Timeline for Proof of Cauchy's Theorem from Group Theory - Generalizable?
Current License: CC BY-SA 3.0
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| May 18, 2015 at 21:15 | comment | added | Geoff Robinson | I presume that you know the proof of Cauchy's Theorem by McKay, which works much the same for Abelian and non-Abelian cases, and is rather similar in spirit: look at ordered p-tuples of group elements $(g_{1},g_{2},\ldots,g_{p}) : g_{1}g_{2}\ldots g_{p} = 1_{G}$. There are $|G|^{p-1}$ such $p$-tuples so a multiple of $p$ when $p$ divides $|G|$. But these are invariant under cycling ( ie the obvious action of $\mathbb{Z}/p\mathbb{Z}$). The fixed points have all "coordinates" equal, but the number is divisible by $p$. But $(1,1,\ldots,1)$ is fixed. So there is another $g$ with $g^{p}=1.$ | |
| May 18, 2015 at 21:00 | history | edited | Ofir Gorodetsky | CC BY-SA 3.0 |
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| May 18, 2015 at 20:56 | comment | added | Ofir Gorodetsky | @KConrad I agree that my question is mathematically awkward, and that there are beautiful proofs for the non-abelian case with new insights. If, by chance, the abelian proof still works, it will be a) a small miracle, and b) deeper than Cauchy's Theorem. I don't want this problem to be solved so we'll have another proof of Cauchy's Thm, I want it to be solved to satisfy my curiosity and my understanding of non-abelian groups. | |
| May 18, 2015 at 20:48 | comment | added | LSpice | As a silly comment, an induction on $\lvert G\rvert$ allows one to use this argument to reduce the proof of Cauchy's theorem to the case where $G = [G, G]$. That sort of condition gladdens the heart of a Lie theorist, but I'm not sure that it's so useful for a finite-group theorist. | |
| May 18, 2015 at 20:43 | comment | added | KConrad | I don't think it is realistic to expect a proof exploiting commutativity to carry over to the non-abelian case. When I first learned about groups, they were all abelian (unit groups mod $m$). For the results (i) order of each $g \in G$ divides $|G|$ and (ii) $g^{|G|} = e$ for all $g \in G$, I had learned a proof of (ii) that exploited commutativity and derived (i) from (ii). To prove the same two results for general finite groups, you prove (i) first via cosets and then derive (ii) from (i). At the end of the day you have the same two theorems, but without commutativity new ideas are needed. | |
| May 18, 2015 at 20:25 | history | asked | Ofir Gorodetsky | CC BY-SA 3.0 |