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I'm slightly confused about the product of the following permutation cycles.

I am given that $s_1 = (1\ 2)$ and $s_2 = (2\ 3)$ where both are generators for the symmetric group $S_3$.

My textbook proceeds by saying that $s_1 s_2 = [3\ 1\ 2]$ and that $s_2 s_1 = [2\ 3\ 1]$, but I can't really see why. To me it looks like the results have been mixed up.

My understanding is that (correct me if I am wrong),

$s_1 s_2 = (1\ 2)\cdot(2\ 3)=(1\ 2)(3)\cdot(1)(2\ 3)=(1\ 2\ 3)$

which on one-line notation is $[2\ 3\ 1]$.

Similarly, I compute the product $s_2 s_1$ as

$s_2 s_1 = (1)(2\ 3)\cdot(1\ 2)(3)= (1\ 3\ 2)=[3\ 1\ 2]$

Clearly, my results are the exact opposites of the textbook's.

What am I missing? At first I thought it was a typo, but Wolfram Alpha agrees with my textbook so there must be something I have misunderstood.

virttop
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  • Check book to see if they compute composition of cycles left-to-right or the other way around. – coffeemath May 05 '19 at 14:20
  • Exactly. I do it from right to left, just like any composition of functions. This is how it is in most books. However some do it from left to right. No idea why. – Mark May 05 '19 at 14:22
  • I agree, usually it is $s_1s_2(i)=s_1(s_2(i))$, so you are correct -see my answer. – Dietrich Burde May 05 '19 at 14:26
  • My textbook has been doing it right to left all the time up until this point, no idea why they didnt in this case. – virttop May 05 '19 at 14:32

1 Answers1

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Usually we apply $(s_1s_2)(i)=s_1(s_2(i))$. Hence $(12)(23)$ applied to $1$, is first applying $(23)$, so fixing $1$, and then moving $1$ to $2$ by $(12)$. Similarly, $2$ is moved to $3$ and then fixed. So in this order, $$ (12)(23)=(123). $$

Dietrich Burde
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  • @Don, are you still confused? Otherwise you could take back the downvote. – Dietrich Burde May 05 '19 at 14:29
  • Ok thanks, I thought there only was one "correct" way of doing it but that definitely explains why it was the opposite. I assume Wolfram Alpha does it left to right then as well. – virttop May 05 '19 at 14:30
  • @user3067688 Yes, sometimes $g\circ f$ is interpreted differently. I think this is not useful. I also have seen $xA$ for matrix multiplication with a vector, which I also don't find useful. – Dietrich Burde May 05 '19 at 14:32