# The form of subgroups in the direct product of groups

I am trying to understand the following fact from the direct product of subgroups.

Suppose that $$G$$ and $$H$$ be two groups. Consider direct product $$G\times H$$. We know that if $$H_1\leq H$$ and $$G_1\leq G$$ then $$G_1\times H_1\leq G\times H$$. But not every subgroup of $$G\times H$$ has this form. For example, in the case $$G=H$$ we can take $$\Delta:=\{(g,g): g\in G\}$$ which is a subgroup of $$G\times G$$ but has not form $$G_1\times H_1$$ where $$G_1$$ and $$H_1$$ are subgroups of $$G$$ and $$H$$, respectively.

But what if $$G\neq H$$? I am not able to come up with something similar.

Would be very grateful if somebody can show how to find the counterexample in this case, i.e. when $$G\neq H$$.

• See my answer at math.stackexchange.com/questions/485512/… for a complete description of such subgroups. – Tobias Kildetoft Oct 29 '18 at 15:24
• If $G$ and $H$ are finite, then all subgroups of $G \times H$ are of the form $G_1 \times H_1$ with $G_1 \le G$, $H_1 \le H$ if and only if $\gcd(|G|,|H|) = 1$. Proof left as exercise. – Derek Holt Oct 29 '18 at 15:32
• @DerekHolt, Thanks a lot for nice example. I have to think about it. Guess that it is not so easy – ZFR Oct 29 '18 at 16:05

To answer your question more specifically, suppose that $$G$$ and $$H$$ are not isomorphic, but they contain isomorphic subgroups, then you can use the same "diagonal construction''.
Take for example, $$S_3\times C_4$$, let $$g$$ be an element of order $$2$$ in $$S_3$$ and $$h$$ be an element of order $$2$$ in $$C_4$$, then consider the subgroup generated by $$gh$$.