1. ## normal subgroup proof

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?

2. Originally Posted by 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?

Hint: if $\displaystyle ord(G)=n$ for some group, then $\displaystyle g^m=1\,\,\,\forall g\in G$

Tonio

3. oooh right. by lagrange the order of the element has to divide the order of the subgroup... so $\displaystyle a^m$ H=eH whether its order is exactly m or not, ie. $\displaystyle a^m$ is in the identity of G/H, which is exactly H. thanks.