Let $H$ and $K$ be groups and $V$ an abelian subgroup of the semidirect product $\ H\rtimes K$. Do there exist abelian subgroups $H^{\prime }\leq H$ \ and $K^{\prime }\leq K$ \ such that $V\cong H^{\prime }\times K^{\prime }$ ?.\ In the case that the answer is "no": Are there any reasonable constraints to $H$ and $K$ such that the answer is "yes"?


closed as off-topic by Derek Holt, Chris Godsil, Jan-Christoph Schlage-Puchta, David Handelman, Mark Sapir Aug 23 '18 at 17:01

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  • $\begingroup$ I voted to close on the grounds that there is a straightforward counterexample of order $8$. $\endgroup$ – Derek Holt Aug 22 '18 at 13:43

No. Let $C_p$ (the cyclic group of order $p$) act on $C_p^p$ by permuting the basis vectors $e_1$, $e_2$, \dots, $e_p$. Writing $\sigma$ for a generator of the $C_p$ that acts, we have $(\sigma, e_1)^p = (0,e_1+e_2+\cdots+e_p) \neq (0,0)$ and $(\sigma,e_1)^{p^2} = (0,0)$, so the subgroup generated by $(\sigma, e_1)$ is cyclic of order $p^2$. But every abelian subgroup of the factors is $C_p$, so a direct product of them would have to be $C_p^k$.

To be more concrete, when $p=2$, the semidirect product $C_2^2 \rtimes C_2$ is dihedral of order $8$, and the dihedral group of order $8$ contains elements of order $4$.


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