$G$ is a finite, simple, nonabelian group. If $|G:H| = p$, then $|\{gHg^{-1}\}|=p$

Question

I want to show that if $G$ is a finite, simple, nonabelian group with $H \leq G$ so that $|G:H| = p$ for $p$ is prime, then $|\{gHg^{-1}\}|=p$, i.e. the number of conjugates of $H$ in $G$ is $p$.

First idea: Now, we know $\displaystyle\bigcap_{g \in G} gHg^{-1} \lhd G$. By simplicity, $\displaystyle\bigcap_{g \in G} gHg^{-1} = \{1\}$ or $G$. If it equals $G$, then $H=G$ and so $|G:H| = 1$, a contradiction. So $\displaystyle\bigcap_{g \in G} gHg^{-1} = \{1\}$. Toward a contradiction, suppose the number of conjugates is not $p$. Then ...?

Second idea: As $G$ is simple, then if $H$ is normal, $H = \{1\}$ or $G$. Since $|G:H| = p$, then $H < G$ and so $H = \{1\}$. Now, for every $g \in G$, we get $gHg^{-1} = \{g \cdot 1 \cdot g^{-1} \} = \{1\}$. Then $|\{gHg^{-1}\}| = |G|$, but it is not known that $G$ is a $p$ group.

Both of these ideas fail to include that $G$ is non-abelian, which leads me to believe there is an issue.

Any suggestions here would be great!

Answer 1

Consider the normalizer of $H$ in $G$: $N=N_G(H)= \{ g \in G \;|\; gHg^{-1}=H\}$.

We always have $H \subseteq N \subseteq G$. Thus $[G:N]$ divides $[G:H]$. Assuming $[G:H]=p$ (prime), we have two options: $[G:N]=1$ or $p$. If $[G:N]=1$, $N=G$ and thus $H$ is normal. This would imply $G$ has a normal subgroup of index $p$. If $G$ is simple, this implies $G$ is cyclic of prime order (thus abelian). Therefore, $[G:N]=[G:H]=p$ so that $N=H$.

Let $G$ act on the set of its subgroups. We know that for any subgroup $H$, its normalizer is its stabilizer. So, by the orbit-stabilizer theorem, the index of the normalizer $[G:N]=p$ is the size of $H$'s orbit (i.e., $H$ has $p$ conjugates).

Answer 2

From $|G:H|=p$ follows $|G|=p|H|$. From $G$ nonabelian follows that $H$ is nontrivial (as otherwise $|G|=p$ would imply $G$ cyclic, hence abelian). Now, $G$ acts by conjugation on the set of its subgroups. Your set is the orbit of $H$ under this action, and the pointwise stabilizer is $N_G(H)$. By the orbit-stabilizer theorem: $$|\{gHg^{-1}\}|\cdot|N_G(H)|=p|H|$$ namely: $$|\{gHg^{-1}\}|\cdot|N_G(H):H|=p$$ Since $|\{gHg^{-1}\}|\ne 1$ (as otherwise $H$ would be a proper nontrivial normal subgroup of $G$, against the simplicity of this latter), necessarily $|\{gHg^{-1}\}|=p$. (Btw, from this follows that $H$ is self-normalizing, namely $N_G(H)=H$.)