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I was studying the Barrington Theorem in https://homes.cs.washington.edu/~anuprao/pubs/CSE531Sp2020/lecture2.pdf when I found a doubt regarding permutations. Particularly, with conjugations.

It is said, in the 2nd page, 4th point, that two cyclic permutations $\pi$ and $\sigma$, always have a permutation $\tau$ s.t. $\tau\pi\tau^{-1} = \sigma$.

I have never studied permutations, and wanted to see why this is true.

The unique thing I have came up is that: $\tau\pi\tau^{-1} = \sigma$ iff $\tau\pi = \sigma\tau$ but I am stack here.

Any clue would be really appreciated.

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2 Answers 2

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Separate the domains of the permutation and the conjugation. If X is a set and $\sigma$ is some permutation of X (take $X=\{1,2,\ldots, n\}$, say), imagine there's a new set of Z of the same cardinality as X and a one-to-one, onto mapping $f: Z\to X$. Consider $f^{-1} \sigma f$. It's a function on Z that first maps everything to X, permutes according to $\sigma$, and maps back along the same "mapping lines" as f. Intuitively, "the result does to Z exactly what $\sigma$ does to X". Working out a few small examples should help.

But be careful: The permutations are conjugate in $S_n$ if they have the same cycle structure. This may not be true in subgroups of $S_n$. Consider $S_4$. The elements (1 2 3) and (1 3 2) of $A_4$ have the same cycle structure, but they are not conjugate in $A_4$.

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As pointed by @D.W., Theorem 2 from the link https://www.planetmath.org/conjugacyclassesinthesymmetricgroupsn has a proof for the existance of such $\tau$.

In essence, consider two cycles $\pi$ = ($a_1$ ... $a_n$) and $\sigma$ = ($b_1$ ... $b_n$). We can build $\tau$ as the permutation that moves each $b_i$ to $a_i$.

Indeed, we see that, $\tau^{-1} \pi \tau(b_i) = \tau^{-1} \pi (a_i) = \tau^{-1}(a_{i+1}) = b_{i+1}$. That is, each $b_i$ is mapped to $b_{i+1}$ as $\sigma$ does.

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