Can we ascertain that there exist an epimorphism $G\rightarrow H?$

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

https://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

(Crossposting my counterexample from MSE here)

Let $G=Q_8\times D_8$, where $Q_8$ is the quaternion group and $D_8$ is the dihedral group of order $8$.

Let $f$ be an isomorphism $$f:G\times G =\left(Q_8\times D_8\right)\times \left(Q_8\times D_8\right)\longrightarrow \left(Q_8\times Q_8\right)\times \left(D_8\times D_8\right).$$ Now, let $\mu$ and $\lambda$ be epimorphisms $$\begin{eqnarray*}\mu:Q_8\times Q_8&\longrightarrow&Q_8 {\small \text{ Y }} Q_8\\ \lambda:D_8 \times D_8&\longrightarrow&D_8 {\small \text{ Y }}D_8\end{eqnarray*}$$ where $A {\small \text{ Y }} B$ denotes the central product of $A$ and $B$. Then $$\mu\times \lambda:\left(Q_8\times Q_8\right)\times \left(D_8\times D_8\right)\longrightarrow \left(Q_8 {\small \text{ Y }}Q_8\right)\times \left(D_8 {\small \text{ Y }}D_8 \right)$$ is an epimorphism. The key is that $D_8{\small \text{ Y }} D_8\cong Q_8{\small \text{ Y }} Q_8$, so if we take an isomorphism $$\phi:D_8{\small \text{ Y }} D_8\longrightarrow Q_8{\small \text{ Y }} Q_8,$$ we can take $H=Q_8{\small \text{ Y }} Q_8$ and form an isomorphism $$1_H\times \phi:\left(Q_8 {\small \text{ Y }}Q_8\right)\times \left(D_8 {\small \text{ Y }}D_8 \right)\longrightarrow \left(Q_8 {\small \text{ Y }}Q_8\right)\times \left(Q_8 {\small \text{ Y }}Q_8 \right)=H\times H.$$ So, all in all, we have $$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} \left(Q_8\times D_8\right) \times \left( Q_8 \times D_8 \right)& \ra{f} &\left(Q_8\times Q_8\right) \times \left( D_8 \times D_8 \right)&\\ & & \da{\mu\times \lambda} & & & & \\ & & \left(Q_8 {\small \text{ Y }}Q_8\right)\times \left(D_8 {\small \text{ Y }}D_8\right) & \ras{1_H\times \phi} & \left(Q_8 {\small \text{ Y }}Q_8\right)\times \left(Q_8 {\small \text{ Y }}Q_8\right) \end{array} $$ and thus an epimorphism $$f(\mu\times\lambda)(1_H\times \phi):G\times G\longrightarrow H\times H.$$ However, $Q_8{\small\text{ Y }}Q_8$ is not a homomorphic image of $Q_8\times D_8$. So this is a counterexample.

(Credit and thanks to Peter Sin for the crucial step in this answer.)

Answer 2

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 3

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 4

[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.