# Where does the abelian part come in?

1. Nov 26, 2007

### rsa58

1. The problem statement, all variables and given/known data
Prove the set H of all solutions satisfying the equation x^n = e form a subgroup of an abelian group G with identity e.

2. Relevant equations

3. The attempt at a solution

e^n = e so e belongs to H.

x^n = e => x^(-n)= e => (x^(-1))^n= e so for any x belonging to H, x inverse belongs to H.

if a^n=e and b^n= e => (ab)^n=e so ab belongs to H.

where does the abelian part come in?

2. Nov 26, 2007

### ircdan

(ab)^n = (ab)(ab)(ab)...(ab)(ab)(ab) =(<--here) (aa...aaa)(bb...bb) = a^nb^n = ee = e

note that more is true, the set of all elements of finite order of an abelian group form a subgroup. the proof is the same as this except showing closure is slightly different, you could do it for practice if u wanted

Last edited: Nov 26, 2007
3. Nov 28, 2007

### rsa58

i'm going to do the practice problem just to make sure i've got it. if a^n=e and b^s=e then (a^n * b^s)=e. then by the group being abelian (ab)^s * a^(n-s)=e. we can assume n>s without loss of generality. therefore (ab)^s = a^s. if we take both sides to the power (n/s) then we have (ab)^n = e. thus ab belongs to G.

4. Nov 28, 2007

### rsa58

by the way i still don't understand why G must be abelian in the first problem.

5. Nov 28, 2007

### HallsofIvy

Staff Emeritus
That was answered in ircdan's first response. If G is not abelian, then (ab)n is NOT, in general, anbn so you cannot use the fact that an= bn= e.