Let 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 $x H$ for some specific $x$ . There are also two right cosets: $H$ and $H y$ .

Now, since $x \notin H$ , it must be the case that $x \in H y$ ; so without loss of generality, $x = y$ .

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

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 $x H$ ; so it is either $h$ or $x h$ , for some $h \in H$ .

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

$x h H (x h)^{- 1} = x h H h^{- 1} x^{- 1} = x H x^{- 1} = H$ .

This completes the proof.