Timeline for Finite subgroups of GL_n(C)
Current License: CC BY-SA 2.5
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 15, 2010 at 21:55 | comment | added | Portland | Thanks Jack for references, I had no knowledge of $M$-groups until recently, that's probably one of the reasons why I found Blichfelt's lemma difficult. | |
| May 15, 2010 at 17:18 | comment | added | Jack Schmidt | Ok, it might be tomorrow before I have time to carefully check my answer for general m or explain Blichfeldt's lemma. If you get bored, general m should follow by looking at Sylow p-subgroups, and Blichfeldt's lemma is in lots of books that talk about M-groups (it is one of the standard steps in showing supersoluble groups are M-groups). I am trying to write a simpler version using the idea that G is a p-group, not just finite. Curtis-Reiner's proof is not too crazy, but it is a bit fancier than needed. Reading 3 proofs sometimes helps make one of them clear. | |
| May 15, 2010 at 16:27 | vote | accept | Portland | ||
| May 15, 2010 at 16:25 | comment | added | Portland | Jack, Yes I'll be very interested to find all possible orders of the subgroups of GLn(C) of exponent m? | |
| May 15, 2010 at 16:08 | comment | added | Jack Schmidt | Let me know if you really want all exponents, but the answer is the same: all m*k where k divides m^(n-1). The witnesses are all diagonal subgroups. This is easier to understand for p-groups, especially if Cauchy's theorem is mysterious. | |
| May 15, 2010 at 16:00 | answer | added | Jack Schmidt | timeline score: 12 | |
| May 15, 2010 at 15:43 | comment | added | Xandi Tuni | @Portland: You still miss it, there does not exist any group of exponent 7 and order 42. | |
| May 15, 2010 at 15:35 | comment | added | Portland | Indeed, first question was obvious, I just edited and posted the question that was of interest. | |
| May 15, 2010 at 15:34 | history | edited | Portland | CC BY-SA 2.5 |
Inital question was obvious, sorry again
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| May 15, 2010 at 15:24 | comment | added | Pete L. Clark | @Xandi: That's not the question. But I agree that maybe it should be the question...let's wait and see what transpires. | |
| May 15, 2010 at 15:24 | comment | added | Xandi Tuni | @Pete: Sure, but why is $(\mathbb Z/7)^{1'000'000}$ not a subgroup of $\mathrm{GL}_{999'999}(\mathbb C)$? I think that's the question. | |
| May 15, 2010 at 15:19 | comment | added | Pete L. Clark | Any finite group $G$ of prime exponent $p$ has $p$-power order: if not, there exists a prime $\ell \neq p$ dividing $\# G$ , and then by Cauchy's theorem, $G$ has an element of order $\ell$, contradiction. | |
| May 15, 2010 at 15:11 | comment | added | Xandi Tuni | you intended to say for $0\leq k \leq n$ I suppose. | |
| May 15, 2010 at 15:08 | history | asked | Portland | CC BY-SA 2.5 |