Simple groups in group theory 2

Question

Problem Let $G$ be a finite group and $H \subset G$ be a subgroup of index $\lvert G:H \rvert =n$.

(a) Show that $\lvert H:(H\cap gHg^{-1})\rvert \leq n$ for all $g\in G$

(b) If $H$ is a maximal subgroup of $G$ and $H$ is abelian, show that $(H\cap gHg^{-1})$ is a normal subgroup of $G$ for all $g\notin H$

(c) Suppose that $G$ is simple. If $H$ is abelian and $n$ is a prime, prove that $H=1$.

I proved part (a), (b) in the problem. I'm stuck with (c).

I know $H$ is a maximal subgroup by using $n$ is a prime and $H\cap gHg^{-1}=1$ since $G$ is simple and (b) in the problem.

Any help is appreciated.

Answer 1

We can assume that $H$ is not normal in $G$, so $H \cap g^{-1}Hg = 1$ for all $g \not\in H$. Then $|G| \le n^2$ by (a).

Groups of order $n^2$ with $n$ prime are abelian, so $|G| = nk$ with $|H|=k < n$.

Now by a standard counting argument, there are $(k-1)n$ non-identity elements in conjugates of $H$, which leaves just $n$ remaining elements, which must form a unique Sylow $n$-subgroup of $G$, which is therefore normal, contradicting simplicity.

Answer 2

Let $C$ be the set of $gGg^{-1}$, $G$ acts on $C$ transitively by conjugation. Let $N$ be the stabilizer of $H$, if $N$ is distinct of $H$ there exists an element $g$ of $N$ not in $H$ such that $gHg^{-1}=H$. 2 implies that $H$ is normal and trivial.

If $N=H, |G|=|H||C|=|H|p$ since the index of $H$ is $p$. Since 2 implies that the intersection of two elements of $H$ is the identity since $G$ is simple, we deduce that $|G|=p|H|-p+1$ since $|H|=|gHg^{-1}|$ and $p=1$ contradiction.