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.


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.


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".

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.