Let \(G\) be a finite group and \(p\) a prime number. Let \(X_p\) be the set of Sylow-\(p\) subgroups of \(G\) and \(n_p\) be the cardinality of \(X_p\). Let \(\mathrm{Sym}(X)\) be the permutation group on the set \(X_p\).
Construct a homomorphism \(\rho: G \to \mathrm{Sym}(X_p)\) with image a transitive subgroup (i.e. with a single orbit).
Deduce that if \(G\) is simple then the order of \(G\) divides \(n_p!\).
Show that for any \(1\leq a \leq 4\) and any prime power \(p^k\), no group of order \(ap^k\) is simple.
Solution
Define the required group action by \[\begin{aligned} \rho: G \to \mathrm{Sym}(X_p) \\ g \mapsto (\gamma_g: P \mapsto gPg^{-1}). \end{aligned}\] The claim is that this action is transitive on \(X_p\). This can be equivalently stated as \[\forall P \in X_p, \exists g \in G, P' \in X_p \mid gP'g^{-1} = P.\]
However, by Sylow 2, all Sylow \(p-\)subgroups are conjugate to each other, and thus this condition is satisfied.
Suppose that \(G\) is simple, so that we have \[H \trianglelefteq G \implies H = \{ e \} \text{ or } H = G.\]
Note that \(\mathrm{Sym}(X_p) = (n_p)!\), and if we have an injective homomorphism \(G \xrightarrow{\phi} \mathrm{Sym}(X_p)\), then \(|G| = |\phi(G)|\), since \(\phi(G) \leq \mathrm{Sym}(X_p)\) will be a subgroup and thus have order dividing \((n_p)!\), which proves the statement.
Using the \(\phi\) defined in (1), we can apply the first isomorphism theorem \[G / \ker \phi \cong \mathrm{im} \phi \leq \mathrm{Sym}(X_p),\] and so it suffices to show that \(\ker \phi = \{e\}\).
Note that since \(\ker \phi \trianglelefteq G\) and \(G\) is simple, we can only have \(\ker \phi = \{e\}\) or \(\ker \phi = G\).
Towards a contradiction, suppose \(\ker \phi = G\).
By definition, we have \[\begin{aligned} \ker \phi &= \{ g\in G \mid \gamma_g = \mathrm{id}_{X_p} \} \\ &= \{ g\in G \mid \forall P \in X_p,~ gPg^{-1} = P \} \\ &= \bigcap_{P \in X_p} N_G(P), \end{aligned}\]
and so the kernel of \(\varphi\) is the intersection of all of the normalizers of the Sylow \(p-\)subgroups.
But this means that \(N_G(P) = G\) for every Sylow \(p-\)subgroup, which means that \(n_p = 1\) and there is a unique \(P\) which must be normal in \(G\). Since \(G\) is simple, this forces \(P\) to be trivial or the whole group.
Towards a contradiction, suppose \(P = G\). Then \(G\) is a \(p-\)group and thus has order \(p^n\). But then \(G\) has normal subgroups of order \(p^k\) for all \(0 < k < n\), contradicting the simplicity of \(G\).
But the only other option is that \(P\) is trivial, whereas we know nontrivial Sylow \(p-\)subgroups exist by Sylow 1.
Thus we can not have \(\ker \phi = G\), and so \(\ker \phi\) is trivial as desired.
Suppose \(\left| G \right| = ap^k\), where \(1 \leq a \leq 4\). Then by Sylow 3, we have \(n_p = 1 \mod p\) and \(n_p\) divides \(a\). If \(a=1\), then \(n_p = 1\), and so \(G\) can not be simple. Moreover, if \(p \geq a\), then \(n_p \leq a\) and \(n_p = 1\mod p\) forces \(n_p = 1\) again.
So we can restrict our attention to \(2 \leq a \leq 4\) and \(p = 2, 3\), which reduces to checking the cases \(ap^k = 2 (3^k), 4 (3^k)\), or \(3 (2^k)\) for \(k\geq 1\).
If \(ap^k = 2(3^k)\), we have \(n_3 = 1 \mod 3\) and \(n_3 \divides 2\), which forces \(n_3 = 1\), so this can not be a simple group.
Similarly, if \(ap^k = 4(3^k)\), then \(n_3 = 1 \mod 3\) and \(n_3\) divides 4, which forces \(n_3 = 1\) and thus \(G\) can’t be simple.
If \(ap^k = 3(2^k)\), then \(n_2 = 1 \mod 2\) and \(n_2\) divides 3, so \(n_2 = 1, 3\). But then \(n_3! = 6\), and if \(k > 1\), we have \(3(2^k) > 6 = n_3!\), so \(G\) can not be simple by the result in (2).
If \(k = 1\), then \(G\) is order 6, so \(G\) is isomorphic to either \(\ZZ_6\) or \(S_3\). The group \(S_3\) is not simple, since \(A_3 \trianglelefteq S_3\), and the only simple cyclic groups are of prime order, so \(\ZZ_6\) is not simple. This exhausts all of the possible cases.