the number of Sylow \(2\)-subgroups must divide \(k\). Consider the elements \(s, sr, sr^2, \ldots, sr^{k-1}\). We claim that any subgroup that contains any two of these elements is not a \(2\)-subgroup. Indeed, for \(0 \le i < j < k\) we have \(sr^isr^j=r^{j-i}\), where \(0 < i-j < k\). But \(r^b\) has \(2\)-power order if and only if \(b\) is a multiple of \(k\). Thus \(\langle r^i, r^j \rangle\) is not a \(2\)-subgroup. Since every \(2\)-subgroup is contained in a Sylow \(2\)-subgroup, there are at least \(k\) disinct such groups each containing one of \(s, sr, sr^2, \ldots, sr^{k-1}\).

Remark: The actual Sylow \(2\)-subgroups themselves are all of the form \(\langle r^{k}, sr^i \rangle\) for \(0 \le i < k\).

We factor \(312=2^3 \times 3 \times 13\). The number \(n_{13}\) of Sylow \(13\)-subgroups must satisfy \(n_{13} \equiv 1 \pmod{13}\) and \(n_{13} | 2^3 \times 3\). The only possibility is \(n_{13}=1\). Thus the Sylow \(13\)-subgroup is unique and therefore normal.

We factor \(2704=2^4 \times 13^2\). Using the elementary divisor description of abelian groups, we can consider the Sylow \(2\)-subgroup and Sylow \(13\)-subgroups separately. The possible partitions of \(4\) are \(4\), \(3+1\), \(2+2\), \(2+1+1\) and \(1+1+1+1\). The possible partitions of \(2\) are \(2\) and \(1+1\). Thus there are \(5\) possibilities for isomorphism classes of Sylow \(2\)-subgroup and \(2\) for Sylow \(13\)-subgroups. We conclude that there are \(10\) isomorphism classes in total.