This is an exercise from textbook Analysis I by Amann/Escher.
Let $G$ be a group and $X$ a nonempty set. Then $G$ acts (from the left) on $X$ if there is a function $$G \times X \rightarrow X, \quad(g, x) \mapsto g \cdot x$$ such that:
GA1: $e \cdot x=x$ for all $x \in X$.
GA2: $g \cdot(h \cdot x)=(g h) \cdot x$ for all $g, h \in G$ and $x \in X$.
Let $\mathrm{S}_{m}$ be the set of all permutations of $\{1,2,\cdots,m\}$. Then $(\mathrm{S}_{m},\circ)$ is a group where $\circ$ is function composition operation.
Show that $$\mathrm{S}_{m} \times \mathbb{N}^{m} \rightarrow \mathbb{N}^{m}, \quad(\sigma, \alpha) \mapsto \sigma \cdot \alpha := \left(\alpha_{\sigma(1)}, \ldots, \alpha_{\sigma(m)}\right)$$ defines an action of $\mathrm{S}_{m}$ on $\mathbb{N}^{m}$.
I am stuck at proving the GA2, i.e. $\pi \cdot (\sigma \cdot \alpha) = (\pi \circ \sigma) \cdot \alpha$.
My attempt:
Clearly, $\operatorname{id}$ is the identity element of $\mathrm{S}_{m}$ and $\sigma \cdot \alpha = \alpha \circ \sigma$.
Because $\operatorname{id} \cdot \alpha = \left(\alpha_{\operatorname{id}(1)}, \ldots, \alpha_{\operatorname{id}(m)}\right) = \left(\alpha_1,\cdots,\alpha_m\right) = \alpha$, GA1 follows.
For $\pi,\sigma \in \mathrm{S}_{m}$ and $\alpha \in \mathbb{N}^{m}$, we have
$$\pi \cdot (\sigma \cdot \alpha) = (\alpha \circ \sigma) \circ \pi = \alpha \circ (\sigma \circ \pi) = (\sigma \circ \pi) \cdot \alpha$$
It is clear that $\sigma \circ \pi = \pi \circ \sigma$ if and only if $(\mathrm{S}_{m},\circ)$ is Abelian, which is not true in general.
Please help me figure out where I am wrong!
Well, its wrong. The definition must be $$\sigma \cdot x = (x_{\sigma^{-1}(1)}, \ldots, x_{\sigma^{-1}(n)}).$$ Then $$\sigma\tau \cdot x = (x_{(\sigma\tau)^{-1}(1)}, \ldots, x_{(\sigma\tau)^{-1}(n)}) = (x_{\tau^{-1}\sigma^{-1}(1)}, \ldots, x_{\tau^{-1}\sigma^{-1}(n)}) = \tau\cdot (x_{\sigma^{-1}(1)}, \ldots, x_{\sigma^{-1}(n)}) =\sigma\cdot(\tau \cdot x). $$