proving a subgroup is a subset of its inverse

Question

My proof states, show that $H\subseteq H^{-1}$
Note: $H^{-1}$={$h^{-1}: h\in H$}
Here is my proof:
Let $h\in H$
We will show that $h\in H^{-1}$
Hence, $h^{-1}\in H$, because $h^{-1}\in H^{-1}$ of the subgroup of H
But $h^{-1}\in H$
Hence, $h\in H^{-1}$
therefore, $H\subseteq H^{-1}$
Is the right?

Answer 1

Your wording isn't straightforward. Suppose $h \in H$. Because $H$ is a subgroup, it contains the inverse of any one of its elements. In particular, $k := h^{-1} \in H$. It then follows that $h = k^{-1} \in H^{-1}$. By the arbitrary nature of $h \in H$ we've proven the inclusion $H \subset H^{-1}$.