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?
Question about proof that $A_n$ is simple for n > 4.
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abstract-algebra
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2It 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
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The element $\sigma$ is chosen to fix the most symbols possible.
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0As per peoplepower's comment, it should say more symbols. – 2012-10-31