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I know the definition of conjugate permutations. $$\exists p \quad p^{-1} \alpha p=\beta$$ So the $\alpha$ and $\beta$ is a pair of conjugate permutations. But can anybody can give some concise, vivid example to describe conjugate permutations? To help me understand the relation of $\alpha$ and $\beta$ intuitively? I'm a newbie in group-theory.

First time here. If there is anything unsuitable in my specification, do tell me please.

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  • $\begingroup$ @ZeroXLR Thanks for your edit.:) $\endgroup$
    – mayi
    Apr 17, 2016 at 13:55

2 Answers 2

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Conjugation of a permutation is the same as renaming the objects on which your group of permutations acts. Therefore, two permutations are conjugate iff they act in the "same way", just on different elements. For instance, in $S_5$, the permutations $(123)(45)$ and $(145)(23)$ are conjugate, since they essentially do the same thing.

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    $\begingroup$ For another example, consider the general linear group where conjugacy is similarity. In that group, consider the diagonalizable matrices similar to some given diagonal matrix (which is not the identity). These all act the same way - scaling-, but on different vectors -their eigenvectors. $\endgroup$ Apr 17, 2016 at 13:57
  • $\begingroup$ @Andrew In that case, the "renaming of objects" I mentioned above is replaced with "change of basis". A conjugation in a general linear group is the same as change of basis. $\endgroup$
    – Arthur
    Apr 17, 2016 at 14:01
  • $\begingroup$ Sorry I use Mathematica to express my comprehension about what you express in answer.Look this picture.As I see,$(123)(45)$ and $(145)(23)$ do a unsame thing?Or can you make a small example to specify the "same thing" you mean? $\endgroup$
    – mayi
    Apr 17, 2016 at 14:02
  • $\begingroup$ I think it might be worthwhile to try to visualize this without the help of Mathematica. The reason that @Arthur says $(123)(45)$ and $(145)(23)$ "do the same thing" is that if you start with $(123)(45)$, and if you interchange the positions of $2$ and $4$ and interchange the positions of $3$ and $5$, then you get $(145)(23)$. Interchanging these positions is the same as conjugating by $(24)(35)$. $\endgroup$
    – Lee Mosher
    Apr 17, 2016 at 14:55
  • $\begingroup$ @mayl The technical term is that the two elements have the same cycle structure, i.e. in disjoint cycle form (the form I've written them in) they both have one two-cycle and one three-cycle. $\endgroup$
    – Arthur
    Apr 17, 2016 at 21:43
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A little more explicitly:

Let's say you have the permutation $\alpha$ with $\alpha(1) = 3$, $\alpha(2) = 1$, $\alpha(3) = 2$. Let's say your are afraid of numbers, so to make this permutation friendly you use letters instead. You relabel $1$ as $a$, $2$ as $b$, $3$ as $c$ to get a new permutation $\beta$ of the letters $\{a,b,c\}$. We get $\beta(a) = c$, $\beta(b) = a$, $\beta(c) = b$.

To figure out more carefully what is going on, define $p:\{a,b,c\} \rightarrow \{1,2,3\}$ by $p(a) = 1,p(b) = 2, p(c) = 3$, so that $p$ is the "translator" between your sets of symbols. You want to know where $\alpha$ should do in this translated alphabet. To find out what should happen to $a$, first translate: $p(a) = 1$. Then apply $\alpha$: $\alpha(1) = 3$. Finally, translate back using $p^{-1}$: $p^{-1}(3) = c$. So the "translated" version of $\alpha$ is the map $\beta = p^{-1} \alpha p$ with $\beta(a) = c$. W get $$\beta(a) = p^{-1}\alpha p(a) = p^{-1} \alpha(1) = p^{-1}(3) = c,$$ $$\beta(b) = p^{-1} \alpha p(b) = p^{-1} \alpha(2) = p^{-1} (1) = a,$$ $$\beta(c) = p^{-1} \alpha p(c) = p^{-1} \alpha (3) = p^{-1} (2) = b$$ which is what we want.

Here we're using two sets of symbols, $\{1,2,3\}$ and $\{a,b,c\}$, but we might as well relabel using the same set of symbols $\{1,2,3\}$ and $\{1,2,3\}$. The same idea works, and then you have the notion of conjugacy in $S_n$: if $\beta = p^{-1} \alpha p$, then $\beta$ is just given by taking $\alpha$ and relabeling the symbols using $p$ as a "translation".

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  • $\begingroup$ Wow,Thanks a lot for your efforts.:) $\endgroup$
    – mayi
    Apr 17, 2016 at 18:23

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