# Do all groups implicitly act on a set?

When recently proving that Inn$(G)$ $\le$ Aut$(G)$ we needed to choose two elements $C_x ,C_y \in$ Aut$(G)$ and show the composition, that is $C_y \circ C_x = C_{xy}$, is in Inn$(G)$ too. To do this we have to appeal to something slightly new, and slightly uncomfortable for me: we start by examining $C_y \circ C_x \;(g)$ . I am immediately struck that we are looking to see how this composition acts on an arbitrary $g \in G$ . Eventually we discover $C_y \circ C_x \;(g) \; = C_{xy}(g)$ .

Questions:

We took two elements in Inn$(G)$ that make no sense without a set $G$ to "act" on. For any group $K$, does $K$ always need a set to "act" on? If so, for instance, what set am I shuffling around when I play with $\mathbb{Z_6}$? Is this what is meant by group actions?

• Yes and no. No, a group doesn't "need" a set to act on. Yes, each group can easily be given a set to act on, for example it acts on itself by Cayley's theorem. – Dan Shved Nov 17 '17 at 7:43
• A group acts on itself (in various ways). – Lord Shark the Unknown Nov 17 '17 at 7:43
• Cayley's theorem says that each group is embedded in a symmetric group and I think thus each group has a set to act on. – josephz Nov 17 '17 at 7:45
• Do these comments agree with @Soarer 's answer , math.stackexchange.com/questions/3547/… ? My intuition also sprang up when I recognized permutations as functions needing a set to "act" on. This also motivated the question. – user4396386 Nov 17 '17 at 9:29
• Permutations always act on a set and every group is isomorphic to a permutation group by Cayley's theorem (this is equivalent to one of the actions in my answer). In that sense, every group implicitly acts on a set, but you can define a group without talking about group actions. – Qudit Nov 17 '17 at 21:22

Groups can be defined independently of any group action by the standard group axioms.

That being said, one way of defining a group action is as a homomorphism $\alpha : G \rightarrow \text{Sym}(X)$ where $X$ is some set. The group $G$ then acts on $X$ by $\alpha(g)(x)$ for each $g \in G$ and $x \in X$.

Every group $G$ automatically acts on itself ($X = G$) by the inner automorphisms (as you noted). Formally, this corresponds to the homomorphism $\alpha : G \rightarrow \text{Sym}(G)$ where $\alpha(g) = C_g$. Another group action is simply the group operation. Formally, this is the homomorphism $\beta : G \rightarrow \text{Sym}(G)$ where $\beta(g)(x) = g x$ for each $g, x \in G$.

Super deep theorem. Given any group $G$ and any set $X$, there exists an action of $G$ on $X$.
Proof: Take the trivial action, $gx=x$ for all $g\in G$ and $x\in X$.