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Dear Stefan Kohl, thank you so much for your answer. Actually, I am working on alternating groups and in all your examples $q^2\nmid n!/m!$. My proofs works for these cases. So, I have to edit my question again. By the way, if $n=1342$ and $m=1327$, then $q=443$ and not $q=269$.
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?
