I am asked: enter image description here

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 $\sigma\gamma(a_j)=\sigma(a_j)=a_i$


$\sigma\gamma\sigma^{-1}(a_i) = \sigma(a_{j+1})$ (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?

  • $\begingroup$ I think there is a mistake in the question I would rather say "[..] in terms of $a_1,\dots, a_r$ and $\sigma$". Here is how I would present the thing let $a\in X$ then there are two cases, either $\sigma^{-1}(a)=a_i$ where $1\leq i\leq r$ or not. In other words, forget about the $j$ index, it confuses things. $\endgroup$ – Clément Guérin Mar 1 '18 at 3:10

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.

  • $\begingroup$ Trust me that is not easier $\endgroup$ – Makogan Mar 1 '18 at 5:27
  • $\begingroup$ @Mako it gets you a formula, namely $(\sigma(a_1)\cdots\sigma(a_n))$, which is easy to see for a transposition. $\endgroup$ – Matt Samuel Mar 1 '18 at 21:19

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