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

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


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

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 $(abcde)$?

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

  • $\begingroup$ 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? $\endgroup$ – Oo3 Mar 25 '12 at 11:15
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    $\begingroup$ @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. $\endgroup$ – joriki Mar 25 '12 at 11:27
  • $\begingroup$ @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 $xgf$? $\endgroup$ – William DeMeo Mar 25 '12 at 21:13
  • $\begingroup$ @William: I see. I've never used GAP, and I've never seen that notation used. I also hadn't noticed that group theorists like to write operations on the right. But it's often surprising how different niches of the mathematical universe develop independent conventions, so you may well be right about your part of the universe :-) $\endgroup$ – joriki Mar 25 '12 at 21:16
  • $\begingroup$ @joriki Yes, and the conventions can sometimes turn an ugly/hard proof into a pretty/easy one. I guess the best convention is "whatever works" ...as long as we're consistent! :) $\endgroup$ – William DeMeo Mar 25 '12 at 21:28

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