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This question already has an answer here:

Does there exist a finite non-abelian $2$-group $G$ such that $G^{\prime}\cong D_8$ or $G^{\prime}\cong Q_8$? By an easy inspection with GAP, I could not find any example!

Any answer or comment will be greatly appreciated!

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marked as duplicate by José Carlos Santos, steven gregory, Ethan Bolker, Rhys Steele, user284331 Mar 27 '18 at 1:18

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I found a proof in George A Duckett, Finite Groups: Questions and Answers, that there is no group whose commutator subgroup is nonabelian dihedral (the page is available on Google Books):

Suppose that $G$ is a group with a dihedral commutator subgroup $G'=D_{2n}$ for $n\ge3$. The subgroup $R$ of rotations is characteristic in $G'$, and therefore normal in $G$, so $G$ acts on $R$ by automorphisms. Since $R$ is cyclic, its automorphism group is abelian, and thus the kernel of this action must contain $G'$. In particular, $G'$ centralizes $R$, so $R\le Z(G')$. But we know that rotations are not centralized by reflections in dihedral groups, so this can't be true.

This is attributed to Alexander Gruber.

It also says that the derived subgroup of SL$_2(3)$ is $Q_8$, but SL$_2(3)$ is not a 2-group.

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  • $\begingroup$ Dear Gerry Myerson, thank you very much for your answer! $\endgroup$ – sebastian Mar 26 '18 at 12:45

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