Let \(G\) be a finite group and \(H<G\) a subgroup. Let \(n_H\) be the number of subgroups of \(G\) that are conjugate to \(H\). Show that \(n_H\) divides the order of \(G\).
Solution
.* Let \[C_H = \{ gHg^{-1} \mid g\in G \}\] be the conjugacy class of \(H\), so \(|C_H| = n_H\).
We wish to show that \(n_{H}\) divides \(|G|\).
Claim 1: \[n_{H} = [G: N_G(H)],\] where \(N_G(H) \leq G\) is the normalizer of \(H\) in \(G\).
Note that if this claim is true, then we can apply Lagrange’s theorem, which states \[A \leq G \implies |G| = [A: G]~|A|,\]
which in this case translates to \[|G| = [N_G(H) : G]~|N_G(H)| = n_H~|N_G(H)|.\]
Since \(n_H\) divides the right-hand side, it must divide the left-hand side as well, which is precisely what we would like to show.
Proof of Claim 1:
The normalizer of \(H\) in \(G\), written \(N_G(H)\), is the largest subgroup of \(G\) containing \(H\) such that \(H \trianglelefteq N_G(H)\), i.e. \[N_G(H) = \{g \in G ~\mid~ gHg^{-1} = H \} \leq G.\]
Now consider \(S\), the set of left cosets of \(N_G(H)\). Suppose there are \(k\) of them, so \[[G: N_G(H)] = |S| \coloneqq k.\]
Then \(S\) can be written as \[S = \{ g_1 N_G(H), ~g_2 N_G(H), ~\cdots, ~g_k N_G(H) \}.\]
where each \(g_i\) is a distinct element of \(G\) yielding a distinct coset \(g_i N_G(H)\). In particular, if \(i\neq j\), then \(g_i \neq g_j\), and \(g_i N_G(H) \not \in g_j N_G(H)\).
In particular, \(S\) acts on \(C_H\), \[\begin{aligned} S &\curvearrowright C_H \\ g_i N_G(H) &\curvearrowright H = g_i H g_i^{-1},\end{aligned}\]
taking \(H\) to one of its conjugate subgroups.
So define \[K \coloneqq \{ g_i H g_i^{-1} \mid 1 \leq i \leq k \}.\]
Note that \(K \subseteq C_H\), and has at most \(k\) elements.
We claim that \(K\) has \(k\) distinct elements, i.e. that each \(g_{i}\) takes \(H\) to a distinct conjugate subgroup. We have \[\begin{aligned} g_{i} H g^{-1}_{i} &= g_{j} H g^{-1}_{j} &\implies \\ g_j^{-1} g_{i} H g^{-1}_{i} g_j &= H &\implies \\ (g_j^{-1} g_{i}) H (g_j^{-1} g_{i})^{-1} &= H &\implies \\ g_j^{-1} g_{i} &\in N_G(H) &\implies \\ g_i &\in g_j N_G(H) &\implies \\ g_i &= g_j,\end{aligned}\]
where the last line follows because we assumed that each coset contains at most one \(g_i\).
Thus \(K\) has \(k\) distinct elements, and so \[= k = |K| \leq |C_H| = n_H.\]
We now claim that \(k \geq n_H\) as well.
Let \(H' \in C_H\) be any subgroup conjugate to \(H\), so \(H' = gHg^{-1}\) for some \(g\in G\). Then \(g = g_i\) for some \(i\), so \(g \in g_{i} N_G(H)\).
Thus \(g = g_{i} n\) for some \(n\in N_G(H)\), but \(n\in N_G(H) \iff nHn^{-1} = H\) by definition, and so we have \[\begin{aligned} H' &= gHg^{-1} \\ &= (g_i n) H (g_i n)^{-1} \\ &= g_i (n H n^{-1}) g_i^{-1} \\ &= g_i H g_i^{-1} \in K.\end{aligned}\]
Since \(H' \in C_H\) was an arbitrary subgroup conjugate to \(H\), this says that \(C_H \subseteq K\) and thus \[n_H = |C_H| \leq |K| = k\]
Thus \[[G: N_G(H)] = k = |M| = |K| = n_H,\] which is what we wanted to show. ◻