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A group $H$ is said to be capable if there is a group $G$ such that $G/Z(G)\cong H$.

It is well known that $G/Z(G)$ can never be cyclic of order $>1$; so cyclic groups are not capable.

More interesting: quaternion group $Q_8$ is not capable: there is no $G$ with $G/Z(G)\cong Q_8$.

Now $Z_p$ (cyclic of order $p$) is not capable; but $Z_p\times Z_p$ is capable.

This raises question, is $Q_8\times Q_8$ capable? In general,

Q. Given any group $H$, is $H\times H$ capable?


Edit: Is there a known example of a finite group $H$ such that $H\times H$ is not capable?

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    $\begingroup$ See here about capable groups from Arturo. $\endgroup$ – Dietrich Burde Dec 27 '16 at 13:37
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    $\begingroup$ From the last theorem quoted in my answer, if $H$ is not capable, and $Z^*(H)\cap [H,H]\neq \{1\}$, then $H\times H$ is not capable. On the other hand, if $Z(H)\cap[H,H]=\{1\}$, then the two-nilpotent product $H\amalg^{\mathfrak{N}_2}H$ would be a witness to the capability of $H\times H$. This would leave cases with $Z^*(H)\neq\{1\}$, $Z^*(H)\cap[H,H]=\{1\}$, but $Z(H)\cap[H,H]\neq \{1\}$ to be considered. $\endgroup$ – Arturo Magidin Dec 30 '16 at 19:37
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Shahriar Shahriari proved in On normal subgroups of capable groups, Arch. Math. (Basel) 48 (1987) no.3 193-198, MR 0880078 (88e:20026), among other things, that if a finite group $G$ contains a normal subgroup $H$ which is either generalized quaternion of order $2^n$, $n\gt 2$, or semidihedral of order $2^n$, $n\gt 3$, then $G$ is not capable. This immediately shows that $Q_8\times Q_8$ cannot be capable (and more generally, gives you lots of examples of groups $H$ such that $H\times H$ is not capable; note that if $H$ is capable, then so is $H\times H$, since any witness $K$ to the capability of $H$ yields $K\times K$ as a witness for the capability of $H\times H$).

Shahriari also proved that if $G$ is finite nilpotent and contains a normal subgroup $H$ which is an extraspecial $p$-groups of order $p^3$ and exponent $p$, $p$ odd, then $G$ is not capable.

Finally, Shahriari proves the following:

Theorem (Prop. 3.2 in the above-cited paper) Let $G$ be a finite group. If $G = QC_Q(G)$ for some $Q\leq G$, and $1\neq M\subseteq Z^*(Q)\cap[Q,Q]$, then $M\subseteq Z^*(G)$; in particular, $G$ is not capable.

Here, $Z^*(Q)$ is the epicenter of $Q$ (similar with $G$). The epicenter is the obstruction to capability: it is the smallest central subgroup of $G$ such that $G/Z^*(G)$ is capable.

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    $\begingroup$ Thank you very much! We are really happy for your help! Given that you are "gone for the foreseeable future" at MSE this is a pleasant surprise. $\endgroup$ – Dietrich Burde Dec 28 '16 at 20:09
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    $\begingroup$ Right up my alleyway, so a brief exception during winter break. $\endgroup$ – Arturo Magidin Dec 28 '16 at 20:14
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In the article "On groups occurring as center factor groups, J. Algebra, 61 (1979)" F. R. Beyl, U. Felgner and P. Schmid present a condition in which the capability of a direct product of finitely many of groups implies the capability of each of the factors. For details see section $6$. So for many classes of groups, the condition that $H\times H$ is capable already implies that $H$ is capable. For example, if we let $G$ be a $p$-group of class at most two and odd prime exponent, then we obtain that if $G$ is a nontrivial direct product, then $G$ is capable if and only if each direct factor is either capable or nontrivial cyclic. There are furthermore articles by Arturo Magidin on this subject.

I suppose one can decide the question on direct products of quaternion groups with these results too, but I did not search "long enough" to find it in the literature. There is also a classification of extra-special capable $p$-groups, see Corollary 8.2 in the paper above.

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  • $\begingroup$ About extra-special $p$-groups; they are capable only when it doesn't contain $Q_8$ or it doesn't contain $Z_{p^2}\rtimes Z_p$ for odd $p$. I don't know how this would help for capability of direct products. But..ok! Thanks for the important information! $\endgroup$ – p Groups Dec 27 '16 at 14:29
  • $\begingroup$ Perhaps - $Q_8\times Q_8$ is not semi-extraspecial, since it has a normal subgroup $N$ of order $2$ in its center such that quotient by $N$ is not extra-special (I followed definition of semi-extraspecial from here Section 1, Line 3 msp.org/ant/2014/8-2/ant-v8-n2-p05-p.pdf $\endgroup$ – p Groups Dec 27 '16 at 15:31
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    $\begingroup$ I just used GAP to compute that the Schur multiplier of $Q_{8} \times Q_{8}$ is elementary abelian of order $16$. $\endgroup$ – Andreas Caranti Dec 27 '16 at 15:38
  • $\begingroup$ Well, how would this help about the question; I have no idea of it; please can you explain a little @Caranti. $\endgroup$ – p Groups Dec 28 '16 at 6:07

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