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 $H$?

Hint: Conside the map $G \to G$ given by $x \mapsto x^{-1}$. Prove that this map is an involution, that is, is its own inverse. Conclude that the map is a bijection and so is a surjection.
