Let G be a group and $ H\leq G $ show $H^{-1}=\{h^{-1} \in G: h\in H\}= H$

Question

Let G be a group and $ H\leq G $ show $H^{-1}=\{h^{-1} \in G: h\in H\}= H$

I want to prove this but I need some hint to start

Answer 1

Often when you have to show that two subsets are equal you would go by proving that each of the sides contains the other side. So lets show that $H^{-1}=H$:

"$\subseteq$": Let $k\in H^{-1}$. By definition there is $h\in H$ such that $k=h^{-1}$. Since $H$ is a subgroup and $h\in H$ then $h^{-1}\in H$, thus $k\in H$.

"$\supseteq$" Let $h\in H$ and define $k:=h^{-1}$. Obviously $k\in H$ and $k^{-1}=h$, therefore $h\in H^{-1}$ because we've expressed it as an inverse of some other element from $H$.