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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.

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The comments in MathSE included the remarks that it's true when $G$ is abelian (by direct checking) and when the epimorphism is an isomorphism (by the Krull-Schmidt theorem). Certainly it's far from the general picture but I wouldn't deny them as "constructive answers". –  Yves Cornulier Oct 28 '12 at 9:36
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I'm wondering about the following kind of dual question: Suppose that $G\times G$ is isomorphic to a subgroup of $H\times H$. Does this imply that $G$ is isomorphic to a subgroup of $H$? (Here again, only finite groups are considered.) Note that we cannot replace subgroup by normal subgroup here: Let $H=A_4$ be the alternating group of order $12$. Then $(C_2\times C_2)\times 1$ is normal in $A_4\times A_4$, but $A_4$ has no normal subgroup of order $2$. –  Peter Mueller Nov 7 '12 at 13:34
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My question in the comment above has an easy negative answer: The alternating group $A_6$ contains $C_2\times C_2$, and also $C_3\times C_3$. So $C_6\times C_6$ is a subgroup of $A_6\times A_6$. However, $C_6$ is not a subgroup of $A_6$. –  Peter Mueller Nov 9 '12 at 16:15
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@Peter Mueller: Since the merging happened it is not surprsing that the link no longer works. For context see tea.mathoverflow.net/discussion/1467/… (in a nutshell, OP wished to have no accepted answer and thus gave-up the upvotes, as technically there was no other solution) –  quid Nov 22 '12 at 19:32
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Looks like a counterexample was just posted on MSE: math.stackexchange.com/questions/221152/… –  fedja Aug 27 '13 at 23:14
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4 Answers

(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 the 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 the 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.)

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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 $H^4$.

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Thanks! This certainly helps. –  Kerry Nov 3 '12 at 1:21
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Is it a good idea to accept an answer which contributes to the question, but which still is far away from answering the question? I think if an appealing question is still open, this should be visible when going through the lists of questions. –  Peter Mueller Nov 9 '12 at 7:24
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Yes, Kerry should uncheck this - a definitive answer may still appear, and there's no obligation to give partial credit. Also, no one has stated that this is an open problem found in the literature, which would be another satisfactory answer. –  Ian Agol Nov 9 '12 at 16:33
    
Hi Agol: I did not do it, but for some reason mathoverflow automatically registered your answer as the accepted one. I do not know how to uncheck it. –  Kerry Nov 10 '12 at 0:02
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It's because of the bounty on the question - the top-voted answer is automatically accepted when the bounty expires. –  HJRW Nov 10 '12 at 19:47
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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.

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Sorry, I think your argument is wrong when $H$ is decomposable. Take for example $G=H=(\mathbb Z/2\mathbb Z)^2$.Take $a= \begin{pmatrix} 1&0\\1&0 \end{pmatrix}$ and $b= \begin{pmatrix} 0&1\\0&1 \end{pmatrix}$ and whatever $c,d$ you like to make the matrix $\begin{pmatrix} a&b\\c&d \end{pmatrix}$ invertible.Then $a+b:G\to H$ is not an epimorphism. –  Erwan Biland Nov 1 '12 at 18:33
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@François: I think you are using the fact that '$a_i : G \rightarrow H_i$ ($i=1,2$) are surjective so $(a_1,a_2) : G \rightarrow H_1 \times H_2$ is surjective', which is wrong. –  Auguste Hoang Duc Nov 2 '12 at 13:08
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Sorry, the argument indeed only works in the case $H$ has trivial center and is indecomposable. I edited my answer accordingly. The decomposable case leads to the following question : does $G \twoheadrightarrow A^2$ and $G \twoheadrightarrow B^2$ imply $G \twoheadrightarrow A \times B$? Again, this holds in the abelian case. –  François Brunault Nov 4 '12 at 10:37
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[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.

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Sullvian suggested to his students on even days one try to prove the theorem, on odd days try to find a counter-example. Either the way one will succeed unless one met a Godel type proposition. I have no idea which category this belongs. –  Kerry Nov 6 '12 at 5:38
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