# Thread: Prove that G is abelian if and only if (ab)^(2) = a^(2) * b^(2)

1. ## Prove that G is abelian if and only if (ab)^(2) = a^(2) * b^(2)

Let a and b be elements of a group G. Prove that G is abelian if and only if (ab)^(2) = a^(2) * b^(2)

2. What have you tried? Where are you getting stuck? (Also, I believe the statement should be that $\displaystyle G$ is abelian $\displaystyle \Leftrightarrow (ab)^2 = a^2b^2 \ \forall a,b \ \in G$)

3. Originally Posted by Defunkt
What have you tried? Where are you getting stuck? (Also, I believe the statement should be that $\displaystyle G$ is abelian $\displaystyle \Leftrightarrow (ab)^2 = a^2b^2 \ \forall a,b \ \in G$)

i got stuck how to do it conversely.$\displaystyle \Leftrightarrow a^2b^2 = (ab)^2 \ \forall a,b \ \in G$

4. consider $\displaystyle a,b \in G$
if $\displaystyle a^2b^2 = (ab)^2$ then $\displaystyle aabb = abab$
since G is a group, $\displaystyle a^{-1}, b^{-1} \in G$
consider $\displaystyle a^{-1}(aabb)b^{-1} = a^{-1}(abab)b^{-1}$ and use the properties of groups and you'll see why G is abelian.

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# (ab)^2 = a^2b^2 abelian

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