Since H is a normal group of G and \(m=(G:H)\) then we have that the order of \(G/H\) is m. Therefore for every a in G we gave that \((aH)^m =H\), since the order of every element divides the order of the group,which implies \(a^m \in H\)

Result

Follows from the fact the assumptions imply \(|G/H|=m\)