Really Easy Group Calculations which i cant do

[SNOOTCHIEBOOCHEE]

Homework Statement

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

Prove that ab=ba

The Attempt at a Solution

I know there is some magical thing you can multiply by that will make things nice... i just can't figure this out.

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

 

[Mark44]

Homework Statement

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

Prove that ab=ba

The Attempt at a Solution

I know there is some magical thing you can multiply by that will make things nice... i just can't figure this out.

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

Some questions to help you get going here.
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.

 

[SNOOTCHIEBOOCHEE]

a has order 5 means that a⁵=1

when i multiply by a² i get b=a²ba³

 

[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?

 

[SNOOTCHIEBOOCHEE]

its not fair dick is too good at math.Thanks.. feel free to lock

 

[HallsofIvy]

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

 

[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...