Subgroups of $G\times H$

Question

Is it true that if $G,H$ are groups, then every subgroup $S$ of $G\times H$ can be written in the form $S\cong S_G\times S_H$, where $S_G$ and $S_H$ are subgroups of $G$ and $H$ respectively? That is, is $S$ isomorphic (not necessarily equal) to a direct product of subgroups of $G$ and $H$?

Notice that I am not asking whether $S=S_G\times S_H$ for some $S_G,S_H$ subgroups of $G,H$; this is obviously false. Wikipedia provides a trivial counterexample: consider the subgroup of $G\times G$ consisting of elements in the form $(g,g)$. This is clearly not a direct product of subgroups of $G$. However, it is isomorphic to a direct product of subgroups of $G$; namely, the product $G\times \{e\}$.

Can someone please help me prove or disprove the above statement?

Answer 1

If you set $G = H = S_3$, the symmetric group of order 6, then consider the subgroup $K$ of elements $(x,y)$ where either both $x$ and $y$ are even or both $x$ and $y$ are odd. It's not hard to see that $K$ is a subgroup of $S_3 \times S_3$ of order 18.

Now if $K$ was to have a decomposition $K \cong G_1 \times G_2$ where $G_1, G_2$ are isomorphic to subgroups of $S_3$, then by consideration of order one of $|G_1|,|G_2|$ would be equal to 6 and the other would be equal to 3. Without loss of generality set $|G_1| = 6, |G_2| = 3$. Let $x$ be an element of $G_1$ of order 2. Then $(x,1)$ must commute with all elements of $G_1 \times G_2$ of form $(1,y)$ for all $y$ in $G_2$. This includes the two non-identity elements of $G_2$ that are of order 3. This provides two commuting elements $(x,1)$ and $(1,y)$ of $G_1 \times G_2$ of orders 2,3, respectively.

However, it is straightforward to check that the elements of order 2 in $K$ do not commute with any of the elements of order 3 in $K$. The elements of $K$ of order two are all of the form $(t_1, t_2)$ where $t_1$ and $t_2$ are both transpositions. Furthermore, all nine such elements are conjugate (as any two conjugates in $S_3$ are conjugate), meaning that for any given element $(t_1,t_2)$, the centralizer is exactly given by $\langle t_1 \rangle \times \langle t_2\rangle$, which is a Klein four subgroup containing no elements of order 3.

Thus, by contradiction, $K$ does not decompose as a direct product of form $G_1 \times G_2$.