Showing that the conjugate of a cycle and a permutation is a cucle fo same length

Question

I am asked:

Which I am attempting to answer as:

Since permutations are bijection from $X \rightarrow X$. Then $a_i = \sigma(a_j)$ for some $j$.

Assume $1 \leq i \leq r$

We get:

$\sigma\gamma\sigma^{-1}(a_i)=\sigma\gamma\sigma^{-1}(\sigma(a_j)) = \sigma\gamma(a_j)$

If $j>n$ then $\sigma\gamma(a_j)=\sigma(a_j)=a_i$

Else,

$\sigma\gamma\sigma^{-1}(a_i) = \sigma(a_{j+1})$ (although if $j=r$ then it's actually $a_1$)

This allows me to show that the final function is a cycle of length $r$ but it's not enough to express $\sigma\gamma\sigma^{-1}$ in terms of $a_1, a_n$

Is there a hint I can use to finalize the proof?

Answer 1

A somewhat easier way to do this is to note that $$(a_1\cdots a_k)=(a_1\cdots a_i)(a_{i}\cdots a_k)$$ Iterating this reduces a cycle to a simple product of transpositions. Now you just have to find a formula for conjugating a transposition.