# Group action definition

The Wikipedia for group action and several posts here have what seems to me to be two, slightly different definitions for a group action.$$\$$ Let G be a group acting on space X.$$\$$ These two definitions are:

1. A group homomorphism $$\varphi$$ from the group to the automorphism group of the space. $$\varphi : G\ \longrightarrow \text{Aut}(X)$$

2. A map that takes an element of the group and an element of the space, and returns an element of the space. $$\alpha : G\, \times X\ \longrightarrow X$$

Which of these is the the group action?$$\$$ Is it a function of one argument or two?$$\$$ What is the return type? Does it return a function (automorphism) or an element of the space?$$\$$ The first definition is in line with what I think of as a representation, ex. from a symmetry group C$$_\text{2v} \longrightarrow$$ GL(V) that returns a matrix, rather than a vector.

I (think) I understand the basic idea of what is going on, but I'd like to know what the actual mathematical definition is.

• They are equivalent. Which one is "the" definition is a matter of preference. Commented Feb 4, 2021 at 3:34
• The short answer is that they are both valid definitions. For a fixed $g\in G$, you can consider the action of $g$ on $X$ as a permutation of $X$. The first definition tells you what the permutation is. The second tells you where $g$ sends a particular $x\in X$. Please read this lovely piece by Tim Gowers on the relation between these two definitions. Commented Feb 4, 2021 at 3:35
• Definition 2 is missing some parts, namely $\alpha(1,x)= x$ and $\alpha(g,\alpha(h,x))=\alpha(gh,x)$ for all $g,h,x$. Without them you would have allowed e.g., $\alpha(G\times X)=\{x\}\subsetneq X$. Commented Feb 4, 2021 at 3:44
• Related Commented Feb 4, 2021 at 3:54
• @DerekHolt The only "plus" I can think of about the second definition is that it mimics for abstract groups the basic properties of the "prototypical" action, namely the natural action of $Sym(X)$ on $X$.
– user870827
Commented Feb 4, 2021 at 9:02

The definitions end up being more or less the same in the sense that if you have a map $$f: G \times X \to X$$ satisfying the second definiton, there exists a group homomorphism (induced by $$f$$) satisfying the first definition (and vice versa).

To show you how this might work, suppose you have $$f: G \times X \to X$$. For each $$g \in G$$, you can define a map $$\phi_g: X \to X$$ by letting $$\phi_g(x) = f(g,x)$$. This will be a bijection since $$\phi_g^{-1}$$ is precisely $$\phi_{g^{-1}}$$. Then, the homomorphism $$G \to \text{Aut}(X)$$ is the one sending $$g \mapsto \phi_g$$. If instead you start with a homomorphism $$f: G \to \text{Aut}(X)$$ and let $$f_g := f(g)$$, we can define a map $$\alpha: G \times X \to X$$ by letting $$\alpha(g,x) = f_g(x)$$. Hence, there is a one-to-one correspondence between maps $$G \times X \to X$$ (which satisfy other conditions as mentioned in the comments) and homomorphisms $$G \to \text{Aut}(X)$$

Though, in my experience, the 2nd definition you mentioned is a more conventional starting place for defining group actions. If you look in many classic Algebra books like the one by Dummit and Foote, they will define a group action as a map $$G \times X \to X$$ satisfying various conditions. However, as you learn more about Abstract Algebra, the difference between these two notions becomes miniscule, and you will start to use both notions of a group action interchangeably.

There's a more general setting for this equivalence. Let $$G$$ be a group and $$X$$ a set; consider:

D1) There is a map: $$\cdot:G\times X\to X$$, $$(g,x)\mapsto g\cdot x$$, such that:

• $$e\cdot x=x$$, for every $$x\in X$$;
• $$g\cdot(h\cdot x)=(gh)\cdot x$$, for every $$g,h\in G, x\in X$$.

D2) There is a group homomorphism $$\varphi\colon G\to\operatorname{Sym}(X)$$.

Claim. D1) $$\iff$$ D2).

Proof.

• D1) $$\Longrightarrow$$ D2). Let's define $$\varphi(g)(x):=g\cdot x$$; then (injectivity):

$$\varphi(g)(x)=\varphi(g)(y)\Rightarrow g\cdot x=g\cdot y\Rightarrow g^{-1}\cdot(g\cdot x)=g^{-1}(g\cdot y)\stackrel{\text{(D1)}}{\Longrightarrow}x=y$$

and (surjectivity), for every $$y\in X$$, $$y\stackrel{\text{(D1)}}{=}\varphi(g)(g^{-1}\cdot y)$$. Therefore, indeed $$\varphi(g)\in\operatorname{Sym}(X)$$ for every $$g\in G$$. Now, for every $$g,h\in G,x\in X$$:

\begin{alignat}{1} \varphi(gh)(x) &= (gh)\cdot x \\ &\stackrel{\text{(D1)}}{=} g\cdot(h\cdot x) \\ &= \varphi(g)(\varphi(h)(x)) \\ &= (\varphi(g)\varphi(h))(x) \\ \end{alignat}

whence $$\varphi(gh)=\varphi(g)\varphi(h)$$.

• D2) $$\Longrightarrow$$ D1). Let's define $$g\cdot x:=\varphi(g)(x)$$; then, for every $$x\in X$$:

$$e\cdot x=\varphi(e)(x)\stackrel{\text{(D2)}}{=}Id_X(x)=x$$

Moreover, for every $$g,h\in G,x\in X$$:

$$(gh)\cdot x =\varphi(gh)(x)\stackrel{\text{(D2)}}{=}(\varphi(g)\varphi(h))(x)=\varphi(g)(\varphi(h)(x))=g\cdot(h\cdot x)$$

$$\Box$$

Therefore, feel free to choose any between D1 and D2 as definition of group action.

What you are seeing here is a special case of the very general concept of currying: For three sets $$A, B, C$$ there is a natural bijection between maps $$A\times B\to C$$ and maps $$A\to \operatorname{Maps}(B,C)$$, where $$\operatorname{Maps}(B,C)$$ denotes the set of maps $$B\to C$$.

The bijection is rather trivial and works as follows: Given a map $$f\colon A\times B\to C$$ we can define for each $$a\in A$$ the map $$f_a\colon B\to C$$ given by $$f_a(b)=f(a,b)$$. This defines a map $$A\to\operatorname{Maps}(B,C)$$ by $$a\mapsto f_a$$. Conversely, from a map $$g\colon A\to\operatorname{Maps}(B,C)$$ you can always define a map $$A\times B\to C$$ by $$(a,b)\mapsto \big(g(a)\big)(b)$$.

These two constructions are inverse to each other and hence establish a bijection between maps $$A\times B\to C$$ and maps $$A\to\operatorname{Maps}(B,C)$$.

In your case $$A=G$$ and $$B=C=X$$ so that maps $$G\times X\to X$$ correspond to maps $$G\to\operatorname{Maps}(X,X)$$. However, you are only interested in the subset of those maps that actually map into $$\operatorname{Sym}(X)$$ (which is the subset of $$\operatorname{Maps}(X,X)$$ consisting of only the bijections) and have the property of being a group homomorphism. In the setting of maps $$G\times X\to X, (g,x)\mapsto g\cdot x$$, these criteria translate to the group action axioms $$1\cdot x=x$$ and $$(gh)\cdot x=g\cdot(h\cdot x)$$ for all $$g,h\in G$$ and $$x\in X$$.