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 $V\cong G\times\{1\}$.

• There's a subgroup of index 2 in $S_3 \times S_3$ that is not of that form. If the groups are finite of coprime order then the answer is yes. Oct 16 '13 at 21:41
• Consider the subgroup $\langle (1,1) \rangle \subseteq (\mathbb{Z}/2\mathbb{Z})^2$. Oct 16 '13 at 22:01
• @BrandonCarter: $\langle (1,1)\rangle \cong \mathbb Z/2\mathbb Z$. Oct 16 '13 at 23:19
• @azimut: My mistake, I thought you were looking for subgroups not equal to a product of subgroups, rather than isomorphic. Oct 17 '13 at 0:55
– user89712
Oct 17 '13 at 20:22

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 $\operatorname{sgn}(\tau\sigma) = -1$, $(\tau,\sigma)\notin V$.)

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.