# Permutations in $A_n$ within a proof of simplicity

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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Just notice that $\tau(a)=\sigma^{-1}(\alpha \sigma \alpha^{-1})(a)=\sigma^{-1}(b)=a$. I have used only the data you have given us. – user21436 Mar 25 '12 at 10:58
(You have told us that $\alpha \sigma \alpha^{-1}(a)=b$; and $\sigma(a)=b \implies \sigma^{-1}(b)=a$) – user21436 Mar 25 '12 at 11:00
In which direction do you compose? From left to right or from right to left? – Oo3 Mar 25 '12 at 11:09
Same as Joriki's style! : ) – user21436 Mar 25 '12 at 13:15

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 $a$.
If I compose left to right I find that $a$ goes in $c$, but, if I compose from right to left I find that $a$ is fixed... What is the correct way? – Oo3 Mar 25 '12 at 11:15
@Oo3: From right to left. The notation is chosen such that $(fg)(x)=f(g(x))$; it would be rather confusing if it were $(fg)(x)=g(f(x))$ instead. – joriki Mar 25 '12 at 11:27
@joriki Just a minor comment re. your comment: doesn't GAP use the left to right convention? e.g. (1,2,3)*(2,3) = (1,3). I guess that was adopted because group theorists like to write operations on the right; $x f g$, in your example. I agree this can be confusing sometimes. e.g. with the left to right convention, how should one interpret the operation $(fg)(x)$? That is, the operation $fg$ applied to $x$. Is it $xfg$ or $xgf$? – William DeMeo Mar 25 '12 at 21:13