$A$ and $B$ are finite groups. $H \leq A \times B$. Can we find some $C \leq A$, $D \leq B$ such that $H \cong C \times D$?
In case the statement is not true: is it true under further assumptions about A and B, such as solvability, nilpotency, etc?
Special cases I can prove:
$A$ and $B$ are abelian (following ideas from another discussion: $G$ finite abelian. $A \times B$ embedded in $G$. Is $G=C \times D$ such that $A$ embedded in $C$, $B$ embedded in $D$?)
$(|A|,|B|)=1$. In this case we even have $H = C \times D$. By using the Chinese remainder theorem for instance.