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Suppose $G_1$ and $G_2$ are two finite groups and let $H\leq G_1\times G_2$.

Is it true that $H=P_1\times P_2$ with $P_i\leq G_i$?

I think so for since $H\subseteq G_1\times G_2$ we must have $H=P_1\times P_2$ with $P_i\subseteq G_i$. We have $P_i\neq \phi$, otherwise $H=\phi$. Indeed, $(1_{G_1}, 1_{G_2})\in H$ for $H\leq G_1\times G_2$. Now, if $a, b\in P_1$ then $$(ab, 1_{G_2})=(a, 1_{G_2})\cdot (b, 1_{G_2})\in H=P_1\times P_2$$ so that $ab\in P_1$. A similar reason would work for the inverse.

I read somewhere this was not true, but I can't see what is wrong in the above argument.

Obs: The group structure on $G_1\times G_2$ is given pointwise.

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Andrea Mori has given you a counterexample, so I'll try to comment on your attempted argument.

It looks like you have shown that if $P_1 \leq G_1$ and $P_2 \leq G_2$, then $P_1 \times P_2 \leq G_1 \times G_2$. Let's see what happens when we try to prove the converse.

Suppose $H \leq G_1 \times G_2$. We aim to find $P_1 \leq G_1$ and $P_2 \leq G_2$ such that $H = P_1 \times P_2$. Then certainly $P_1$ must contain the set $S_1 = \{x \in G_1 : (x,y) \in H \text{ for some } y \in G_2\}$ and $P_2$ must contain $S_2 = \{y \in G_2 : (x,y) \in H \text{ for some } x \in G_1\}$. In fact, one can show these sets $S_1, S_2$ are subgroups of $G_1$ and $G_2$, respectively. We shouldn't take anything extraneous in $P_1$ and $P_2$, so let's just try letting $P_1$ and $P_2$ being these subgroups, i.e., \begin{align*} P_1 &= \{x \in G_1 : (x,y) \in H \text{ for some } y \in G_2\}\\ P_2 &= \{y \in G_2 : (x,y) \in H \text{ for some } x \in G_1\} \, . \end{align*}

Certainly $H \subseteq P_1 \times P_2$: given $(x,y) \in H$, then $x \in P_1$ and $y \in P_2$ by definition, so $(x,y) \in P_1 \times P_2$. But is it true that $H \supseteq P_1 \times P_2$? Given $(x,y) \in P_1 \times P_2$, then there exist $v \in G_2$ and $u \in G_1$ such that $(x,v) \in H$ and $(u,y) \in H$. But there is no reason that $(x,y) \in H$, and in general this is false. For instance, in the case of the diagonal subgroup $\Delta=\{(g,g)\mid g\in G\}$ from Andrea Mori's answer, we have $P_1 = P_2 = G$, so $P_1 \times P_2 = G \times G$, which is much bigger than $\Delta$.

Here's a way to think about this geometrically. The diagonal subgroup is like the line $y = x$ in the plane. If we take $P_1$ and $P_2$ to be the projections of the line onto the $x$- and $y$-axes, does taking $P_1 \times P_2$ give us the line back? No, of course not: we get the entire plane.

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The answer is no

For instance the diagonal subgroup $$ \Delta=\{(g,g)\mid g\in G\}<G\times G $$ is never a product of subgroups in each factor.


Remark: The answer remains negative also if $G_1\not\simeq G_2$. Consider for instance the case where there exists a prime $p$ dividing both $|G_1|$ and $|G_2|$. Let $C_1<G_1$ and $C_2<G_2$ subgroups of order $p$ (they exist because of Cauchy's theorem). Then $$ C_1\times C_2\simeq\Bbb F_p\times\Bbb F_p $$ where $\Bbb F_p$ is (the additive grouo of the) field with $p$ elements. But so $C_1\times C_2$ has a structure of $2$-dimensional $\Bbb F_p$-vector space and thus has got $p-1$ subgroups of order $p$ which are not products, namely all the lines generated by vectors not "parallel to the axes".

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  • $\begingroup$ All right, but what was the problem with my argument above? $\endgroup$ – PtF Nov 4 '14 at 19:52
  • $\begingroup$ Well, try to repeat your argument for the diagonal subgroup $\Delta$ and your mistake should be immediately visible. $\endgroup$ – Andrea Mori Nov 4 '14 at 19:54

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