Let $G$ be a group with $H < G.$ I'd like to show that if $H$ has finite index in $G$, then $H$ has finitely many conjugates in $G$.
One of the standard proofs I've seen begins by recognizing that $[G:H] < \infty$ means $G/H=\{g_1H,...,g_nH\}$ for some n. So any $g \in G$ has a coset $gH$ that must satisfy $gH=g_iH$ for some $1 \leq i \leq n$. We can then use this fact to deduce that any conjugate of $H$ in $G$ must satisfy $gHg^{-1}=g_iHg_i^{-1}$. Hence there are precisely n unique conjugates.
This is all great, but my question is:
Can I also proceed in the following way?
Namely . . .
Since $H < N_G(H) < G$, we have the tower
$$[G:H]=[G:N_G(H)][N_G(H):H],$$
and we know that the number of conjugates of $H$ in $G$ is given by $[G:N_G(H)]$. Now the tower says
$$[G:N_G(H)]=\frac{[G:H]}{[N_G(H):H]}.$$
But $[G:H]$ is finite, and $[N_G(H):H] \geq 1$ can never be zero. So $[G:N_G(H)]$ must be finite, and hence there are finitely many conjugates of $H$ in $G$. $\Box$
It seems alright to me unless maybe these towers only work for finite groups.