# Quick questions for test tomorrow

1. Dec 2, 2007

### rsa58

1. The problem statement, all variables and given/known data
1- Sometimes my teacher writes the proof for when a set is a subgroup by saying the following: since e(identity) belongs to G then G is not empty. He then puts if a(b^(-1)) belongs to G then the rest of the conditions are satisfied. Why is this so? is it because if a and b belong to G then we can take a=e and we have the inverse. and then a(b^(-1))^(-1) belongs to G so ab belongs to G?

2- H is normal in G. m=(G) show a^m belongs to H for all a belonging to G. Proof: any aH belonging to G/H has the property that (aH)^m = e. And then the rest i get. my question is why is (aH)^m = e. Should G/H be cyclic for this to be the case? I know if we have a finite group then we can say a^m =e for some m in Z+. but yeah.

2. Relevant equations

3. The attempt at a solution

Last edited: Dec 2, 2007
2. Dec 2, 2007

### Dick

You seem to get the first one. For the second one, G/H is a group of order m. So the power m of any element of that group is the identity. aH is an element of that group, so (aH)^m=eH. (aH)^m=(a^m)H=eH. Now show a^m is an element of H and don't do it by 'cancelling' the H. For which g in G is it true that gH=H?

3. Dec 3, 2007

### rsa58

yeah, i know, but you didn't answer the question. obviously for any g belonging to H we have gH=H. but i don't get why any element in G/H taken to power (order G/H) is the identity element. can you prove it for me? i thought this was only for cyclic groups.

4. Dec 3, 2007

### Dick

The order of any element g divides the order of G. Since the group generated by g is a subgroup of G. So |G|=k*m, where m is the order of g. Since g^m=e, g^(|G|)=e. So an element of a group taken to the power of the order of that group is e. For all groups, not just cyclic.