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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)

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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 $Z = Z({\rm SL}_2(5))$.

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  • $\begingroup$ 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
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    $\begingroup$ 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

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