If (ab)^2 = a^2 b^2, then ab = ba for a, b in group G.

[mcwang719]

This is my first semester dealing with proofs and it's a little difficult for me. The question at hand is as follows:

Prove that if (ab)^2 = a^2 b^2 in a group G, then ab = ba.

What I did was this:

. . .(ab)^2 = a^2 b^2 ---> a^2 b^2 = a^2 b^2

Then I multiplied on the left by a^-1 to get:

. . .(a^-1)(a)ab^2 = (a^-1)(a)b^2 ----> ab^2 = ab^2

Then I multiplied on the left again by b^-1 to get:

. . .(b^-1)(b)ba = a(b^-1)(b)b ----> ab = ba

The problem is I don't know if I did this right or if I'm even on the right track. Am I showing what the problem is asking?

Thank you!

 

[pka]

Yes, that works.

 

[stapel]

What I did was this:

. . .(ab)^2 = a^2 b^2 ---> a^2 b^2 = a^2 b^2

I'm not sure how you went from the left-hand side of the arrow to the right-hand side. I mean, the right-hand side it true, but fairly irrelevant: of course a²b² is equal to a²b²!

This might be a little more what they'd had in mind:

. . . . .Given: (ab)² = a²b²

. . . . .Expand the left-hand side:

. . . . .(ab)² = (ab)(ab) = abab

. . . . .Then the "given" actually means:

. . . . .abab = a²b² = aabb

. . . . .abab = aabb

Now do your multiplication on the left (and on the right). :wink:

Eliz.

 

[mcwang719]

Thanks a lot!