Proof of Conjugate Cycles Property of Permutations

[Kiwi1]

Prove:

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

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

Assume \pi (a_1) is a member of the cycle. Then:

\pi \alpha \pi ^{-1}(\pi (a_1))=\pi (a_2)

and

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

and finally

\pi (a_{s+1})=\pi \alpha \pi ^{-1}(\pi (a_s))=\pi(a_1) (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.