Finite groups: $H \leq A \times B$. Is $H \cong C \times D$ for some $C \leq A$, $D \leq B$? $A$ 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.
 A: You might be interested in the expository article "Subgroups of direct products of groups, ideals and subrings of direct products of rings, and Goursat's lemma" by Anderson and Camillo.  A couple of excerpts:



A: It's not possible if you want the isomorphism to be compatible with the product structure.
Namely, choose $A=B$ and $H = \lbrace (a,a) : a\in A\rbrace$ the "diagonal subgroup". Clearly, the subgroup $H$ does not have the form $H = C\times D \subseteq A\times A$.
A: 
No, a subgroup of a direct product need not be a direct product of subgroups, even up to isomorphism, even if the factors are almost abelian.

Examples are plentiful.  Here is a small nearly-abelian example (nilpotent and solvable and hamiltonian) and a small easy to write down example:


*

*Take A=B to be quaternion of order 8. A×B has 15 maximal subgroups of order 32, 9 of which are directly indecomposable, and so no such C or D exist.  Hence nilpotent of class at most 2 with all subgroups normal (just barely not-abelian) is not sufficient.

*Take A=B to be non-abelian of order 6. A×B has a maximal subgroup H of order 18 that is directly indecomposable, so again no such C or D exist. H can be generated by (1,2,3), (4,5,6) and (1,2)(4,5).  Its elements of order 2 are all of the form (a,b)(c,d) where a≠b in {1,2,3} and c≠d in {4,5,6}, and so all of them are self-centralizing, and so none of them can be part of a proper direct factor (which is centralized by the entire other non-identity factor).  One of the direct factors has to have even order, and so that one has to be all of H.
