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I thought I understand how to multiply cyclic permutation, until I tried to determine what are the conjugates of $(12)$ in the symmetric group of three elements.

In order to find the conjugate class of $(12)$, what I need to do is to find the elements from $S_3$ which conjugate with $(12)$. When i tried to do so, I came up with these:

$(1)(12)(1)^{-1}=(12)$

next,

$(12)(12)(12)^{-1}=(12)(12)(21)=(12)(1)=(12)$

but when I tried to compute conjugation with $(13)$, I came to this:

$(13)(12)(13)^{-1}=(13)(12)(31)$ I don't understand how the answer is $(23)$.

Thank you.

Salech Alhasov
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    What do you think the product $(13)(12)(13)$ should be instead? And why? – hmakholm left over Monica Jan 23 '16 at 19:42
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    By the way, the quick way to conjugate a permutation in cycle notation is $$ \sigma;(a_1,a_2\cdots a_n)(b_1,b_2\cdots b_m);\sigma^{-1} = (\sigma(a_1),\sigma(a_2)\cdots \sigma(a_n))(\sigma(b_1),\sigma(b_2)\cdots \sigma(b_m)) $$ – hmakholm left over Monica Jan 23 '16 at 19:44
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    I thought it should be (13), because the evaluation in 1, gives me 1; and evaluation in 2 gives 3 and evaluation in 3 gives me 2. SOLVED. Thank you @HenningMakholm! – Salech Alhasov Jan 23 '16 at 19:47
  • z @Salech, so we agree that the result in two-line notation is $({}^1_1,{}^2_3,{}^3_2)$. But in cycle notation that is $(23)$ (or $(32)$), not $(13)$. – hmakholm left over Monica Jan 23 '16 at 19:49
  • @Henning, absolutely. I don't know what happened to me. I did that competition 10 times and somehow convinced myself that I get $(13)$ all the time. Thank you again. – Salech Alhasov Jan 23 '16 at 19:52

2 Answers2

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Let me add there's a general formula for the conjugates of a permutation, when the permutation is expressed as a product of cycles, which in the present case is: $$\sigma(1\,2)\sigma^{-1}=(\sigma(1)\,\sigma(2)).$$

Bernard
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A multiplication of permutations is just composition. And $(13)$ means the permutation that maps 1 to 3 and 3 to 1 (and 2 to 2, as it is not mentioned).

Composition works right to left. So $(13)(12)(31)$ maps 1 to 3 (first step), 3 stays the same under the second $(12)$, and finally $(13)$ maps 3 back to 1. So the product maps 1 to 1.

Similarly 2 maps to 2 under (13), then 2 to 1 under $(12)$ and next 1 to 3 under $(13)$. So 2 maps to 3.

Of course then 3 has to map to 2 (to be a permutation), but you check in the same way as above.

So the product is $(23)$, the permutation that sends 2 to 3, 3 to 2 and 1 to 1.

Henno Brandsma
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