Let $G,H$ be finite groups. Suppose we have an epimorphism $$G\times G\rightarrow H\times H$$ Can we find an epimorphism $G\rightarrow H$?
Can we ascertain that there exists an epimorphism $G\rightarrow H$?
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group-theory
finite-groups
abstract-algebra
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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 $G$... – 2012-10-25
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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. – 2012-10-26
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0@JacobSchlather: Doesn't that follow from the [Krull-Schmidt theorem](http://en.wikipedia.org/wiki/Krull–Schmidt_theorem#Krull.E2.80.93Schmidt_theorem)? – 2012-10-26
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0@commenter Ah, right. Well, that's no help. – 2012-10-26
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0I'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 Krull-Schmidt there is an isomorphism $G\rightarrow H$? What am I missing here? – 2012-10-26
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0@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. – 2012-10-26
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0@Olivier: I think your argument only works for finitely generated abelian groups. – 2012-10-26
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7@Jacob, commenter: the Krull-Schmidt 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! – 2012-10-26
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3@DerekHolt: I was tacitly assuming that $G$ and $H$ are finite groups (so Krull-Schmidt 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. – 2012-10-26
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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! – 2012-10-27
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0I once asked the question @JacobSchlather asked:http://math.stackexchange.com/questions/128465/direct-products-of-infinite-groups. 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. – 2012-11-02
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2http://mathoverflow.net/questions/110857/can-we-ascertain-that-there-exist-an-epimorphism-g-rightarrow-h – 2012-11-03
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4@DerekHolt yes, and it is a non trivial theorem of Archimedes-Gauss-Gromov that finite groups are finitely generated ^^ – 2012-11-03
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3@OlivierBégassat You surely mean something different, as finite groups are a fortiori finitely generated. – 2012-11-17
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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. – 2012-11-18
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2@HagenvonEitzen It was a joke ^^ – 2012-11-20
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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. – 2012-11-29
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1@Omar I'm afraid I had read the question incorrectly. I understand now what the trouble is. – 2012-11-29
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6[Updated MO link](http://mathoverflow.net/questions/114139/can-we-ascertain-that-there-exist-an-epimorphism-g-rightarrow-h) – 2012-12-19
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3@AlexanderGruber: I think this is a bona-fide 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. – 2013-04-02
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1why this problem is "answered" now? I could not find it in the "unanswered" category.... – 2013-04-12
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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. – 2013-05-04