Why are two permutations conjugate iff they have the same cycle structure?

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?

Answer

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 σ and τ are permutations.

Claim. Let ρ=τστ1 (multiplication corresponding to composition of functions). If σ(i)=j, then ρ(τ(i))=τ(j). In particular, the cycle structure of ρ is the same as the cycle structure of σ, replacing each entry a with τ(a).

Proof. ρ(τ(i))=τστ1τ(i)=τσ(i)=τ(j). QED.

Conversely, suppose that σ and ρ have the same cycle structure. List the cycles of σ above the cycles of ρ, aligning cycles of the same length with one another. Now interpret this as the two-line presentation of a permutation, and call it τ; then τστ1=ρ by the claim.

For example, if σ=(1,3,2,4)(5,6) and ρ=(5,2,3,1)(6,4), then write
132456523164
Then we let τ be the permutation 15, 32, 23, 41, 56, and 64. Then by the claim above, τστ1=ρ. (Note. As Gerry Myerson notes, if we aren’t working in all of Sn, we may not have τ in whatever subgroup we happen to be working in; so there is an implicit assumption for the “if” part that we are working in Sn).

Attribution
Source : Link , Question Author : Han Solo , Answer Author : Arturo Magidin

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