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Let $p$ be an odd prime and $G$ a group of order $p^3$. Prove that the $p$-th power map $x \mapsto x^p$ is a homomorphism $G \rightarrow G$.

The abelian case is easy. Suppose $G$ is non-abelian group. If $x \in G$ or $y \in G$ has order $p$, then $\phi(xy)=\phi(x)\phi(y)$. Thus, we can assume that $x,y$ are order of $p^2$.

That's what I did. I don't have any tools. I need your help.

egreg
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Guillermo
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    How does the proof go if $x$ or $y$ has order $p$? – Greg Martin Oct 09 '14 at 07:06
  • the center $Z=Z(G)$ is a nontrivial normal subgroup (we assume it is $\neq G$). so let $x,y \notin Z$ and let $$ 1 \neq c := xyx^{-1}y^{-1} \Rightarrow xy=cyx $$ Show that $c\in Z$ and multiply by $x$ and iterate to get formulas for $x^ny$ and $x^ny^n$ – Blah Oct 09 '14 at 07:26
  • https://math.stackexchange.com/q/1132336/688539 – Clemens Bartholdy Mar 10 '22 at 00:51

1 Answers1

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Hint: If $G$ is not abelian, $|Z(G)|=p$. Then, notice that $xy=yxh$ for some $h \in Z(G)$. Therefore, you can write $(xy)^p=x^py^p h^q$ for some integer $q$; it turns out that $q$ is divisible by $p$, hence $h^q=1$.

Seirios
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