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Let $S_3$ be the symmetric group on $\lbrace 1,2,3\rbrace.$ Decide how many elements who commutate with $(23)$

Permutation naturally commutes itself, with it's inverse and with the identity permutation. So that's 3.

And then I'm insecure. What shall I do next?

amWhy
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soetirl13
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1 Answers1

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First of all, the permutation $(23)$ is the inverse of itself so you actually found only $2$ permutations that commute with it so far, not $3$. Next, $S_3$ is a very small group, it has only $6$ elements. So you can just check every element in a direct way.

Mark
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  • How is it a inverse of itself? the inverse is (32) ? – soetirl13 Nov 25 '18 at 20:08
  • $(23)$ and $(32)$ is the same permutation. It maps $2$ to $3$, $3$ to $2$, and $1$ to itself. It's just two different ways to write the same permutation. – Mark Nov 25 '18 at 20:09
  • Cool. Ok, so far we have 2. And then I look at {1,2,3} and therefore get, (12)(21)(23) that is 3.. So 5? – soetirl13 Nov 25 '18 at 20:13
  • It depends what is your knowledge in group theory. The set of permutations that commute with $(23)$ is actually a subgroup of $S_3$, so its number of elements must divide the order of $S_3$ by Lagrange's theorem. So it can't be $5$ elements. Just check all the permutations and find what the answer will be. – Mark Nov 25 '18 at 20:23
  • Ahh 2. :))))) thank you – soetirl13 Nov 25 '18 at 21:06