Let $~$H$~$ be a Subgroup of the Group $~$G$~$, of [index_of_subgroup index] $~$2$~$. Then $~$H$~$ is a Normal subgroup of $~$G$~$.

# Proof

We must show that $~$H$~$ is closed under conjugation by elements of $~$G$~$.

Since $~$H$~$ has index $~$2$~$ in $~$G$~$, there are two left cosets: $~$H$~$ and $~$xH$~$ for some specific $~$x$~$. There are also two right cosets: $~$H$~$ and $~$Hy$~$.

Now, since $~$x \not \in H$~$, it must be the case that $~$x \in Hy$~$; so without loss of generality, $~$x = y$~$.

Hence $~$xH = Hx$~$ and so $~$xHx^{-1} = H$~$.

It remains to show that $~$H$~$ is closed under conjugation by *every* element of $~$G$~$.
But every element of $~$G$~$ is either in $~$H$~$, or in $~$xH$~$; so it is either $~$h$~$ or $~$xh$~$, for some $~$h \in H$~$.

- $~$hHh^{-1}$~$ is equal to $~$H$~$ since $~$hH = H$~$ and $~$Hh^{-1} = H$~$.
- $~$xh H (xh)^{-1} = xhHh^{-1} x^{-1} = xHx^{-1} = H$~$.

This completes the proof.