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The question is a bit too long for the title so I will elaborate here.

Let $j \geq 7$ be an odd integer and let

$$ \alpha = (1,2,\ldots j), \ \ \beta = (1,2,3) $$

be cycles within the alternating group $A_j$. Show that $\alpha^k \beta \alpha^{-k}$ commutes with $\beta$ if $3 \leq k \leq j-2$, but not for $k = j-1$.

This question comes from these lecture notes, particularly this fact is used in Corollary 4.6 without proof.

I've tried arguing this via some kind of combinatorial argument myself, as well as direct calculation, but it has not been very successful as of yet.

Thanks in advance.

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2 Answers 2

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There's an error in your transcription of the result given in those lecture notes, which is:

In the alternating group $A_j$ for odd $j \geq 7$, let $ \alpha = (1 2\ldots j)$ and $ \beta = (1 2 3) $. Then $\alpha^k \beta \alpha^{-k}$ commutes with $\beta$ if $3 \leq k \leq j-3$, but not if $k = j-2$.

In the following proof, I adopt the convention that the permutations in a product act from right to left. Then \begin{align} \alpha^k\beta\alpha^{-k} &=\left(\alpha^k\left(1\right),\ \alpha^k\left(2\right),\ \alpha^k\left(3\right)\right)\\ &=\cases{\left(k+1,\ k+2, \ k+3\right)&if $\ 1\le k\le j-3$,\\ \left(j-1,\ j, 1\right)&if $\ k= j-2$,\\ \left(j,\ 1, 2\right)&if $\ k= j-1$ .} \end{align} Thus,

  • if $\ 3\le k\le j-3\ $, then $\ \{1,2,3\}\cap\{ k+1,k+2,k+3\}=\emptyset\ $, and $\ \beta\alpha^k\beta\alpha^{-k}=(1,2,3)( k+1,k+2,k+3)=\alpha^k\beta\alpha^{-k}\beta\ $,
  • if $\ k=j-2\ $ then $$\ \beta\alpha^k\beta\alpha^{-k}=(1,2,3)(j-1,j,1)=(j-1,j,2,3,1)\ ,\text{ but}\\ \alpha^k\beta\alpha^{-k}\beta=(j-1,j,1)(1,2,3)=(1,2,3,j-1,j)\ . $$
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  • $\begingroup$ Thanks for spotting that error. I'm curious about the very first step where you write the product as a single cycle - is there an intuitive reason as to why we can do that? $\endgroup$
    – jpmacmanus
    Nov 6, 2019 at 18:33
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    $\begingroup$ If $\ \gamma\ $ and $\ \pi\ $ are any permutations, and $\ \pmatrix{c_1&c_2&\dots&c_r} $ is a cycle of $\ \gamma\ $, then $\ \pi\gamma\pi^{-1}\hspace{-0.3em}\left(\pi(c_j)\right)=\pi(c_{j+1})\ $ for $\ j=1,2,\dots,r-1\ $ and $\ \pi\gamma\pi^{-1}\hspace{-0.3em}\left(\pi(c_r)\right)=\pi(c_1)\ $, so $\ \pmatrix{\pi\left(c_1\right)&\pi\left(c_2\right)&\dots&\pi\left(c_r\right)} $ must be a corresponding cycle of $\ \pi\gamma\pi^{-1}\ $. $\endgroup$
    – lonza leggiera
    Nov 7, 2019 at 3:08
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Imagine a dial with the numbers $1,...,j$ written in clockwise order. There is a marker next to the dial which is initially aligned with the number $1$. The operation $\alpha^k$ spins the dial clockwise $k$ notches, $\beta$ erases the first three numbers clockwise from the position of marker and rewrites them in a different order and $a^{-k}$ spins the dial anticlockwise.

If $3\le k\le j-3$ then, whether we do the sequence of operations $\alpha^k\beta\alpha^{-k}\beta$ or the sequence $\beta\alpha^k\beta\alpha^{-k}$, the dial will be back to its original position and there will be two triples of numbers whose order has been scrambled. If $k=j-2$ then the two places where the erase and rewrite operations are performed overlap and you can work out for yourself that $\alpha^k\beta\alpha^{-k}\beta\ne\beta\alpha^k\beta\alpha^{-k}$

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  • $\begingroup$ This insight was very helpful, thank you. This is a very clever way of looking at the problem. $\endgroup$
    – jpmacmanus
    Nov 6, 2019 at 18:28

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