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.