Prove that $H$ is normal.

Question

Let $G$ be a finite group and let $H$ be a subgroup of $G$ with the property that no other subgroup of $G$ has order $|H|$. Prove that $H$ is normal.

My attempt:

To show that $H$ is normal, I need to show that for any $g,h \in G$, $ghg^{-1}\in H$.

Let the order of $G=n$ and let the order of $H=k$. If $H$ is a subgroup of $G$ then $k$ divides $n$ and the identity element $e$ is both $G$ and $H$. Also for any $a,b \in H$, $ab^{-1}\in H$.

Since $k$ divides $n$, $n=qk$ for some integer $q$.

$a^{n}=a^{qk}=e$...

(I don't think this proof is getting me anywhere. I think I need a hint about what does "$H$ being a subgroup of $G$ such that no other subgroup of $G$ has order $|H|$" imply...)

Answer 1

Hint: for any $\;x\in G\;$ , we have that $\;H^x:=x^{-1}Hx\;$ is a subgroup of $\;G\;$ of order $\;|H|\;$ .

Answer 2

HINT: For $g\in G$ let $H^g=\{ghg^{-1}:h\in H\}$, and show that $H^g$ is a subgroup of $G$.

Answer 3

Note that $gHg^{-1}$ must be a subgroup of $G$. It also has size $|H|$. But there is only one of these groups, namely $H$ itself. Thus, $gHg^{-1}=H$.

Answer 4

You can say more: $H$ is characteristic. In fact, an automorphism of $G$ has to send $H$ in itself, because it is the only subgroup with order $|H|$; so it is also normal (in particular, it is fixed by an inner automorphism).