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I have heard that two permutations are conjugate if they have the same cyclic structure. Is there an intuitive way to understand why this is?

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Have you read a proof of this fact, and found it to be unintuitive? –  Jonas Meyer Jun 28 '11 at 3:56

4 Answers 4

It's much like with linear transformations: conjugating a matrix amounts to a "change of basis", a translation from one basis to another, but similar matrices still represent the same linear transformation.

Conjugating by a permutation amounts to "translating" into new labels for the elements being permuted, so "similar permutations" (conjugate permutations) must represent the same underlying "shuffling" of the elements of the set, just under possibly different names.

Formally: Suppose that $\sigma$ and $\tau$ are permutations.

Claim. Let $\rho = \tau\sigma\tau^{-1}$ (multiplication corresponding to composition of functions). If $\sigma(i)=j$, then $\rho(\tau(i)) = \tau(j)$. In particular, the cycle structure of $\rho$ is the same as the cycle structure of $\sigma$, replacing each entry $a$ with $\tau(a)$.

Proof. $\rho(\tau(i)) = \tau\sigma\tau^{-1}\tau(i) = \tau\sigma(i) = \tau(j)$. QED.

Conversely, suppose that $\sigma$ and $\rho$ have the same cycle structure. List the cycles of $\sigma$ above the cycles of $\rho$, aligning cycles of the same length with one another. Now interpret this as the two-line presentation of a permutation, and call it $\tau$; then $\tau\sigma\tau^{-1}=\rho$ by the claim.

For example, if $\sigma=(1,3,2,4)(5,6)$ and $\rho=(5,2,3,1)(6,4)$, then write $$\begin{array}{cccccc} 1&3&2&4&5&6\\ 5&2&3&1&6&4 \end{array}$$ Then we let $\tau$ be the permutation $1\mapsto 5$, $3\mapsto 2$, $2\mapsto 3$, $4\mapsto 1$, $5\mapsto 6$, and $6\mapsto 4$. Then by the claim above, $\tau\sigma\tau^{-1}=\rho$. (Note. As Gerry Myerson notes, if we aren't working in all of $S_n$, we may not have $\tau$ in whatever subgroup we happen to be working in; so there is an implicit assumption for the "if" part that we are working in $S_n$).

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Warning: the permutations are conjugate $\bf in\ S_n$ if they have the same cycle structure. This may not be true in subgroups of $S_n$. For example, $A_4$ is the alternating group on 4 symbols, it consists of the even permutations in $S_4$. The elements $(1\ 2\ 3)$ and $(1\ 3\ 2)$ of $A_4$ have the same cycle structure, but they are not conjugate in $A_4$. That is, there are elements $g$ in $S_4$ such that $g^{-1}(1\ 2\ 3)g=(1\ 3\ 2)$, but there is no such element in $A_4$.

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Thank you for your answer, sir. How can we see that for a cycle we have the equation $r(i_1,i_2,\dots,i_k)r^{−1}=(r(i_1),r(i_2),\dots,r(i_k))$? –  Student Jun 22 '14 at 13:30
Why don't you post that as a question, instead of hiding it in a comment on an old answer? –  Gerry Myerson Jun 22 '14 at 23:08
Very well, I'll ask my question in a separate post. I "hid" it in a comment becouse of this thread. –  Student Jun 23 '14 at 5:49

The intuitive way to see this is to realize that "conjugation" in a permutation group is the same as "renaming". Take some permutation; conjugate it by (1 2), the permutation that swaps 1 and 2; what's the result? Calculate a few examples, and you'll see that the result is the same as the original permutation with 1 and 2 changing roles.

Another good way to understand this is to separate the domains of the permutation and the conjugation. If $A$ is a set and $\sigma$ is some permutation of the objects of $A$ (take $A=\{1,2,\ldots, n\}$ for example), imagine there's a new set $Z$ of the same cardinality as $A$ and a one-to-one, onto mapping $f:Z\to A$. What is $f^{-1} \sigma f$? It's a function on $Z$ which first maps everything to $A$, permutes according to $\sigma$, and maps back along the same "mapping lines" as $f$. It should be relatively obvious that the result "does to $Z$ exactly what $\sigma$ does to $A$". Again, working out a few small examples should help.

So, conjugation in $S_n$ is the same thing only when $Z$ happens to be the same set as $A$; the "names" and the "objects" are one and the same.

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Suppose $\rho=\pi\sigma\pi^{-1}$, for any $m\in Z$, we have $\rho^m=\pi\sigma^m\pi^{-1}$, i.e. $\rho^m\pi=\pi\sigma^m$. For a cycle $(i,\sigma(i),\ldots,\sigma^{r-1}(i))$, we have $$(\pi(i),\pi\sigma(i),\ldots,\pi\sigma^{r-1}(i)) =(j,\rho(j),\ldots,\rho^{r-1}(j))$$ where $j=\pi(i)$. This is intuitive, isn't it?

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