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Some time ago I wrote a message about a proof for the simplicity of $A_n$ in the case $n \geq 5$, taken form Bhattacharya's book. After some time I read that proof again and found something that I can't explain. First I write the beginning of the proof (we have to prove that $A_n$ is simple).


Proof. Suppose $H$ is a normal subgroup of $A_n$. We first prove that $h$ must contain a $3$-cycle. Let $\sigma \neq e$ be a permutation in $H$ that moves the least number of integers in $n$ Being an even permutation, $\sigma$ cannot be a cycle of even lenght. Hence, $\sigma$ must be a $3$-cycle or have a decomposition of the form

$ (1)\quad \sigma = (a b c \cdots)\cdots $

or

$ (2)\quad \sigma = (a b)(c d) \cdots , $

where $a$, $b$, $c$, $d$ are distinct. Consider the first case (1). Because $\sigma$ cannot be a $4$-cycle, it must move at least two more elements, say $d$ and $e$. Let $\alpha = (c d e)$. Then

$ \alpha\sigma\alpha^{-1} = (c d e)(a b c \cdots)\cdots(e d c) = (a b d \cdots)\cdots . $ Now let $\tau = \sigma^{-1}(\alpha\sigma\alpha^{-1})$. Then $\tau(a)=a$, $\;$ [CUT]


Why is $\tau(a) = a$ in the case that $\sigma$ is the $5$-cycle $(abcde)$?

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    Same as Joriki's style! : )2012-03-25

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Since both $\alpha\sigma\alpha^{-1}$ and $\sigma^{-1}$ map $a$ to $b$, applying one and then the inverse of the other maps $a$ to $b$ and back to $a$.

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    @ArturoMagidin Yes, thanks! Not all group theorists fit the stereotype. :) I was merely trying to explain/understand/justify GAP's adoption of the convention. Thanks for the modules example too.2012-03-26