Talk:Cauchy's theorem (group theory)

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Mathematics rating:	Start Class	Mid Priority	Field: Algebra
Please update this rating as the article progresses, or if the rating is inaccurate. Click to show/hide comments. Please add to or update the comments to suggest improvements to the article. Someone Else: 18:28, 28 June 2007 (EST) At the end of the first paragraph of the proof, I'm not clear why hn/px has order p since (hn/px)p=hnxp=xp=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.)

[edit] 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)

[edit] 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)

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)