If $ghg^{-1}\in H$ for all $g\in G$ and $h\in H$, prove that $gH=Hg$ for $g\in G$.

Question

If $ghg^{-1}\in H$ for all $g\in G$ and $h\in H$, prove that $gH=Hg$ for $g\in G$.

I am not sure where to begin with this proof. I would assume we use set equality to prove it, but I don't know where to begin with that either...

Do I say that since $ghg^{-1}\in H$ for all $g\in G$ and $h\in H$, then $gHg^{-1}=H$?

Answer 1

If $x \in gH$, then $x=gh$ for some $g \in G$ and some $h \in H$. Then we get

$xg^{-1}=ghg^{-1} \in H$, hence $x=xg^{-1}g \in Hg$.

We have shown: $gH \subseteq Hg$.

It is now your turn to show: $gH \supseteq Hg$.