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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 $G\neq H$.

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    $\begingroup$ See my answer at math.stackexchange.com/questions/485512/… for a complete description of such subgroups. $\endgroup$ – Tobias Kildetoft Oct 29 '18 at 15:24
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    $\begingroup$ 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. $\endgroup$ – Derek Holt Oct 29 '18 at 15:32
  • $\begingroup$ @DerekHolt, Thanks a lot for nice example. I have to think about it. Guess that it is not so easy $\endgroup$ – ZFR Oct 29 '18 at 16:05
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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 $gh$.

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