How to think of group actions?

Question

I am a little confused on how exactly I should be thinking of an action on a group. I have been trying to read up on it and came across Timothy Gower's blog which I think does a good job explaining it. However, as I started my first homework problem it said prove that the additive group Z acts on itself by z$\cdot$a = z + a for all $a, z \in Z$. My problem comes with the fact Gower's says you can think of an action as a permutation of the set, however for z > 2 you can no longer replicate the set with the action because your set (the image of your action) will start at 1 + z, thus not including 1,2,...,z so how can it be a permutation? I would appreciate help clearing this up. Thanks.

Answer 1

Recall that a group $G$ is said to act on a set $X$ if each $g \in G$ can be thought of as a bijection $X \to X$ (and we need that the identity $e \in G$ is the trivial bijection and that these maps satisfy associativity $(gh)(x) = g(h(x))$ for $x \in X$). Any bijection of a set can be thought of as a rearrangement or a permutation of the set $X$, so what this is saying is that each element $g$ of the group $G$ gives us a way to permute all of the elements of $X$.

Let's now consider your example. We are saying that the integers $\mathbb{Z}$ act on themselves as follows: given an element $a \in \mathbb{Z}$, we get the map $x \mapsto x + a$. That is, the group element $a \in \mathbb{Z}$ ''permutes'' the integers by shifting everything by $a$. Said differently, $1$ gets sent to $1+a$, $2$ gets sent to $2+a$, $3$ gets sent to $3+a$, and so on. But every integer gets hit and it only gets hit by one element, so ''adding $a$'' is indeed a bijection of $\mathbb{Z}$.