Proof: if then hence but thus: a similar argument works for
now suppose and let be the normal subgroup of G with in order to show that , we only need to prove that .
let we want to show that commutes with let since H is normal, there exists an integer such that thus: which gives us
we have for some since we may assume that now by the Lemma we must have
and thus therefore i.e.
Exercise: (very easy!) make the problem interesting by extending this result to a class of finite groups!