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

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I 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
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@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! – Derek Holt Oct 26 '12 at 10:15
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@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. – commenter Oct 26 '12 at 10:51
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@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
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Updated MO link – user1729 Dec 19 '12 at 11:53
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2 Answers

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Consider $\begin{matrix}1&2&3\\4&5&6\\7&8&9\end{matrix}\mapsto\begin{matrix}4&9&2\\3&5&7\\8&1&6\end{matrix}$ and $\begin{matrix}1&1&2&3\\1&1&2&3\\4&4&5&6\\7&7&8&9\end{matrix}\mapsto\begin{matrix}4&9&2\\3&5&7\\8&1&6\end{matrix}$. Similar maps are never possible when $|H|<3$ or $G\times G$ has an appropriate distribution of free elements (i.e., pre-images of no element of $H\times H$), always possible when $|H|>2$ and $G\times G$ has no free elements.

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This should definitely be expanded on, it's not clear what role the matrices you've written down are supposed to play. – Jim Apr 5 at 15:27
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But now that I understand that: here we start with a map $\phi$ from $G\times G$ to $H\times H$. There are no free elements in $G\times G$, as every $(g,g')$ is in the preimage of $\phi(g,g')$...I am having a hard time following your (brief) argument. – julien Apr 5 at 15:36
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Thank you: that is a helpful simplification; and my claim is that the answer to the question is "Yes" if $|H|<3$. Otherwise the answer is "not necessarily"; and the map I offer as counterexample is $(1,1)\mapsto(3,2)$, $(1,2)\mapsto(1,3)$, $(1,3)\mapsto(2,1)$, $(2,1)\mapsto(1,1)$, $(2,2)\mapsto(2,2)$, $(2,3)\mapsto(3,3)$, $(3,1)\mapsto(2,3)$, $(3,2)\mapsto(3,1)$, $(3,3)\mapsto(1,2)$. – mindless oaf Apr 5 at 22:02
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Can someone explain what these maps are, and how they relate at all to the question? – Steve D Apr 8 at 5:53
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@dustanalysis Could you elaborate further? For example, do you think my suspicion that mind is unfamiliar with group theory is unreasonable? If so, what's at least one indication you see that this answer has something to do with groups? Note mind him or herself says there is a likely possibility they don't understand the question (also see Steve's comment). And - if an apparently useless and unclear answer remains so after a user has been given ample chance to clarify (and get help in clarifying), does the answer not deserve a downvote? If not, what kinds of answers deserve downvotes? – anon Apr 12 at 4:34
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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? – julien Apr 5 at 1:06
Perhaps you are confusing "indecomposable" and "simple"? – Steve D Apr 8 at 6:00
You mean semisimple algebraic group? – user32240 Apr 10 at 9:35
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 well-known 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 at 18:03
Your 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 at 4:32
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