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?

  • $\begingroup$ 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 $(mod r)$. $\endgroup$ – Geoff Robinson May 30 '15 at 14:32
  • $\begingroup$ 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. $\endgroup$ – pjs36 May 30 '15 at 16:10

Yes, you are correct; the convention used is often clear from the context. It can be tricky sometimes though, like when group theory (which tends to compose maps right-to-left) gets used in semigroup theory (which tends to compose maps left-to-right).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.