IDA: Interactive Document on Algebra

The choice X = {1, ..., n} fixes the set X under consideration. Suppose someone chooses a different numbering of the elements in X. How do we compare two permutations of X with respect to these two numberings?

There is a permutation h of X, which changes our numbering in the new one; so h can be used as a change of names. We describe a given permutation g with respect to the new numbering as follows. First, we apply the `back-transformation' h^-1 to our own numbering, then we apply g, and, finally, we use h again to translate back to the other numbering. As a formula, with respect to the new numbering, the transformation g `reads' hgh^-1. The map g -> hgh^-1 is called conjugation with h. The cycle decomposition of g yields a nice way to calculate the effect of conjugation with a permutation h:

Lemma

Let h be a permutation in S_n.

For every cycle (a₁, ...,a_m) in S_n, h (a₁, ..., a_m) h^-1 = (h(a₁), ..., h(a_m)).
If g₁, g₂, ..., g_t are in S_n, then h(g₁ ··· g_t)h^-1 = (hg₁h^-1) ··· (hg_th^-1).
In particular, if c₁, ..., c_t are (disjoint) cycles, then h(c₁ ··· c_t)h^-1 is the product of the (disjoint) cycles hc₁h^-1, ..., hc_th^-1.

It follows that any two conjugate permutations (one permutation can be obtained from the other by conjugation) have the same cycle structure. The converse also holds.

Proposition

Two elements g and h in S_n have the same cycle structure if and only if there exists a k

S_n with khk^-¹ = g.