The symmetric group $S_n$ acts naturally on the set $\{1,2,\dots, n\}$. Is this a left or a right action?

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    $\begingroup$ It depends on your conventions for the order of a function and its argument and for the order of composing functions. It should be a straightforward check of definitions. $\endgroup$ – Jonas Meyer Feb 19 '11 at 22:05
  • $\begingroup$ Left, at least by the conventions I'm used to. (In all fairness, though, I'm not good with left/right things.) $\endgroup$ – Asaf Karagila Feb 19 '11 at 22:06

First of all, there is no essential difference between a left or right action. A left action $\lambda: G \times S \to S$ becomes a right action $\rho: S \times G \to S$ by setting $\rho(s,g) = \lambda(g^{-1},s)$.

Next, it really depends on how you think of $S_{n}$, how you realize it and on your conventions. If you think of $S_{n}$ as bijections $\pi: \mathbf{n} \to \mathbf{n}$ with composition, and if you adhere to the convention that functions take arguments on the right, then the action is on the left: $(\sigma \circ \pi)(k) = \sigma(\pi(k))$.


If you think of permutations as functions, then the action is on the left as indicated by Theo.

There is, however, a slight wrinkle when you write your permutations as products of cycles. Some books/people like to perform the multiplication of cycles "left to right", as you are reading the cycles, while others do it right to left. For example, if $\rho=(1,2,3,4)$, $\sigma=(1,2,3)$, and $\tau=(1,4)(2,3)$, and you write $\rho\sigma\tau$, then there are two ways of computing it: $$(1,2,3,4)(1,2,3)(1,4)(2,3)$$ Working "left-to-right", you'd say:

$1$ maps to $2$, then $2$ maps to $3$, then $3$ maps to $2$, so $1\mapsto 2$. $2$ maps to $3$, $3$ maps to $1$, $1$ maps to $4$, so $2\mapsto 4$. $3$ maps to $4$, then $4$ maps to $1$, so $3\mapsto 1$; and finally $4$ maps to $1$, $1$ to $3$, and $3$ to $2$, so $4\mapsto 2$. Thus, $\rho\sigma\tau = (1,2,4,3).$

I'm not in my office, so I cannot confirm, but I believe Herstein's "Topics in Algebra" uses this convention.

On the other hand, you could read them right-to-left, as a composition of functions. In that case, you would say:

$1$ maps to $4$, then $4$ is fixed, then $4$ to $1$, so $1$ is fixed. $2$ maps to $3$, then $3$ to $1$, then $1$ to $2$, so $2$ is fixed. $3$ maps to $2$, then $2$ to $3$, then $3$ to $4$, so $3\mapsto 4$. And $4$ maps to $1$, then $1$ to $2$, then $2$ to $3$, so $4\mapsto 3$. Therefore, $\rho\sigma\tau = (3,4)$.

If compute your products of cycles in the former manner, then your action is a right action. If you compute them in the latter, then your action is a left action, just like Theo described it.

  • $\begingroup$ Thanks Arturo, I was about to add something to the same effect, I'm glad you did it. The same applies of course to the matricial notation $\sigma = \begin{pmatrix} 1 & 2 & 3 & \cdots & n-1 & n \\\ \sigma(1) & \sigma(2) & \sigma(3) & \cdots & \sigma(n-1) & \sigma(n)\end{pmatrix}$. There are two ways of defining the composition and depending on that, the action will be on the left or right. $\endgroup$ – t.b. Feb 19 '11 at 22:36
  • $\begingroup$ @Theo: No problem; it's something by abstract algebra professor made a big point about when I learned group theory, because it is so easy to get utterly confused when reading about permutations if you don't know which convention is being used. So I make a point of it when I teach, and so it's something I have near the front of my mind. $\endgroup$ – Arturo Magidin Feb 19 '11 at 22:43
  • $\begingroup$ So did mine... I don't have Herstein, and most books I own use the second convention you described (e.g. Jacobson, Basic Algebra). E.H. Connell Elements of abstract and linear algebra uses the first convention. You can download it from here: math.miami.edu/~ec/book $\endgroup$ – t.b. Feb 19 '11 at 23:03

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