Prove that for every odd prime number $p$ there is a non-commutative group of order $p^3$ such that $a^p = e$, $\forall a \in G$.
Prove that for every odd prime number $p$ there is a non-commutative group of order $p^3$ such that...
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0hint: can you write down a non-trivial map from $\mathbb{Z}/p$ to Aut$((\mathbb{Z}/p)^2)$? – 2012-12-09
2 Answers
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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0Ah, got it now. Thanks very much! – 2012-12-09
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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1Nicely done, my friend! +1 – 2013-11-19