Subgroups written as products Suppose a finite group $G$ is the product of two of its proper subgroups $G=AB$. Assume also that $A\lhd G$ and that $A,B$ have relatively prime orders. Isn't it true that any subgroup $H$ of $G$ can be written as $H=(H\cap A)(H\cap B)$?
 A: No. For example, $H$ may be a nontrivial conjugate of $B$. If $B$ is not normal such a conjugate always exists and by hypothesis $H \cap A$ will be trivial and $H \cap B$ will be strictly smaller than $H$.
For an explicit example, take $G = S_3, A = A_3$, let $B$ be the subgroup generated by a transposition, and let $H$ be the subgroup generated by another transposition. Then $(H \cap A)$ and $(H \cap B)$ are both trivial. 
A: While Qiaochu's answer settles the original question, it might be worth noting that every subgroup $H$ of $AB$ under these hypotheses can be written in the form $(H \cap A)(H \cap B^{x})$ for some $x \in A.$ Let $\pi$ be the set of prime divisors of $A$. Then $H/(H \cap A)$ is isomorphic to a subgroup of $B$, so is a $\pi^{\prime}$-group. By the Schur-Zassenhaus theorem, we have $H = (H \cap A)C$ for some subgroup $C$ of $H$ with $(H \cap A) \cap C = 1.$ By Schur-Zassenhaus again, we have $C^{g} \leq B$ for some $g \in G.$ Write $g = ab$ for some $a \in A, b \in B.$ Then $C^{a} \leq B.$ Hence $H^{a} = (H \cap A)^{a}C^{a} \leq (H^{a} \cap A)(H^{a} \cap B) \leq H^{a}.$ Setting $x = a^{-1}$, conjugating by $x$ gives $H = (H \cap A)(H \cap B^{x}).$
