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

  • 7
    $\begingroup$ 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 $G$... $\endgroup$ – Olivier Bégassat Oct 25 '12 at 23:09
  • 7
    $\begingroup$ @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! $\endgroup$ – Derek Holt Oct 26 '12 at 10:15
  • 5
    $\begingroup$ @DerekHolt yes, and it is a non trivial theorem of Archimedes-Gauss-Gromov that finite groups are finitely generated ^^ $\endgroup$ – Olivier Bégassat Nov 3 '12 at 18:13
  • 6
    $\begingroup$ Updated MO link $\endgroup$ – user1729 Dec 19 '12 at 11:53
  • 7
    $\begingroup$ I already put a bounty on this. No luck. I think it's time to make this a millenium problem to replace the Poincaré conjecture. $\endgroup$ – Julien May 4 '13 at 2:11

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 $$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} \left(Q_8\times D_8\right) \times \left( Q_8 \times D_8 \right)& \ra{f} &\left(Q_8\times Q_8\right) \times \left( D_8 \times D_8 \right)&\\ & & \da{\mu\times \lambda} & & & & \\ & & \left(Q_8 {\small \text{ Y }}Q_8\right)\times \left(D_8 {\small \text{ Y }}D_8\right) & \ras{1_H\times \phi} & \left(Q_8 {\small \text{ Y }}Q_8\right)\times \left(Q_8 {\small \text{ Y }}Q_8\right) \end{array} $$ 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 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 $Q_8 {\small \text{ Y }} Q_8\cong D_8 {\small \text{ Y }} D_8 \not\cong Q_8 {\small \text{ Y }} D_8$.

  • 1
    $\begingroup$ This 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? $\endgroup$ – user641 Aug 27 '13 at 22:25
  • 9
    $\begingroup$ @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. $\endgroup$ – Alexander Gruber Aug 28 '13 at 0:48
  • 1
    $\begingroup$ Can you explain why $Q_8 {\small \text{ Y }}Q_8$ is not a quotient of $Q_8 \times D_8$ ? $\endgroup$ – user10676 Aug 29 '13 at 13:44
  • 2
    $\begingroup$ @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. $\endgroup$ – user641 Aug 29 '13 at 18:01
  • 2
    $\begingroup$ @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. $\endgroup$ – user641 Aug 29 '13 at 18:50

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.

  • $\begingroup$ What is your working definition of semisimple group? One of these? $\endgroup$ – Julien Apr 5 '13 at 1:06
  • $\begingroup$ Perhaps you are confusing "indecomposable" and "simple"? $\endgroup$ – user641 Apr 8 '13 at 6:00
  • $\begingroup$ You mean semisimple algebraic group? $\endgroup$ – Bombyx mori Apr 10 '13 at 9:35
  • 1
    $\begingroup$ 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. $\endgroup$ – anon Apr 12 '13 at 4:32
  • 1
    $\begingroup$ Either 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. $\endgroup$ – anon Apr 12 '13 at 4:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.