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.