Timeline for Finite Homomorphic images of infinite products of finite solvable groups
Current License: CC BY-SA 3.0
28 events
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| Nov 21, 2016 at 23:58 | comment | added | Nazih Nahlus | @ Y. Cor. Again, Thank you very much Yves for your help. In fact, I found a slight variation of your proof that avoids the 2-generation theorem (such proof was sent to you by email ) which I will post as a comment on MathOF in few weeks. Recall that Yilong Yang has found another proof using the covering properties of finite groups introduced by his paper in 2016. | |
| Nov 21, 2016 at 15:05 | comment | added | Nazih Nahlus | To be more precise (in my last comment), the last line should be: [the subdirect product subgroup A (in a finite product G of finite simple groups) has the additional property that its canonical projections on each S^{k_S} is surjective in order to get A=G.] | |
| Nov 21, 2016 at 14:19 | comment | added | Nazih Nahlus | @ Y Cor. Thank you very much. I was esp. worried about the factors S^{k_S} when k_S > 1. But you have the projection of the subgroup P above is surjective on every M_S or S^{k_S). So in the generalization of the elementary fact (you mentioned above) to the case of n factors while allowing isomorphic factors, the subdirect product has been augmented with all S^{k_S} to get the desired 'whole" group. | |
| Nov 21, 2016 at 2:44 | comment | added | YCor | No it does not boil down to the 2-generated fact, which was used to produce this family of $d+2$ elements. Choosing such a family for all $S$ produces a subgroup $P$ of $\prod_SM_S$ whose projection on $M_S\simeq S^{k_S}$ is surjective for all $S$. So I'm using the fact that this forces $P$ to be dense. This is quite elementary: it boils down to the case of a finite product. (It's a generalization of the elementary fact that if $S,T$ are non isomorphic simple groups, then the only subgroup of $S\times T$ whose both projections to $S$ and $T$ are surjective, is equal to $S\times T$.) | |
| Nov 21, 2016 at 1:59 | comment | added | Nazih Nahlus | @Y Cor, I do not see clearly the last step in your proof that "we can choose a family of d+2 elements in \prod M_S whose S-component generates M_S, we obtain a generating subset for a dense subgroup of \prod M_S". It seems to boil down to every finite product of finite simple groups is generated by 2 elements which is not clear at all. Can you please explain! | |
| Nov 14, 2016 at 1:13 | vote | accept | Nazih Nahlus | ||
| Nov 7, 2016 at 0:28 | comment | added | Nazih Nahlus | You are absolutely right. Btw, George Bergman has already "generalized" the work of Serre to finite dimensional nilpotent Lie algebras at math.berkeley.edu/~gbergman/papers/pro-np2.pdf . Moreover, we both answered the analogous conjectures for Lie algebras over infinite fields (of char not 2, 3) at math.berkeley.edu/~gbergman/papers/prod_Lie1.pdf | |
| Nov 6, 2016 at 3:01 | comment | added | YCor | No, it's not true. The hypothesis in Saxl-Wilson is necessary: if $S$ is a simple group and $I$ is infinite, then there is a non-continuous surjective homomorphism from $S^I$ to $S$ (e.g., given as limit along an ultraproduct). In my argument I don't "delete" factors. The above homomorphism is trivial in restriction to any factor. But it's nontrivial, say, on the diagonal. | |
| Nov 6, 2016 at 1:58 | comment | added | YCor | If you click YCor you get my public profile, which includes a link to my web page so that you can have my real name there, in case you need to quote me. | |
| Nov 5, 2016 at 19:50 | comment | added | Nazih Nahlus | @ Y. Cor. I just checked that you are in the top 0.17 % this quarter in MathOverflow. Congratulations. WOW. Amazing! | |
| Nov 5, 2016 at 19:44 | comment | added | Nazih Nahlus | @Y. Cor. Many Many Thanks to you . But at least we need to greatly thank you in acknowledgments. Shall we say Y.Cor of MathOVerflow ??? | |
| Nov 5, 2016 at 19:31 | comment | added | YCor | Thanks Nazih, but I'm happy to contribute through MathOverflow and not more. Good luck! | |
| Nov 5, 2016 at 18:45 | comment | added | Nazih Nahlus | @ Y.Cor, Thank you. AGAIN: myself and Yilong Yang, have already invited you above to be a 3rd author in our present paper on such questions & related ones on inverse limits, quasisimple, ..etc. Are you interested ? | |
| Nov 5, 2016 at 17:36 | comment | added | YCor | Yes one can avoid this 2-generated stuff by appealing directly to the Saxl-Wilson theorem (which anyway also relies on the classification of finite simple group). | |
| Nov 5, 2016 at 17:23 | comment | added | Nazih Nahlus | @ Y. Cor. I guess we may be able to skip step 2 (thus avoiding the very deep theorem on 2 generators of simple groups) and work directly on the more general situation of inverse-limits of finite products of (non-abelian) finite simple groups. (Here the directed inverse system is assumed to be surjective). This is plausible by the uniquemess of the Remak-Schmidth decompositions of such groups. | |
| Nov 4, 2016 at 14:06 | comment | added | Nazih Nahlus | @Y Cor. Big Thank you. The case of simple groups was solved recently by Yilong Yang by using the results of his paper in J.Group Theory. 2) At present, Yilong and myself are writing a joint paper on such questions. 3) My self & Yilong like to invite you to join us as a 3rd author in our present paper on such questions & related questions 4) Can you please email me at [email protected]. ManyThanks. | |
| Nov 3, 2016 at 16:39 | comment | added | YCor | @HJRW thanks! it's a bad typo because the conclusion that $F$ is isomorphic to a quotient of $G$ by a closed normal subgroup would be tempting. But I think there are (nontrivial) counterexamples. | |
| Nov 3, 2016 at 16:37 | history | edited | YCor | CC BY-SA 3.0 |
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| Nov 3, 2016 at 9:32 | comment | added | HJRW | In the last line of (1) above, I think you meant $H$ instead of $G$. In any case, $H$ isn't mentioned in the conclusion. | |
| Nov 3, 2016 at 4:27 | history | edited | YCor | CC BY-SA 3.0 |
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| Nov 3, 2016 at 0:35 | comment | added | YCor | Your question whether every quotient of a product of finite simple groups is is isomorphic to a quotient by a closed normal subgroup has a positive answer. The reason is that every finite subset of a product of simple groups is contained in a closed subgroup that has a finitely generated dense subgroup and itself is isomorphic to a product of simple groups (this latter fact can be checked by hand, once we know that each simple finite group is 2-generated). Then one uses my argument which shows more precisely that all large enough $H$ admit the given group as quotient. | |
| Nov 3, 2016 at 0:18 | comment | added | YCor | The explicit Nikolov-Segal statement (the one I refer to) is that any homomorphism from a profinite group with a dense finitely generated subgroup, to a finite group, is continuous. Equivalently, in such a group, all finite index subgroups are open. It's a 2007 paper in Annals of Math. | |
| Nov 2, 2016 at 23:55 | comment | added | Nazih Nahlus | Thanks. Do Nikolov-Segal paper(s) show explicitly the statement of the above result & similar results in profinite groups ? like what if each G_i is a finite simple group, then One would expect a finite homomorphic image of G =\prod G_i to be a product of simple groups each is isomorphic to some G_i. Are you referring to their paper in 2012 ? Big THANK you. | |
| Nov 2, 2016 at 23:35 | comment | added | YCor | Yes. en.wikipedia.org/wiki/Profinite_group | |
| Nov 2, 2016 at 23:14 | comment | added | Nazih Nahlus | Wow. It looks Great!. I need to check carefully the Nikolov-Segal Theorem. Question: Are you taking the pro-discrete topology on G (that is, the topology obtained by putting the discrete topology on each G_i) ?. | |
| Nov 2, 2016 at 20:57 | history | edited | YCor | CC BY-SA 3.0 |
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| Nov 2, 2016 at 18:01 | history | edited | YCor | CC BY-SA 3.0 |
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| Nov 2, 2016 at 17:55 | history | answered | YCor | CC BY-SA 3.0 |