# Really Easy Group Calculations which i cant do

1. Oct 12, 2008

### SNOOTCHIEBOOCHEE

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

let a,b$$\in$$G Assume that a has order 5 and that $$a^{3}b=ba^{3}$$

Prove that ab=ba

3. The attempt at a solution

I know there is some magical thing you can multiply by that will make things nice... i just cant figure this out.

if tried left and right multiplying by a2 and a-2 with no prevail.

2. Oct 12, 2008

### Staff: Mentor

What does it mean that a has order 5?
Why did you pick a2 to multiply by?
What did you get when you multiplied by a2?

One other thing: at some point you'll want to use the fact that a3b = ba3.

3. Oct 13, 2008

### SNOOTCHIEBOOCHEE

a has order 5 means that a5=1

when i multiply by a2 i get b=a2ba3

4. Oct 13, 2008

### Dick

If you multiplied on the other side you must have gotten b=a^3ba^2 as well. So that's a^3ba^2=a^2ba^3. Now what? Can't you see an 'ab=ba' in there somewhere?

5. Oct 13, 2008

### SNOOTCHIEBOOCHEE

its not fair dick is too good at math.

Thanks.. feel free to lock

6. Oct 13, 2008

### HallsofIvy

Staff Emeritus
Dick practiced a lot. The rest of us are trying to catch up.

7. Oct 13, 2008

### boombaby

Well, I find it is often possible to prove it by a series of equation as ab=.....=....=ba, which is clear and forthright.
e.g. ab=aeb=aa^5b= a^3(a^3b)=a^3(ba^3)=(a^3b)a^3=(ba^3)a^3=baa^5=bae=ba
But it may still need magics to prove more challenging problem....