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.