Prove that $H\cap gHg^{-1}$ is normal

This problem is from a Ph.D Qualifying Exam from algebra:

Let $G$ be a finite group and let $H \le G$ with index $[G:H]=n$.

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

2. Now suppose that $G$ is simple. If $H$ is abelian and $n$ is prime, prove that $H$ is trivial.

My attempt: Let $G$ act on $(G/H)^2$ by componentwise left multiplication. Then the stabilizer of the element $X=(H,g^{-1}H)\in (G/H)^2$ is precisely $H\cap gHg^{-1}$. Therefore, $|GX|=[G:(H\cap H^g)]=[G:H][H:(H\cap H^g)]\le n^2$, so we conclude that $[H:(H\cap H^g)]\le n$.

Now I'd like to exploit this property to solve 1, so I tried as follows:

Let $G$ act on $GX$ just as on $(G/H)^2$, then the action induces a homomorphism $\phi:G\to S_{GX}$. Clearly, $\ker(\phi)\le H\cap H^g$, but I have no idea how to show the reverse inclusion.

For 2, I tried as follows:

Since $H$ has prime index in $G$, $H$ is a maximal subgroup of $G$. Therefore, by 1, $H\cap H^g =1$ for all $g\notin H$. Then how should I prove that $H=1$ from this fact?

Any hints or advice will help a lot! Thanks!

• If $H \ne H^g$ then $H \cap H^g$ is central in both $H$ and in $H^g$, and hence in $G$. On the other hand if $g \not\in H$ and $H = H^g$ then $H \unlhd G$. – Derek Holt Jun 19 '18 at 7:42
• Thanks for hint! I finally figured out 1, and I'm struggling with 2 now. – bellcircle Jun 19 '18 at 7:55
• You've worked out that $|G|\le n^2$ can $|G|=n^2$? then use Sylow's theorems, with $n$ dividing $|G|$ – Robert Chamberlain Jun 19 '18 at 9:17
• @RobertChamberlain Do you mean for the question 2? – bellcircle Jun 19 '18 at 10:21
• @RobertChamberlain Oh, I also figured out question 2 thanks to your question. I appreciate it! – bellcircle Jun 19 '18 at 10:25

1. If $H=H^g$ for all $g\notin H$, then $H\cap H^g = H$ is normal in $G$.
If $H\neq H^g$ for some $g\notin H$, note that every element in $H\cap H^g$ commutes with all elements in $H$ and $H^g$, so it also commutes with all elements in $\langle H,H^g \rangle=G$ by maximality of $H$, so $H\cap H^g \le Z(G)$, so it is normal in $G$.
2. Since $H$ has a prime index in $G$, $H$ is a maximal subgroup of $G$. Therefore, by 1 and the simplicity of $G$, $H\cap H^g=1$.
Also we know that $|G|=[G:H\cap H^g]\le n^2$, and $n$ is a divisor of $|G|$. Thus we can let $|G|=nk,\:1\le k\le n$.
If $k=n$, then $|G|=n^2$ and $G$ becomes abelian and has a subgroup of order $n$, and clearly such a subgroup is normal, which is a contradiction.
If $k<n$, then there exists unique Sylow $n$-subgroup of $G$, and it must be normal. By simplicity of $G$, $k$ must be 1. Thus $|G|=n$ and $[G:H]=n$ implies that $H$ is trivial.