TITLE: Subgroups of semidirect product
QUESTION [5 upvotes]: If a group $G$ is a direct product of two subgroups $U$ and $V$ such that $\lvert U \rvert$ and $\lvert V \rvert$ are coprime then any subgroup of $G$ is a direct product of subgroups of $U$ and $V$. 
Now if $G$ is only a semidirect product, $\lvert U \rvert$ and $\lvert V \rvert$ still coprime, what can we say about the subgroups of $G$? Do they have a similar form?

REPLY [5 votes]: Let $G = U \rtimes V$, where $U$ and $V$ are finite groups of coprime order, and let $H \le G$.
So, if $\pi$ denotes the set of primes dividing $|U|$, then $U = O_\pi(G)$, and $U$ consists of the set of all $\pi$-elements of $G$. So $O_\pi(H) \le U $ and clearly $U \cap H \le O_\pi(H)$, so $O_\pi(H) = U \cap H$.
Now $H/O_\pi(H) = H/(H \cap U) \cong UH/U$ is a $\pi'$-group so, by the Schur-Zassenhaus Theorem, $O_\pi(H)$ has a complement $C$ in $H$ and hence $H = (H \cap U) \rtimes C$. 
Now, by the conjugacy part of the Schur-Zassenhaus Theorem (of which the proof depends on the Feit-Thompson Theorem that all groups of odd order are solvable, so we are using that here!), all complements of $U$ in $CU$ are conjugate, so $C$ is contained in a conjugate $V^g$ of $V$ and hence $H = (H \cap U) \rtimes (H \cap V^g)$.