Proof of Conjugate Cycles Property of Permutations

[Kiwi1]

Prove:

Let [tex]\alpha = (a_1,...,a_s)[/tex] be a cycle and let [tex]\pi[/tex] be a permutation in Sn. Then [tex]\pi \alpha \pi ^{-1}[/tex] is the cycle [tex](\pi(a_1), ... \pi(a_s))[/tex]

My attempt.
[tex](\pi \alpha \pi ^{-1})^s = (\pi \alpha^s \pi ^{-1})=e[/tex] so if this thing is a cycle and its length divides s.

Assume [tex]\pi (a_1)[/tex] is a member of the cycle. Then:

[tex]\pi \alpha \pi ^{-1}(\pi (a_1))=\pi (a_2)[/tex]

and

[tex]\pi \alpha \pi ^{-1}(\pi (a_i))=\pi (a_{i+1})[/tex] for [tex] 1 \leq i < s[/tex]

and finally

[tex]\pi (a_{s+1})=\pi \alpha \pi ^{-1}(\pi (a_s))=\pi(a_1)[/tex] (validating the assumption)

I think I have all of the components of a proof here. But how can I make it more rigorous?

 

[Fallen Angel]

Hi,

You got it, but if you want to see another way of writing it, I would have do the following.

Let $s_{1},\ldots,s_{n}$ be all the elements on $\mathcal{S}_{n}$.

If $s_{i}\neq a_{j}$ for any $j$, then $\pi\alpha\pi^{-1}(s_{i})=s_{i}$.

If $s_{i}=a_{j}$ for some $j$, then $\pi\alpha\pi^{-1}(s_{i})=a_{j+1}$ (Here we denote $a_{1}=a_{s+1}$).

And that's all.