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Talk:Cauchy's theorem (group theory)

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253: 243: 222: 331:"Cauchy's theorem is generalised by Sylow's first theorem, which implies that if pn is any prime power dividing the order of G, then G has a subgroup of order pn." This is not correct - Sylow's first theorem states that there's a subgroup of order pn only for the highest power of p dividing the order of G, not for any power of p as stated. I don't know how to edit here (and don't want to ruin it more), just wanted to point this out at least. 191: 406:
Someone Else: 12:20, 1 July 2007 (EST) I fixed the proof a couple of days ago. I don't think this entry qualifies as a stub anymore, as it now includes a full proof. I'm dropping the PlanetMath citations, as it also doesn't really incorporate information unique to PlanetMath. The statement of the
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The two "Examples" sections read like exercises from a problem sheet. I don't see what they add to the article. They are currently marked for improvement with citations, but I would prefer to *replace* them with citations, if they remain in the article at all.
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At this point in the proof, H is a finite abelian group of order relatively prime to p, so there are integers u,v such that 1 = |H|u + pv. By Lagrange, a^(|H|u) is the identity, so a=a^1=a^(pv), so let b=a^v.
309: 147: 44: 501:. Perhaps it would be better to substitute the straightforward proof of the abelian case owned by Kenneth Shum on planetmath. Either that or add details to this proof. 618: 299: 79: 613: 407:
theorem is ubiquitous in algebra and the proof here is not the proof from PlanetMath (which is actually a much simpler proof, so I kept the link.)
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By following the link to the Sylow theorems, you find a statement of Sylow's first theorem that is consistent with the one given here.
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Someone Else: 18:28, 28 June 2007 (EST) At the end of the first paragraph of the proof, I'm not clear why
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Someone Else: 18:28, 28 June 2007 (EST) At the end of the first paragraph of the proof, I'm not clear why
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on Knowledge. If you would like to participate, please visit the project page, where you can join
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I found that part of the proof unobvious as well, and it also wasn't obvious to me why
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Last edited at 16:27, 1 July 2007 (UTC). Substituted at 01:51, 5 May 2016 (UTC)
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I don't understand this: "It is easily checked that for every element
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