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

**rainyice** · March 23rd 2010, 12:32 PM

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)

**Defunkt** · March 23rd 2010, 12:54 PM

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

**rainyice** · March 23rd 2010, 07:03 PM

Originally Posted by Defunkt

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

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

**Haven** · March 23rd 2010, 10:43 PM

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

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