Let $G$ be a central extension of the group $K$ by the simple non-abelian group $H$ ($K$ is the normal subgroup). If we know that this extension is non-split, is it true that the order of $K$ must divide the Schur multiplier of the group $H$ ? (Note that all groups are finite)


No, $K$ can be as large as you like. Let $G$ be the direct product of ${\rm SL}_2(5)$ (the covering group of $H=A_5$) and any abelian group $L$, and let $K = Z \times L$, where $Z = Z({\rm SL}_2(5))$.

  • $\begingroup$ Wonderful! Thanks alot. What about the case in which $(|K|,|M(H)|)=1$? I mean whether in this case we can conclude that the extension has to split? $\endgroup$ – Tina Sep 22 '13 at 15:33
  • 2
    $\begingroup$ Yes, if $H$ is simple (in fact $H$ perfect is enough) and $(|K|,|M(H)|)=1$ then the extension has to split. The commutator subgroup is a complement. $\endgroup$ – Derek Holt Sep 22 '13 at 17:56
  • $\begingroup$ Thanks a million for your help. $\endgroup$ – Tina Sep 23 '13 at 7:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.