prove or disprove

**mandy123** · October 25th 2008, 07:16 AM

Let G be an Abelian group. Prove or disprove that
H={g^2|g is a member of G}
is a subgroup of G.

**PaulRS** · October 25th 2008, 07:41 AM

Indeed it's true, take $a,b\in G$ then $a^2*b^2=a*a*b*b=a*b*a*b=(a*b)*(a*b)=(a*b)^2$ (*)

Where we've used the fact that G is abelian and that the associativity holds.

So if $a'\in H$ and $b' \in H$ then by (*) we have $a'*b' \in H$

Also the identity belongs to H since $e^2=e$

And each element has an inverse, say we have $a' \in H$ then $a'=a^2$ for some $a \in G$

Now $a$ has an inverse, namely $a^{-1} \in G$

So note that $a^2*a^{-2}=a*a*a^{-1}*a^{-1}=e$ thus $a^{-2}=(a^{-1})^2\in H$ is the inverse of a'

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