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Let $G$ be a group non-abelian group of order $p^3$, where $p$ is a prime number, prove that:

$\fbox{1}$ $|Z(G)|=p$

$\fbox{2}$ $Z(G)=G'$

$\fbox{3}$ $\frac{G}{Z(G)}\cong \frac{\mathbb{Z}}{p\mathbb{Z}}\times \frac{\mathbb{Z}}{p\mathbb{Z}} $

$G'=\langle \,\{ [x,y] \mbox{: }x,y\in G \} \, \rangle$ commutator subgroup of $G$ where $[x,y]=xyx^{-1}y^{-1}$

Any hints would be appreciated.

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For (1) show that if G/Z is cyclic then G is abelian. To prove this we use coset represntatives $t^i$. Hence the center Z can not have index p so it has index $p^2$. Therefore the center has order $p$. –  i. m. soloveichik Jul 20 '12 at 0:33
    
Do you know any facts about finite $p$-groups? –  Dylan Moreland Jul 20 '12 at 1:01
    
What have you tried? –  lhf Jul 20 '12 at 1:22
    
For (2), show that if $G/N$ is abelian, then $[G,G]\subseteq N$. Conclude that $[G,G]\subseteq Z(G)$. –  Arturo Magidin Jul 20 '12 at 2:38
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Your acceptation rate would make a commutator blush... –  DonAntonio Jul 20 '12 at 2:53
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1 Answer 1

I hope these hints help somewhat.

For (1), show that if $G/Z(G)$ is cyclic, then $G$ is abelian. Moreover, recall that nontrivial $p$-groups have nontrivial centers, which follows from the class equation.

For (2), Consider the projection of the generators $xyx^{-1}y^{-1}$ into the quotient $G/Z(G)$. Since $G/Z(G)$ has order $p^2$, it is abelian (this can be proved much like part 1). Conclude that $G'\leq Z(G)$. Since $G$ is nonabelian, $|G'|\neq 1$, so $|G'|\geq p$.

For (3), you know by part (1) that $G/Z(G)$ has order $p^2$, and by part (2), it is abelian. So it is isomorphic to either $\mathbb{Z}/p^2\mathbb{Z}$ or $\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$. The first choice gives a contradiction, so throw it out.

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