If $|G|=p^3$ and $G$ is not abelian show that $G'=Z(G)$


Since $|G|=p^3$ then $G$ is solvable and let $$1\leq G^{(n-1)}\leq...\leq G^{(n)}=G,(1)$$ be its derived series.We know that $G^{(n-1)}$ is an abelian subgroup of $G$ so $|G^{(n-1)}|=p$ or $p^2$ (since $G$ is not abelian).Hence we have that $\bigg|\dfrac{G}{G^{(n-1)}}\bigg|=p$ or $p^2$ so $\dfrac{G}{G^{(n-1)}}$ is abelian $\Rightarrow G'\leq G^{(n-1)}$ so $(1)$ is $1\leq G' \leq G,(2)$.

Furthermore, $Z(G)\not=1$ (since $G$ is a $p-$group) and $\dfrac{G}{Z(G)}$ is abelian $\Rightarrow G'\leq Z(G)$

Is this correct so far? How could I continue?


Everything you've written is correct, though in the derived series you would have $G'=G^{(n-1)}$ by definition.

You can actually prove this in three small steps. First that $G'\le Z(G)$ you have done.

Second show $|Z(G)|=p$, to do this suppose not and consider the quotient (if quotient by $Z(G)$ is cyclic you can show that $G$ is abelian)

Third show $G'$ is non-trivial, for which you need only show there are elements $x,y\in G$ with $[x,y]\ne 1$

  • $\begingroup$ ok, thank you very much! I had forgotten that if $G/Z(G)$ is cyclic then $G$ is abelian . For the last step it suffices to say that $G'=1$ iff $G$ is abelain, right? $\endgroup$ – giannispapav May 21 '18 at 11:48

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