If H is a normal subgroup of G, and (G:H)=m, show that is in H for all a in G.

i know this means that G/H has order m. i started the proof as follows:

H=(aH)(aH)...(aH) (m times)

aH must have order less than or equal to m. if its equal to m, then it gives back the identity in G/H (H), which means =h for some h, and is therefore in H, and the proof is done.

BUT does the order of aH necessarily have to be m?