Timeline for Large(ish) finite non-abelian subgroups of $\operatorname{GL}_n \mathbb C$ for $n>70$
Current License: CC BY-SA 4.0
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| Nov 8 at 7:12 | vote | accept | Fetchinson0234 | ||
| Nov 7 at 15:26 | history | edited | Fetchinson0234 | CC BY-SA 4.0 |
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| Nov 7 at 9:05 | history | edited | Fetchinson0234 | CC BY-SA 4.0 |
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| Nov 7 at 5:34 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
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| Nov 7 at 2:22 | review | Close votes | |||
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| Nov 6 at 21:31 | history | edited | Fetchinson0234 | CC BY-SA 4.0 |
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| Nov 6 at 21:00 | history | edited | Fetchinson0234 | CC BY-SA 4.0 |
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| Nov 6 at 20:18 | vote | accept | Fetchinson0234 | ||
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| Nov 6 at 20:17 | comment | added | Fetchinson0234 | @HJRW That's true, updated the original post. | |
| Nov 6 at 20:17 | history | edited | Fetchinson0234 | CC BY-SA 4.0 |
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| Nov 6 at 17:46 | comment | added | HJRW | I think one detail in some parts of the above discussion is slightly wrong. The “icosahedral group” (ie $A_5$) is not a subgroup of $SU(2)$; it’s a subgroup of $SO(3)\cong SU(2)/\{\pm I\}$. Its preimage in $SU(2)$ is the binary icosahedral group, which has order 120 and has a non-trivial abelian normal subgroup. | |
| Nov 6 at 14:42 | history | edited | Fetchinson0234 | CC BY-SA 4.0 |
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| Nov 6 at 14:35 | history | edited | LSpice | CC BY-SA 4.0 |
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| Nov 6 at 14:35 | comment | added | Geoff Robinson | OK, I deleted it, but now I have undeleted it. | |
| Nov 6 at 14:34 | history | edited | LSpice | CC BY-SA 4.0 |
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| Nov 6 at 14:14 | comment | added | Fetchinson0234 | @GeoffRobinson I think I've seen an answer from you, which was helpful (although without an explicit construction), but now it's gone, do you plan to post it again or at least put it into a comment? | |
| Nov 6 at 14:13 | history | edited | Fetchinson0234 | CC BY-SA 4.0 |
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| Nov 6 at 14:10 | comment | added | Fetchinson0234 | @DenisT Thanks, I'll add this to the original post. | |
| Nov 6 at 14:08 | comment | added | Denis T | @Fetchinson0234 There's a reasonable queue which returns icosahedral group on SU(2): the largest finite subgroup without normal abelian subgroups. | |
| Nov 6 at 13:43 | comment | added | Sean Eberhard | Anyway the Collins paper is here: degruyter.com/document/doi/10.1515/JGT.2007.032/html?lang=en | |
| Nov 6 at 13:42 | comment | added | Sean Eberhard | If $n$ is even, $\mathrm{SU}(n)$ contains a copy of $S_{n+1}$ as well as a copy of $\mathrm{SL}_2(5)^{n/2} \rtimes S_{n/2}$ (since $\mathrm{SL}_2(5)$ embeds in $\mathrm{SU}(2)$). The former group has no abelian normal subgroup at all, while the latter group has a normal abelian subgroup of index $60^{n/2} (n/2)!$. I believe the point about $n = 70$ is just that $(n+1)! > 60^{n/2} (n/2)!$ for $n > 70$ but not for $n \le 70$. | |
| Nov 6 at 13:00 | history | edited | Fetchinson0234 | CC BY-SA 4.0 |
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| Nov 6 at 12:50 | comment | added | Wojowu | What are the "well-known" results for $n=70$ that you mention? | |
| Nov 6 at 12:46 | history | edited | Fetchinson0234 | CC BY-SA 4.0 |
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| Nov 6 at 12:11 | answer | added | Geoff Robinson | timeline score: 6 | |
| Nov 6 at 11:52 | history | edited | Fetchinson0234 | CC BY-SA 4.0 |
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| Nov 6 at 11:36 | comment | added | Geoff Robinson | For finite subgroups of ${\rm GL}(n,\mathbb{Q}),$ there is a finite subgroup of order $2^{n}n!$, while by an argument of H. Blichfeldt, any finite subgroup of ${\rm GL}(n,\mathbb{Q})$ has order dividing $(2n)!$. | |
| Nov 6 at 11:32 | comment | added | YCor | @SeanEberhard on the rational case: your link points to a discussion of finite subgroups of maximal order in GL(n,Q), which is quite hard. But proving that there is a bound on orders of finite subgroups is very elementary. First they are conjugate into GL(n,Z) (easy, finding an invariant lattice) and then the bound follows using a torsion-free finite index subgroup, e.g. the 3rd congruence subgroup. This gives a bound $<3^{n^2}$ on the orders of finite subgroups of GL(n,Q), which of course is not optimal. | |
| Nov 6 at 11:25 | comment | added | Geoff Robinson | For ${\rm SL}(2,\mathbb{C})$, any finite primitive subgroup $G$ does have $[G:Z(G)] \leq 60.$ | |
| Nov 6 at 11:21 | comment | added | Geoff Robinson | You can only bound ( in terms of $n$) the index of a maximal Abelian normal subgroup of ${\rm GL}(n,\mathbb{C})$ in general (this is basically Jordan's theorem). If you want to bound the order of a finite subgroup of ${\rm SL}(n,\mathbb{C})$ you have to look at (irreducible and) primitive subgroups, in which case it can be done ( again, a bound in terms of $n$). | |
| Nov 6 at 10:03 | comment | added | Fetchinson0234 | So then what's the definition for the subgroup of SU$_2$ which has the answer: the icosahedral group. I thought that's maximal in some sense. | |
| Nov 6 at 9:53 | comment | added | Sean Eberhard | Clearly not: that is the point of my previous comment. On the other hand if you restrict to rational entries the situation changes. See mathoverflow.net/questions/15127/… | |
| Nov 6 at 9:52 | comment | added | Fetchinson0234 | What do you mean here? I thought there is a maximal order for a non-abelian finite subgroup of any SU$_n$. | |
| Nov 6 at 9:48 | comment | added | Sean Eberhard | $\mathrm{SU}(2)$ contains arbitrarily large dihedral groups. | |
| Nov 6 at 9:44 | history | edited | Fetchinson0234 | CC BY-SA 4.0 |
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| Nov 6 at 9:41 | history | edited | YCor | CC BY-SA 4.0 |
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| Nov 6 at 9:34 | history | edited | Fetchinson0234 | CC BY-SA 4.0 |
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| Nov 6 at 9:28 | history | asked | Fetchinson0234 | CC BY-SA 4.0 |