Question

Let $G,H$ be finite groups. Suppose we have a epimorphism $G\times G\rightarrow H\times H$. Can we find an epimorphism $G\rightarrow H$?

A fellow graduate student asked me this question during TA sessions. Baffled, I asked this question on mathstackexchange [site][1], received some positive votes but no answer. According to him he has been running a software check on small order groups for days, and still have not find any counter example. So I venture to ask in here. It 'feels' unlikely to be true, yet we cannot find a proof or a counter example.

This is a repost of

http://mathoverflow.net/questions/110857/can-we-ascertain-that-there-exist-an-epimorphism-g-rightarrow-h

by the request of the moderator from meta.mathoverflow. The original post will be merged with this post.

Answer 1

Here's an observation about a possible minimal counterexample. Suppose one has an epimorphism $\varphi:G\times G\twoheadrightarrow H\times H$. Then we have two maps $\varphi_1:G\to H\times H$ such that $\varphi_1(g)=\varphi(g,1)$, and $\varphi_2: G\to H\times H$ such that $\varphi_2(g)=\varphi(1,g)$. We then have $\varphi(g_1,g_2)=\varphi_1(g_1)\cdot \varphi_2(g_2)$. Then clearly $ker(\varphi_1)\times ker(\varphi_2) \subset ker(\varphi)$. So $(ker(\varphi_1)\cap ker(\varphi_2) )\times (ker(\varphi_1)\cap ker(\varphi_2) ) \subset ker(\varphi)$. Let $G'=G/(ker(\varphi_1)\cap ker(\varphi_2) )$. Then the map $\varphi$ factors through the map $G\times G \to G'\times G'$. Clearly then if $G$ does not admit a surjection to $H$, then neither does $G'$. So for a minimal counterexample, we must have $ker(\varphi_1)\cap ker(\varphi_2)=1$.

This gives some insight to a minimal possible counterexample. Consider the map $\varphi_1\times \varphi_2: G \to H\times H\times H\times H$. Then $ker(\varphi_1\times \varphi_2)=ker(\varphi_1)\cap ker(\varphi_2)=1$, so we have an embedding $G\hookrightarrow H^4$. So a minimal counterexample $G$ must embed in $H^4$.

Answer 2

As explained in the comments, the result is true if $H$ is abelian.

Here is an argument which shows that the result is true in the somewhat orthogonal case where $H$ has trivial center [EDIT] and is indecomposable [/EDIT].

Write the epimorphism $G \times G \to H \times H$ as $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ with $a,b,c,d : G \to H$. Let $A,B,C,D$ be the respective images of $a,b,c,d$ in $H$.

The groups $A$ and $B$ commute elementwise in the sense that $xy=yx$ for every $x \in A$ and $y \in B$. Moreover, they generate $H$ by assumption. So we have an exact sequence

\begin{equation*} 1 \to A \cap B \to A \times B \to H \to 1 \end{equation*}

and similarly for $C,D$. Note that $A \cap B$ commutes with $A$ and $B$, so it must lie in the center of $H$, thus it should be trivial. Therefore $H=A \times B = C \times D$. It follows that $A=\{e\}$ or $B=\{e\}$, thus $a$ or $b$ is surjective.

The same argument also works in some cases where the center of $H$ is not trivial, for example when $H$ is a group of order $p^3$ with $p$ prime.

Answer 3

[Slightly too long for a comment, so I post it community wiki answer.]

The kernel of the epimorphism $\quad\varphi : G\times G \to H\times H\quad$ is a normal subgroup of $G\times G$, for which by an easy calculation one can show that

$$N_{-}:=[\pi_1(N), G]\times [\pi_2(N), G] \le N \le \pi_1(N)\times \pi_2(N)$$

with $\pi_i$ the projection on the $i$-th coordinate. As $G$ acts trivially on $\pi_i(N)/[\pi_i(N), G]\;$, $N/N_{-}$ is central in $(G\times G)/N_{-}\;$. [This might be the motivation for Yves' second comment.]

Similar statements hold for the preimages $\varphi^{-1}(H \times 1)$ and $\varphi^{-1}(1 \times H)$ , one can also play around with Goursat's lemma, but I'm still undecided if I should rather try to prove or disprove the question.