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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.

  • 3
    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 $p$.2012-07-20
  • 0
    Do you know any facts about finite $p$-groups?2012-07-20
  • 0
    For (2), show that if $G/N$ is abelian, then $[G,G]\subseteq N$. Conclude that $[G,G]\subseteq Z(G)$.2012-07-20
  • 2
    Your acceptation rate would make a commutator blush...2012-07-20

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