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**platinumpimp68plus1** If H is a normal subgroup of G, and (G:H)=m, show that $\displaystyle a^m$ is in H for all a in G.

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

$\displaystyle a^m$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 $\displaystyle a^m$=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?