1
$\begingroup$

I'm a bit confused about this proof that $A_n$ is simple for $n > 4$. The conclusion of the second paragraph is not clear to me. Why is that $\sigma'\sigma^{-1}$ fixing fewer symbols than $\sigma$ a contradiction?

  • 2
    It appears to be a typo. What is meant is ***more*** symbols are fixed by $\sigma'\sigma^{-1}$, since $2$ is fixed as well as all fixed points of $\sigma$.2012-10-31

1 Answers 1

1

The element $\sigma$ is chosen to fix the most symbols possible.

  • 0
    As per peoplepower's comment, it should say more symbols.2012-10-31