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"?



Is any abelian subgroup of a semidirect product isomorphic to a direct product of abelian subgroups?
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 $4$.


