# Homework Help: Show that a Group (G, *) definied by a condition is Abelian

1. Jan 3, 2009

### JPC

1. The problem statement, all variables and given/known data

(G, *) is a group (where * is a law)
And for all 'i' belonging to {2, 3, 4}, for all (x, y) belonging to G2
(x * y) ^ i = (x^i) * (y^i)
(where ^ is the law : to the power of)

Question : Show that G is an Abelian (commutative) group

2. Relevant equations

3. The attempt at a solution

we have never done any questions of that sort yet, all i can say is that
"(x * y) ^ i = (x^i) * (y^i)" shows that the law '^' (to the power of) is distributive over the law '*' for 'i' belonging to {2, 3, 4}
But then i don't know where to go

Any help or directions would be appreciated, thank you :)

2. Jan 3, 2009

### tiny-tim

Hi JPC!

My inclination would be to start by saying …

if it's not Abelian, then there are a and b with ab - ba ≠ 0,

so let ab - ba = c, and …

3. Jan 4, 2009

### JPC

oh, i should have precised, here "*" is not necessarily the multiply law, it can be any law that respects the given conditions

i tried with your method, putting to the square after, but then i end up working with 4 laws at the same time :D

Thank you for the indication anyways, i have to find the little trick inside that exercise now

4. Jan 5, 2009

### Sisplat

Hello Jpc,

In the case that i = 2,

(x*y)*(x*y) = x*x*y*y

by associativity:

x*(y*x)*y = x*(x*y)*y

and then just left and right multiply by x inverse and y inverse respectively.

This proves that G is Abelian, since we have shown * to be commutative.

If you want to show this for the more complicated cases where i is only allowed to take the value 3 or only allowed to take the value 4, then you would need to rephrase your question.

5. Jan 7, 2009

### JPC

yes thank you Sisplat, that works very well :)
I was looking for very complicated things, and i did a little confusion, thinking that the "^i" was associated to the multiply law and not the * law

And yes, you were right, there were too many data given, to confuse you probably

Thanks again