Is any subgroup of a direct product isomorphic to a direct product of subgroups?

This question arose in the context of this problem:

Let $G$ and $G'$ be groups and $V$ a subgroup of $G\times G'$. Do there exist subgroups $H \leq G$ and $H'\leq G'$ such that $V \cong H\times H'$?

In the case that the answer is "no": Are there any reasonable constraints to $G$ and $G'$ such that the answer is "yes"? For finite abelian groups, the answer appears to be yes.

Clarification

My question is about the subgroup $V$ being isomorphic to a direct product of subgroups. This condition is strictly weaker than being equal to a direct product of subgroups: For the "equal" version, the classical counterexample is the diagonal subgroup $V = \langle (1,1)\rangle \leq G\times G$ where $G$ may be any group but the trivial one. However, this doesn't provide a counterexample to my question, since $V\cong G\times\{1\}$.

Solution 1:

This is based on the comment of @Derek Holt:

A counterexample is given by the subgroup $$V = \{(g,h) \in S_3\times S_3 \mid \operatorname{sgn}(gh) = +1\}$$ of $S_3\times S_3$.

Reason:

Since $V$ is the kernel of the group epimorphism $$S_3\times S_3\to \{\pm 1\},\quad (g,h)\mapsto \operatorname{sgn}(gh),$$ $V$ is indeed a subgroup and $[S_3\times S_3 : V] = 2$. So $\lvert V\rvert = 18$.

Assume that $V$ is isomorphic to the direct product of two subgroups of $S_3$. Then by its order, up to swapping the factors there is only the possibility $V\cong A_3\times S_3$. Now the group $A_3\times S_3$ contains elements of order $6$ (for example $((1,2),(1,2,3))$, while $V$ does not. (Up to swapping the components, an element of order $6$ in $S_3\times S_3$ is a pair consisting of a transposition $\tau$ and a $3$-cycle $\sigma$. Because of $\operatorname{sgn}(\tau\sigma) = -1$, $(\tau,\sigma)\notin V$.)

Contradiction.

If $G$ and $G'$ are finite groups of coprime order, then the claim is true. In this case we can even get equality in the statement $V \cong H\times H'$, see this answer by user.