Lemma: both equations and have the unique solution

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 theLemmawe must have

and thus therefore i.e.

Exercise: (very easy!) make the problem interesting by extending this result to a class of finite groups!