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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. – Jack Schmidt Sep 17 '12 at 21:58

You know that if $G$ is a group and $x,y\in G$ be two arbitrary elements of it, then $$[x,y]=x^{-1}y^{-1}xy$$ is called a commutator of $G$. All such these elements in $G$, when $x,y\in G$ construct a well-known subgroup called the commutator subgroup $G'$. Now, there are two lemmas which can be proved by induction on $n$:

Lemma 1. Let $G$ is a group and let the element $[x,y^{-1}]$ can be commute with $x$ and $y$, then: $$[x,y^{-n}]=[x,y^{-1}]^n$$

and by using it we have the second:

Lemma 2. Let $G$ is a group and let the element $[x,y^{-1}]$ can be commute with $x$ and $y$, then: $$(xy)^n=x^ny^n[x,y^{-1}]^{\frac{n(n-1)}{2}}$$

What we have is:

$|G|=p^3$ leads $Z(G)=G'$ and $|Z(G)|=|G'|=p$ since $G$ is non- abelian finite group.

$Z(G)=G'$ means the element $[x,y^{-1}]$ is central and then by lemma 2 you will have: $$(xy)^p=x^py^p[x,y^{-1}]^{\frac{p(p-1)}{2}}=x^py^p\big([x,y^{-1}]^p\big)^{\frac{(p-1)} {2}}$$ $p$ is an odd prime number so $\frac{(p-1)} {2}$ is an integer and from $|G'|=p$ we have $$[x,y^{-1}]^p=1$$ This means to us that $$(xy)^p=x^py^p$$ This is what you have been searching for.

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Let $x,y \in G$. As $G/C(G)$ is commutative, $xyC(G) = yxC(G)$, so $yx = xyz$ for some $z \in C(G)$. We have by induction $y^kx = xy^kz^k$ as: \[ y^kx = y^{k-1}yx = y^{k-1}xyz = xy^kz^{k-1}z = xy^kz^k \] No we prove by induction that $(xy)^k = x^ky^kz^{\frac 12k(k-1)}$, which holds since \[ (xy)^k = (xy)^{k-1}xy = x^{k-1}y^{k-1}xyz^{\frac 12(k-1)(k-2)} = x^ky^kz^{\frac 12(k-1)(k-2) + (k-1)} = x^ky^kz^{\frac 12k(k-1)} \] So $(xy)^p = x^py^pz^{\frac 12p(p-1)}$. If $p$ is odd, then the exponent is a multiple of $p$ (which is the order of $C(G)$), hence $z^{\frac 12 p(p-1)} = 1$,and so $(xy)^p = x^py^p$.

If $p$ is even, this needn't hold true: In the quaternion group of order $8 = 2^3$, we have $(ij)^2 = k^2 = -1 \ne 1 = (-1)^2 = i^2j^2$.

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