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Suppose $G$ is a finite group, $H \triangleleft G$, such that $H$ is simple and $Var(H) = Var(G) = Var(\frac{G}{H})$ (Here $Var(G)$ stands for minimal group variety containing $G$). Does that imply that $G \cong H \times \frac{G}{H}$?

If $H \cong C_p$ for some prime $p$, then $G$ is an abelian group of exponent $p$ for some prime $p$, which results $G \cong C_p^n$ for some natural $n$. So by classification of abelian finite groups $H$ is a direct factor of $G$. So $G \cong H \times \frac{G}{H}$.

However I do not know what to do here in non-abelian case.

Chain Markov
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1 Answers1

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The answer is Yes.

Suppose $G$ is a finite group, $H\lhd G$ is a simple, normal subgroup, and $Var(H) = Var(G)$. (I do not assume that $Var(H) = Var(\frac{G}{H})$.)

The case where $H$ is abelian is handled in the problem statement. (That argument does not require $Var(H) = Var(\frac{G}{H})$.) For the case where $H$ is nonabelian, choose $N\lhd G$ maximal for $H\cap N = \{1\}$. By the maximality of $N$, $G/N$ is subdirectly irreducible. Also, $G/N$ contains a subgroup $HN/N$ isomorphic to $H$ in its monolith. In particular, $G/N$ has nonabelian monolith, and $|H|\leq |G/N|$.

Every subdirectly irreducible group with nonabelian monolith which belongs to $Var(H)$ is isomorphic to a section of $H$ (a quotient of a subgroup of $H$) according to Theorem 10.1 of Commutator Theory for Congruence Modular Varieties. But the only section $S$ of $H$ that can satisfy $|H|\leq |S|$ is $H$ itself, so $G/N\cong H$. This shows that $N$ is a normal complement to $H$, and $G\cong H\times N$.

Keith Kearnes
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