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

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

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.

. . .(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!

2. Yes, that works.

3. Originally Posted by mcwang719

. . .(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 a<sup>2</sup>b<sup>2</sup> is equal to a<sup>2</sup>b<sup>2</sup>!

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

. . . . .Given: (ab)<sup>2</sup> = a<sup>2</sup>b<sup>2</sup>

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

. . . . .(ab)<sup>2</sup> = (ab)(ab) = abab

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

. . . . .abab = a<sup>2</sup>b<sup>2</sup> = aabb

. . . . .abab = aabb

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

Eliz.

4. Thanks a lot!

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