# Conjugation of permutations: left-to-right versus right-to-left.

In the group $S_n$ I usually use the fact that if $(a_1 a_2 \dots a_r) \in S_n$ is an r-cycle and $\sigma \in S_n$ then $\sigma (a_1 a_2 \dots a_r)\sigma^{-1} = (\sigma(a_1)\sigma(a_2) \dots \sigma(a_r))$. My question is: does this use the convention that permutations act from left-to-right, or right-to-left?

Here's a proof for this identity:

Let $\rho \in S_n$ be such that $\rho (a_i) = \rho(a_{i+1 \text{ (mod } r)})$, in other words $\rho = (\sigma(a_1) \sigma(a_2) \dots \sigma(a_r))$.

Then $a_i \overset{\sigma}{\longmapsto} \sigma(a_i) \overset{\rho}{\longmapsto} \sigma(a_{i+1}) \overset{\sigma^{-1}}{\longmapsto} a_{i+1} \implies \sigma^{-1}\rho\sigma=(a_1 a_2 \dots a_r)$ and $\sigma(a_1 a_2 \dots a_r)\sigma^{-1} = (\sigma(a_1) \sigma(a_2) \dots \sigma(a_r)) = \rho$.

Here I have used the convention that permutations act from right-to-left. However, in most other situations I prefer to read permutations from left to right - this is probably the most common convention among group theorists (see here).

For consistency, it seems I should be using $\sigma(a_1 a_2 \dots a_r)\sigma^{-1} = (\sigma^{-1}(a_1) \sigma^{-1}(a_2) \dots \sigma^{-1}(a_r))$ instead of the other version. Am I correct, or have I made a misunderstanding somewhere?

• If you apply (and compose) permutations from the left, then it is clear that $\sigma(a_{1}\ldots a_{r})\sigma^{-1}[\sigma(a_{i})] = \sigma(a_{i+1})$ where the subscript $i+1$ is read $(mod r)$. – Geoff Robinson May 30 '15 at 14:32
• I've always let the rightmost permutation go first, so that $(1\ 2)(1\ 3) = (1\ 3\ 2)$. For conjugation, I would always apply $\sigma^{-1}$ on the left; i.e., $\sigma^{-1}(a_1\ a_2\ldots a_r)\sigma$. I believe this is the most consistent "functions act from the left" notation with $f(x)$ and $(f \circ g)(x) = f(g(x))$ rather than $(x)f$. But, I constantly just need to remind myself how it all works. – pjs36 May 30 '15 at 16:10