Timeline for Universal group such that every finite group is a quotient
Current License: CC BY-SA 4.0
12 events
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| Nov 20, 2023 at 21:09 | history | edited | Immanuel van Santen | CC BY-SA 4.0 |
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| Nov 20, 2023 at 21:06 | comment | added | Immanuel van Santen | Thanks YCor, I see my error: it is not clear that the homomorphism $\rho |_N \colon N \to \prod_{F \in \mathscr{F} \setminus \mathscr{F}_S} F$ is trivial. | |
| Nov 20, 2023 at 15:50 | comment | added | YCor | @SeanEberhard I think the answer is no, but I can't prove it at the moment. | |
| Nov 20, 2023 at 15:42 | comment | added | Sean Eberhard | @YCor Is there a countable quotient of the direct product of all finite groups that covers infinitely many alternating groups? | |
| Nov 20, 2023 at 10:52 | comment | added | YCor | (...) Then Saxl-Wilson [A note on powers in simple groups, M.P. Cambridge Ph. Soc. 122 (1997), 91–94] proved that nontrivial homomorphisms $G\to A_n$ are continuous (their result apply to any product of nonabelian finite simple group with order tending to infinity), hence nontrivial on $A_n$. Hence for $n\ge n_0$ they don't factor through $H$. Thus $A_n$, for $n\ge n_0$, is not a quotient of $H$. Note that these SW and T papers rely on the classification. | |
| Nov 20, 2023 at 10:48 | comment | added | YCor | Well, here's a proof (using the Thomas 1999 paper [T] above as well as another paper of Saxl-Wilson). Namely I prove: if $G=\prod F_n$ (product of distinct finite simple groups ) surjects onto a countable group $H$, then there $H$ surjects onto the alternating group $A_n$ for only finitely many $n$. Proof: write $A=\prod_{n\ge 5}A_n<G$; let $K$ be the kernel of $G\to H$, and $L=K\cap A$. Then $L$ has countable index in $A$. By [T], $L$ is open in $A$. So $L$, and hence $K$, contains $A_n$ for say $n\ge n_0\ge 5$. (...) | |
| Nov 20, 2023 at 9:12 | comment | added | YCor | So I believe you definitely need some nontrivial fact relying on the classification of finite simple groups. One would like an intermediate lemma, of the spirit: for every normal subgroup of $\prod F_i$, there exists an ideal of subsets such that it consists of sequences whose support belongs to the ideal... but even this is too optimistic. For instance, start from the product of all alternating groups $A_n$, $n\ge 5$. Let $g$ be an element consisting of one 3-cycle in each $A_n$. Then the normal subgroup generated by $g$ is not everything, although $g$ has full support in $\prod A_n$. | |
| Nov 20, 2023 at 9:02 | comment | added | YCor | This is in addition false, because you allow abelian finite simple groups. However this might become true if you allow only non-abelian ones. Related results can be found in: Simon Thomas, Infinite products of finite simple groups II, J. Group Theory 2, 401-434) proved the following a sequence of finite nonabelian simple groups $F_n$ satisfies that $\prod F_n$ has no non-open subgroup of countable index iff the rank of $F_n$ tends to infinity. This however doesn't apply to the sequence of all ones taken once — but the non-open subgroups this produces are not normal. | |
| Nov 20, 2023 at 8:53 | comment | added | YCor | This proof sounds incorrect to me. Indeed if $\eta$ is a nonprincipal ultrafilter, then the set of sequences $(g_i)$, $g_i\in F_i$ such that $\lim_{i\to\eta}g_i=1$ (that is, such that $\{i:g_i=1\}\notin \eta$) is a normal subgroup (that's the kernel of the canonicap map onto the ultrafilter) and is not a subproduct. | |
| Nov 20, 2023 at 8:00 | history | edited | Immanuel van Santen | CC BY-SA 4.0 |
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| S Nov 19, 2023 at 10:16 | review | First answers | |||
| Nov 19, 2023 at 10:33 | |||||
| S Nov 19, 2023 at 10:16 | history | answered | Immanuel van Santen | CC BY-SA 4.0 |