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Prove that for every odd prime number $p$ there is a non-commutative group of order $p^3$ such that $a^p = e$, $\forall a \in G$.

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hint: can you write down a non-trivial map from $\mathbb{Z}/p$ to Aut$((\mathbb{Z}/p)^2)$? – user29743 Dec 9 '12 at 7:25
up vote 1 down vote accepted

Here is the outline, you fill in the details:

Note that $\text{Aut}(\mathbb{Z}_p^2)\cong\text{GL}_2(\mathbb{F}_p)$, and since $|\text{GL}_2(\mathbb{F}_p)|=(p^2-1)(p^2-p)$ we see in particular, $p\mid |\text{Aut}(\mathbb{Z}_p)|$. Thus, there exists a non-trivial homomorphism $\varphi:\mathbb{Z}_p\to\text{Aut}(\mathbb{Z}_p^2)$. Consider then $G=\mathbb{Z}_p^2\rtimes_\varphi\mathbb{Z}_p$. Prove that this is the group you seek.

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How exactly is $\text{GL}_2(\mathbb{F}_p)$ defined? We haven't learned about it. However, I think I can try to prove that the group $G$ that you provided is the group I seek. – chubbycantorset Dec 9 '12 at 7:30
$2 \times 2$ matrices with nonzero determinant over $\mathbb{F}_p$, who you may know by his psuedonym $\mathbb{Z}_p$ or $\mathbb{Z}/p\mathbb{Z}$. The $\mathbb{F}$ is to emphasize that it is a field. – andybenji Dec 9 '12 at 7:31
It's just $2\times 2$ invertible matrices over the field of $p$ elements. – Alex Youcis Dec 9 '12 at 7:32
And how exactly does it imply that there exists a nontrivial homomorphism $\varphi:\mathbb{Z}_p\to\text{Aut}(\mathbb{Z}_p^2)$ ? – chubbycantorset Dec 9 '12 at 7:39
Because you can just send the generator of $\mathbb{Z}_p$ to the element an element of order $p$ in $\text{Aut}(\mathbb{Z}_p^2)$--one exists by Cauchy's theorem. – Alex Youcis Dec 9 '12 at 7:40

If $|G|=p^3$ where $p$ is an odd prime, then it is no hard seeing that $G'=Z(G)$ is of order $p$. Also there is a good Lemma saying that for this group (I mean $G$) we have $$Im(f)\subseteq Z(G), ~~~f:G\to G, x\mapsto x^p $$ Now consider we have the first theorem for isomorphism and so $$G/ker(f)\cong Im(f)\subseteq Z(G)$$ so $|G/ker(f)|=1$ or $|G/ker(f)|=p$. The first one tells us that every non trivial element of the group may be of order $p$.

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Nicely done, my friend! +1 – amWhy Nov 19 '13 at 13:57

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