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

7I don't have a complete answer, but it looks like it should be true whenever $H$ is abelian. For then we have an induced epimorphism from $(G\times G)^{ab}\to H\times H$ with $(G\times G)^{ab}\simeq G^{ab}\times G^{ab}$, and then by the structure theorem for abelian groups we can deduce that the structures of $G^{ab}$ and $H$ must be such that there is an epimorphism from the first one to the second, and thus an epimorphism from $G$ to $H$. I don't know wether it holds for non abelian $H$, but it might be worthwhile to look for counter examples with simple non abelian groups $G$... – Olivier Bégassat Oct 25 '12 at 23:09

1I wonder if the answer is known to the easier question if $G\times G$ is isomorphic to $H \times H$ then $G \cong H$ is known. – JSchlather Oct 26 '12 at 00:02

@JacobSchlather: Doesn't that follow from the [KrullSchmidt theorem](http://en.wikipedia.org/wiki/Krull–Schmidt_theorem#Krull.E2.80.93Schmidt_theorem)? – commenter Oct 26 '12 at 00:31

@commenter Ah, right. Well, that's no help. – JSchlather Oct 26 '12 at 00:42

I'm sure this is a dumb question, but can anybody explain why this question isn't equivalent to Jacob's comment? Since $G$ and $H$ are finite they are Hopfian, so every epimorphism is an isomorphism right? So by KrullSchmidt there is an isomorphism $G\rightarrow H$? What am I missing here? – Alexander Gruber Oct 26 '12 at 01:51

@AlexanderGruber taking composition series for $G$ and $H$, we get composition series for $G\times G$ and $H\times H$ of double length, and featuring the same factors as those for $G$ and $H$ twice. So by uniqueness of those factors up to permutation, we get that $G$ and $H$ have the same factor groups for their composition series, but that is insufficient to show they are isomorphic. – Olivier Bégassat Oct 26 '12 at 02:10

@Olivier: I think your argument only works for finitely generated abelian groups. – Derek Holt Oct 26 '12 at 10:12

7@Jacob, commenter: the KrullSchmidt Theorem requires assumptions on the groups, such as ACC and DCC on normal subgroups. In fact there exists an abelian group $A$ such that $A$ is isomorphic to $A \times A \times A$ but not to $A\times A$. So if we take $B=A\times A$, then $A \times A \cong B \times B$ with $A \not\cong B$. But this does not answer the question, and at the moment I have no idea what the answer is! – Derek Holt Oct 26 '12 at 10:15

3@DerekHolt: I was tacitly assuming that $G$ and $H$ are finite groups (so KrullSchmidt applies) because this is a hypothesis in the question (and I think so did Olivier). Nevertheless I think even an answer with infinite groups would be interesting. – commenter Oct 26 '12 at 10:51

3@Kerry: since it seems that nobody here can answer your problem, you could try asking it on the MathOverflow site, which is intended for more difficult (research level) problems. I would be very interested to know the answer! – Derek Holt Oct 27 '12 at 10:10

I once asked the question @JacobSchlather asked:http://math.stackexchange.com/questions/128465/directproductsofinfinitegroups. The answer agrees with what Derek Holt says. But, someone gave a link to a paper by ALS Corner, which might be of interest to you. To be honest, I never looked at the paper, so I don't know if it answers the epimorphism case. But, may be you will find it useful. – Rankeya Nov 02 '12 at 21:53

2http://mathoverflow.net/questions/110857/canweascertainthatthereexistanepimorphismgrightarrowh – Rasmus Nov 03 '12 at 01:43

5@DerekHolt yes, and it is a non trivial theorem of ArchimedesGaussGromov that finite groups are finitely generated ^^ – Olivier Bégassat Nov 03 '12 at 18:13

3@OlivierBégassat You surely mean something different, as finite groups are a fortiori finitely generated. – Hagen von Eitzen Nov 17 '12 at 18:09

5@HagenvonEitzen I think Olivier Bégassat might have been joking since the pairwise intersections of the lifetimes of Archimedes, Gauss, and Gromov are all empty. – Mark S. Nov 18 '12 at 15:20

2@HagenvonEitzen It was a joke ^^ – Olivier Bégassat Nov 20 '12 at 23:50

1@AlexanderGruber, you can't have meant "every epimorphism is an isomorphism right?". The projection onto the first factor $G\times G \to G$ is a simple example of an epismorphism that is (usually) not an isomorphism. – Omar AntolínCamarena Nov 29 '12 at 20:10

1@Omar I'm afraid I had read the question incorrectly. I understand now what the trouble is. – Alexander Gruber Nov 29 '12 at 20:56

6[Updated MO link](http://mathoverflow.net/questions/114139/canweascertainthatthereexistanepimorphismgrightarrowh) – user1729 Dec 19 '12 at 11:53

3@AlexanderGruber: I think this is a bonafide research problem (if you look at the MO version of this question, [Ian Agol](http://en.wikipedia.org/wiki/Ian_Agol) has given this problem some thought, but to no real avail). However...I believe that there *must* be a solution in the literature somewhere. It is a rather natural question and it is so very easy to form! I wonder if someone like Peter Neumann might know the answer? Or perhaps someone could ask the group pub forum. I shall not ask Peter Neumann, nor shall I ask the group pub forum. Both avenues are overly daunting. – user1729 Apr 02 '13 at 15:02

1why this problem is "answered" now? I could not find it in the "unanswered" category.... – Bombyx mori Apr 12 '13 at 05:52

7I already put a bounty on this. No luck. I think it's time to make this a millenium problem to replace the Poincaré conjecture. – Julien May 04 '13 at 02:11
2 Answers
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 $$\phi:D_8{\small \text{ Y }} D_8\longrightarrow Q_8{\small \text{ Y }} Q_8,$$ then 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 $$\newcommand{\ra}[1]{\kern1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern1.5ex} \newcommand{\ras}[1]{\kern1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} \left(Q_8\times D_8\right) \times \left( Q_8 \times D_8 \right)& \ra{f} &\left(Q_8\times Q_8\right) \times \left( D_8 \times D_8 \right)&\\ & & \da{\mu\times \lambda} & & & & \\ & & \left(Q_8 {\small \text{ Y }}Q_8\right)\times \left(D_8 {\small \text{ Y }}D_8\right) & \ras{1_H\times \phi} & \left(Q_8 {\small \text{ Y }}Q_8\right)\times \left(Q_8 {\small \text{ Y }}Q_8\right) \end{array} $$ 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.
Appendix.
Credit and thanks to Peter Sin for his help with the crucial step in this answer.
See Prop. 3.13 of these notes ("The Theory of $p$groups by David A. Craven", in case the link breaks again) for a proof that $Q_8 {\small \text{ Y }} Q_8\cong D_8 {\small \text{ Y }} D_8 \not\cong Q_8 {\small \text{ Y }} D_8$.
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1This is great! Can I ask if you attempted an exhaustive search via GAP (or its ilk), or if you first narrowed down to using extraspecial groups via some theory? – Aug 27 '13 at 22:25

9@SteveD I did try a MAGMA search last year, but it didn't turn up with anything. (Finding normal subgroup lattices and checking isomorphisms are both very slow processes, especially for groups of high rank. You hit a brick wall at about $G=16$.) I couldn't think of a counterexample, so I assumed it was true and tried to prove it every so often on the weekend, until finally I showed the problem to Peter Sin, who had the idea to look at central products. – Alexander Gruber Aug 28 '13 at 00:48

1Can you explain why $Q_8 {\small \text{ Y }}Q_8$ is not a quotient of $Q_8 \times D_8$ ? – user10676 Aug 29 '13 at 13:44

2@user10676: By considering orders, the size of the kernel of such a quotient would be $2$. Thus the kernel is in the center. The center of $Q_8\times D_8$ is easy to find, and its order $2$ subgroups are just as easy. You can then check the three possible quotients. – Aug 29 '13 at 18:01

2@user10676: there are three central elements of order $2$: one lives in the $Q_8$ factor, one in the $D_8$ factor, and one in neither. Quotienting by either of the first two gives a group which *still* decomposes as a direct product, and that is not true for $Q_8 {\small \text{ Y }}Q_8$. The third gives the quotient $Q_8 {\small \text{ Y }}D_8$, as you remarked. – Aug 29 '13 at 18:50

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If $G$ and $H$ are both semisimple then we can decompose into a product of irreducible groups. So let $G=G_1^{(j_1)}\times ...\times G_m^{(j_m)}$ where each $G_i^{(j_i)}$ is indecomposable and $(j_i)$ is the multiplicity of $G_i$ in $G$. We can do similar for $H$. By an extension of Schur's Lemma the only homomorphisms from $G\times G\cong G_1^{(2n_1)}\times...\times G_k^{(2n_k)} \to H\times H \cong H_1^{(2j_1)}\times..\times H_m^{(2j_m) } $take $G_i$ to $H_j$ where $G_i$ and $H_j$ are isomorphic. So if we have a surjective homomorphism from $G\times G \to H\times H$ then for each $H_i^{(j_i)}$ in there is a $G_m^{(n_m)}$ with $H_i$ isomorphic to $G_m$ and $j_m \le n_m$. So clearly we can construct a homomorphism from $G \to H$ by mapping said $H_i\to G_m$ and everything else to the identity.
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What is your working definition of semisimple group? [One of these](http://planetmath.org/semisimplegroup)? – Julien Apr 05 '13 at 01:06



Based on the language you employ and how they apply (or don't apply) to this question, you seem to be confusing groups with group representations. By your reasoning, as both $S_3$ and $S_4$ are "irreducible" since neither can be written as direct products of nontrivial groups (this is actually called "indecomposable"), there cannot be an epimorphism $S_4\to S_3$. However, there is a wellknown onto group homomorphism $S_4\to S_3$. (Let $S_4$ act on a set of four elements, and consider the induced action on the space of set partitions of shape (2,2), of which there are three.) – anon Apr 11 '13 at 18:03

1Your comment on *mind*'s answer indicates you might have known all along you were talking about representations, even though you call $G$ and $H$ groups. Or perhaps you didn't realize Schur's doesn't apply to groups and mixed up terminology between groups and representations. I can't say, as you didn't respond to one of the three questions directed to you on this answer despite their being straightforward questions asking you simply for the meaning of your terms, and having ample time to respond. – anon Apr 12 '13 at 04:32

1Either way, the fact that Schur's lemma easily solves this question for representations of finite groups as opposed to finite groups themselves should be an interesting fact to a number of people. Thus, if you were to edit this answer so that it is no longer **misleading** and update it with a disclaimer that you are advertising related mathematics even if not directly answering the question (I'd be a hypocrite to downvote for such a practice; I do this often myself), then I'd feel this deserved an upvote. – anon Apr 12 '13 at 04:33

... hum okay.... I'm edit this post Wednesday. I've got too much homework due till then to care. – Squirtle Apr 14 '13 at 19:20