# Index two subgroup of group is normal

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 $$x \not \in H$$, 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.