Given a group $G$ and subgroup $H<G$ and $g \in G$ such that $gHg^{-1} \subset H$ prove

Question

My problem is the following -

Given a group $G$ and subgroup $H<G$ and $g \in G$ such that $gHg^{-1}\subset H$ prove that if H is a finite group $|H|< \infty$ then $gHg^{-1} = H$

what I tried -

I'm trying to show that given some $h\in H$ then $h\in gHg^{-1}$.

I assume for contradiction that $h$ is not in that group.

However, I know $ghg^{-1}\in H$.

Then I think I need to use $h^2$ or $(ghg^{-1})^2$,

but I'm not getting anywhere so I'm not sure..

Any help would be appreciated.

Answer 1

The map $h \to ghg^{-1}$ is one to one since $ghg^{-1} = gh'g^{-1} \implies h = h'$. Then $|H| \le |gHg^{-1}|$. Since $gHg^{-1} \subset H$, we have the result.

Answer 2

Consider the $\phi: H \to gHg^{-1} $ such that $\phi (h)= ghg^{-1} $. The map is one to one. Moreover the order of $ H $ is the order of $gHg^{-1} $. we are done.

Answer 3

$H$ is a normal subgroup of $ \{ g \in G | gHg^{-1} \subset H \} $ so $gHg^{-1} = gg^{-1}H=eH = H$