Talk:Cauchy's theorem (group theory)

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First Paragraph

"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. 79.176.35.201 (talk) 08:57, 9 July 2009 (UTC)[reply]

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. — Preceding unsigned comment added by 138.92.106.134 (talk) 19:39, 15 January 2015 (UTC)[reply]

Proof and planetmath

Someone Else: 18:28, 28 June 2007 (EST) At the end of the first paragraph of the proof, I'm not clear why h^n/px has order p since (h^n/px)^p=hⁿx^p=x^p=h^-1.

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 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.) — Preceding unsigned comment added by 75.69.94.54 (talk)

Unclear Proof

I don't understand this: "It is easily checked that for every element a in H there exists b in H such that b^p = a," Why should this be true? 82.208.57.182 (talk) 21:44, 9 April 2008 (UTC)[reply]

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. JackSchmidt (talk) 01:06, 10 April 2008 (UTC)[reply]

I found that part of the proof unobvious as well, and it also wasn't obvious to me why h₂x should have order p. 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. — Preceding unsigned comment added by 86.175.249.158 (talk) 18:05, 30 September 2012 (UTC)[reply]

Assessment comment

The comment(s) below were originally left at Talk:Cauchy's theorem (group theory)/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

Someone Else: 18:28, 28 June 2007 (EST) At the end of the first paragraph of the proof, I'm not clear why h^n/px has order p since (h^n/px)^p=hⁿx^p=x^p=h^{-1.

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 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.)}

Last edited at 16:27, 1 July 2007 (UTC). Substituted at 01:51, 5 May 2016 (UTC)

Not a textbook

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. Danielittlewood (talk) 19:07, 1 April 2024 (UTC)[reply]