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Let $G$ be a central extension of the group $K$ by the group $H$. 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 $H$?

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All the studying groups are finite. – Tina Jul 4 '13 at 14:54
I suspect that you meant "does the order of $K$ divide the Schur multiplier of $K$"; here $K$ is the quotient, $H$ is the subgroup, the way you have described the setup. The answer is "No". Once you have found one such $G$, take $G\times A$ for any abelian group $A$. This is a non-split central extension of $K$ by $H\times A$. – Alex B. Jul 4 '13 at 16:28
Dear Alex, thank you for your answer. I should mention that I mean $G/K=H$ and $G=K.H$, where $H$ is a non-abelian simple group. Does the order of $K$ divide the Schur multiplier of $H$? – Tina Jul 4 '13 at 16:48
You are welcome! There was a typo in the first sentence of my comment, I meant "does the order of $H$ divide the Schur multiplier of $K$". – Alex B. Jul 4 '13 at 16:50
Just a quick clarification: there are two standards for naming extensions, unfortunately. In one of them, "extension of $K$ by $H$" means that $K\triangleleft G$ and $G/K\cong H$. But in the other standard, "extension of $K$ by $H$" means that $H\triangleleft G$ and $G/H\cong K$ (hence Alex B.'s mistaking your meaning above). I would suggest always specifying which one you mean. – Arturo Magidin Jul 4 '13 at 18:18

Just so we agree on the setup, you have an exact sequence $$ 1\rightarrow K\rightarrow G\rightarrow H\rightarrow 1, $$ where $K\leq Z(G)$, and you assume that the extension is non-split. You are asking whether it is true that $K$ divides the Schur multiplier of $H$.

The answer is "no", you can make $K$ arbitrarily large.

Indeed, having found one such extension, take $G\times A$ for an abelian group $A$. This is an extension of the form $$ 1\rightarrow K\times A\rightarrow G\times A\rightarrow H\rightarrow 1, $$ and clearly $K\times A\leq Z(G\times A)=Z(G)\times A$. Moreover, the extension still doesn't split, since the map $G\times A\rightarrow H$ factors through $G\rightarrow H$.

That's precisely why in the theory of Schur multipliers you have to demand in addition that $K\leq G'$, and the above example shows that you cannot drop this condition. The above example had $(G\times A)' = G'$. So we no longer had $K\times A\leq G'$, even if this was true for $K$ itself.

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According to a paper by Dixon and Zalesski(Finite imprimitive linear groups of prime degree, J.Algebra (276)340-370), we have "Suppose that $H$ is a perfect group and consider $Z_p$ as a $ZH$-module with trivial action. Then $H^2(H,Z_p)\neq0$ that is, there exist non-split central extensions of $Z_p$ by $H$ if and only if $p$ divides the order of the Schur multiplier $M(H)$ of $H$." So in my question, if $H$ is a simple group and $K=Z_p$, then my conclusion is right. But, I am interested in the other cases for $Z_p$. Can I replace $Z_p$ with abelian groups or some other groups? – Tina Jul 7 '13 at 9:59

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