Prove that a subgroup of index two is normal.
Solution
Let \(H\le G\) with \([G:H]=2\).
1. For \(g\notin H\), \(gH=G\setminus H=Hg\).
The left cosets partition \(G\) into two sets, one of which is \(H\); since \(g\notin H\), \(gH\neq H\), so \(gH=G\setminus H\). The same argument with right cosets gives \(Hg=G\setminus H\).
2. \(gH=Hg\) for every \(g\in G\).
For \(g\in H\), \(gH=H=Hg\); for \(g\notin H\), use step 1.
3. Q.E.D.
Multiplying \(gH=Hg\) on the right by \(g^{-1}\) gives \(gHg^{-1}=H\) for every \(g\in G\), so \(H\trianglelefteq G\).