let and let be the cosets of where if then we can add to the set of the generators of and hence we may

assume that for all now for any and we have and so for some and

call this result now let the claim is that obviously we only need to prove that :

let then where are not necessarily distinct. then by for some again applying we get:

for some continuing this way we'll eventually get for some and some but then and hence so i.e.