Let $G$ be a group and $H = \{x^{-1}\mid x\in G\}$. Show $G=H$. 
Let $G$ be a group and $H = \{x^{-1}\mid x\in G\}$. Show $G=H$.

I have showed that $H \subseteq G$. 
Can somebody give me a hint to show how an arbitrary $x \in G$ also belongs in $H$?

Based on the Axiom of Inverse Element, if $a\in G$, then $b\in G$ if $a* b=b* a=e$, where $*$ represents the groupg operation. Hence the rest of the proof goes as follows:
$$a\in G\implies a^{-1}\in G\implies (a^{-1})^{-1}\in H\implies a\in H\implies G\subseteq H$$
