Talk:Symmetric Group Center Trivial

$h$ only permutes $b,c$. By choice $c \neq g(b)$ and $g(b) \neq b = g(a)$ as $a,b$ are distinct. As for the second part... really? You're going to say something as pretentious as '4/10. See me after class.' without taking more than 3 seconds to observe that $h(b)=c$ implies $gh(b)=g(c)$?

But why $hg(b) = g(b)$? And why the mixing of notations? Or in this contect does $h(b)$ not mean the same as $hb$? And does $c \ne b, g(b)$ mean "$c \ne b$ and $c \ne gb$"? And where does it say that $h(b)=c$? Or are you mixing the notation for group product and one-line permutations?--Matt Westwood 06:57, 26 February 2010 (UTC)

In my opinion, a good proof is one which provides all of the necessary details but forces the student to take a moment and think something over to ensure understanding. A logician such as yourself might disagree and so be it - feel free to leave the proof as it is or even delete this comment. I was only attempting to shorten an unnecessarily long proof.

But what do I know. You're the boss. Deleted the original and replaced it with your stuff. The fact that I can't understand it is my fault, of course. --Matt Westwood 06:57, 26 February 2010 (UTC)

I can't follow the "alternative proof" given.

Why does $hg(b) = g(b)$ and why does $g(c) = gh(b)$?

4/10. See me after class. --Matt Westwood 20:02, 23 February 2010 (UTC)

Okay I think I understand now. There are several different notations used in the "terse" proof as given. $a, b, c$ are elements of $\N_n$. Then $g, h$ are elements of $S_n$ (i.e. permutations on $\N_n$. Finally, $(b c)$ is the transposition of $b$ and $c$.

The "terse" proof, then, is entirely sound. It just does not fit in with the philosophy of ProofWiki, which is not called "ObscureHintInCrypticSymbolsWiki".

Thx Arthur - it makes sense now. I should have looked at it longer, but at the time it was posted I wasn't in the mood. --prime mover 16:42, 23 June 2011 (CDT)