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Map is given, for any element $x \mapsto x^p$. I proved that the image of this map is contained in center of $G$. It is of course true when $G$ is commutative. So assume that is is not that case, $G/C(G)$ must be isomorphic to $Z/pZ\times Z/pZ$. Let the generators $aC(G)$, $bC(G)$. $C(G)=\langle c\rangle$. By using the definition of $C(G)$, for $x \in G$, write an combinations of $a,b,c$ and it is easy to show that $x^pC(G)=C(G)$ this is what i found. but still, It does not give an answer to my first question i.e. i want to show that $x^py^p=(xy)^p$

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    When p=2 things go wrong. In general when G is finite, $x \mapsto x^n$ is a homomorphism from G to its center for $n=[G:Z(G)]$. This is $n=p^2$ though, and so is a pretty trivial map in this case.2012-09-17

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