# Conjugating Subgroups…

There is a corollary in my textbook that states,

The number of conjugates of $H \subset G$is $[G: N_G(H)]$, the index of the normalizer of H in G.

We know that $N_G(H)=\{x \in G| xHx^{-1} = H \}$

I'm a bit confused about what this is saying. So $[G: N_G(H)]$ is the number of disctinct elements of the form $gN_G(H)$ where $g \in G$, right? But I have some trouble understanding how that is related to the number of conjugates in H. So I was wondering if anybody could clarify it for me.

Also, there is another corrolary that states,

Let H be a subgroup of G. THen $H \subset N_G(H)$. In particular, the number of conjugates of H in G divies [G:H].

From the previous corrolary, we know that the number of conjugates of H is $[G:N_G(H)]$. So how does that divide [G:H]. Again, I'm having some trouble understanding this corrolary too...

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For the second question, note that $[G:H] = [G:N_{G}(H)][N_{G}(H):H].$ – Geoff Robinson Feb 13 '13 at 22:40

Let $\{x_i\mid i\in I\}$ be a set of coset representatives for $N_G(H)$ in $G$. For any $y\in G$, there are unique $i\in I$ and $x\in N_G(H)$ such that $y=g_ix$. Then $yHy^{-1} = (g_ix)H(g_ix)^{-1} = g_ixHx^{-1}g_i^{-1} = g_iHg_i^{-1}$. So every conjugate of $H$ is equal to some $g_iHg_i^{-1}$, and these are all distinct (left as easy exercise).
For the second part, use Geoff Robinson's comment (in general, if $H\leq K\leq G$, then $|G:H| = |G:K||K:H|$, which is an easy consequence of Lagrange's Theorem).
Let $G$ act on the set of conjugates of $H$ by, well, conjugation, that is $g \in G$ sends $H^{t}$ to $H^{tg}$. There is only one orbit, which consists of the set of all conjugates of $H$. And you can check that the stabilizer is indeed the normaliser of $H$ in $G$.