# Composition of permutations in $S_{3}$

This is quite embarassing but I've been revising an algebra text and I fail to get past through a supposedly easy detail.

Take the symmetric group of permutations $S_{3}$ and its two elements: $(1 2 3)$ and $(1 3 2)$. Now any multiplication table tells you that $(1 2 3) \circ (1 2 3) = (1 3 2)$ and $(1 3 2) \circ (1 3 2) = (1 2 3)$. I just can't figure out why it is true.

For example, $(1 2 3) \circ (1 2 3)$: we first send $1$ to $2$, then $2$ to $3$ by the $2^{nd}$ permutation. Secondly, we send $2$ to $3$ and then $3$ to $1$, which leads to $(1 2 3) \circ (1 2 3) = (2 3 1) = (1 2 3)$. What am I doing wrong? Did I misunderstand the notation?

Thank you.

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## 2 Answers

I think your issue is that you're using the same symbols to notate the elements being shuffled and the order in which you do the shuffling.

In this example I'll use ABC as the three elements, and use (123) to mean that whatever is in position 1 goes to position 2, position 2 goes to position 3, and position 3 goes to position 1.

perm      pos1   pos2   pos3
none       A      B      C
(123)      C      A      B
(123)      B      C      A


starting over:

perm      pos1   pos2    pos3
none       A      B       C
(132)      B      C       A


In both cases, you end up with BCA. Therefore they are the same. Hope this helps.

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Thanks. I knew it was something pretty stupid. Your answers cleared it up. – johnny Aug 18 '11 at 14:12
Don't worry about it, it's not stupid. My friend did research on change ringing (involving S10, S12) and after a few hours of staring at permutations she'd miss silly things like this. – user863680 Aug 18 '11 at 14:14

Your way to do it is totally true. The only mistake you made is that you have shown that $1$ is mapped to $3$ under $(123)\circ (123)$, $3$ to $2$ and $2$ to $1$, but did'nt understood how this is written as a cycle.

So if you apply three times $(123)\circ (123)$ starting from, says, $1$, you get $1\Rightarrow 3 \Rightarrow 2\Rightarrow 1$, and $(123)\circ (123)=(132)$.

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