Let $H$ be a subgroup of the group $G$, of 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 $x\notin H$, it must be the case that $x\in Hy$; so without loss of generality, $x=y$.

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

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 $h\in H$.

• $hH{h}^{-1}$ is equal to $H$ since $hH=H$ and $H{h}^{-1}=H$.

• $xhH(xh{)}^{-1}=xhH{h}^{-1}{x}^{-1}=xH{x}^{-1}=H$.

This completes the proof.