Why are these cycles disjoint and of equal length?

[Hill]

TL;DR

In the conjugation of disjoint cycles, how do we know that the new cycles are also disjoint and of the same length as the original ones?

Given

In the following derivation

how do we know that

?

 

[pasmith]

It is a fact of basic group theory that conjugation preserves the order of an element (because conjugation is an isomorphism from a group to itself). The order of a cycle is the same as its length: the identity (generally written as a single 1-cycle) has order 1. a transposition (1 2) has order 2, etc.

For a cycle $c$, the objects which appear in $h^{-1} c h$ are exactly those objects $x$ such that $h^{-1}(x)$ appears in $c$. (This follows from considering which elements are fixed by a cycle, and therefore do not appear in its cycle notation. Every element which is not fixed must appear in the cycle notation.) Note that $h^{-1}$ is a bijection, so it maps disjoint subsets to disjoint subsets.

 

[Hill]

It is a fact of basic group theory that conjugation preserves the order of an element (because conjugation is an isomorphism from a group to itself). The order of a cycle is the same as its length: the identity (generally written as a single 1-cycle) has order 1. a transposition (1 2) has order 2, etc.

For a cycle $c$, the objects which appear in $h^{-1} c h$ are exactly those objects $x$ such that $h^{-1}(x)$ appears in $c$. (This follows from considering which elements are fixed by a cycle, and therefore do not appear in its cycle notation. Every element which is not fixed must appear in the cycle notation.) Note that $h^{-1}$ is a bijection, so it maps disjoint subsets to disjoint subsets.

Of course. Thank you!