# Intuition behind group action on a set

In Algebra Chapter 0 the definition of a group action on a set is given as: An action of a group $G$ on a set $A$ is a set function $P:G\times A\rightarrow{A}$ such that $P(e_G,a)=a$ and $P(gh,a)=P(g,P(h,a))$ $\forall g,h \in G\,\forall a\in A$ the first condition makes sense but I am unsure of the purpose of the second condition, it seems rather strange. My thinking is it somehow preserves some aspect of $G$ but I am unsure. If someone could explain the reasoning or intuition behind the 2nd requirement that would be great.

• It's not clear what your second condition says. Did you mean $P(g,P(h,a))=\text{something}$? ${}\qquad{}$ – Michael Hardy Feb 20 '15 at 2:22
• I imagine it should be $P(g, P(h, a)) = P(gh, a)$, corresponding to the usual definition that $(gh)(a) = g(h(a))$. – pjs36 Feb 20 '15 at 2:25
• Yes sorry for the confusion, I just edited it. – Christian LaPointe Feb 20 '15 at 2:25

I assume your second condition is meant to say $P(g,P(h,a))=P(gh,a)$. The purpose of this is to say that the group action is "associative" in some sense. It is far easier to see if we write $g\cdot a$ for $P(g,a)$. Then the condtions become $e_G\cdot a=a$ and $g\cdot(h\cdot a)=(gh)\cdot a$ for all $g,h\in G$ and $a\in A$. Hopefully this makes the intuition clearer.

• Yes! Thank you, the change of the notation makes it abundantly clear now. – Christian LaPointe Feb 20 '15 at 2:37

My philosophy is that groups are to group actions as potential energy is to kinetic energy. You might recall reading that originally, historically groups were first conceived of as sets of functions that acted on some thingie (points in space, numbers in a number system, whatever). The more pointed term for this is that a group is a set of symmetries. As such, whenever an object exhibits symmetry, in other words if there are transformations that preserve it, then these transformations can be composed and inverted. That's where the idea of a group comes from. This idea was then abstracted into just a set with an associative binary operation having inverses and identity.

Group actions recover the "function" nature of group elements; they get to go hunting out in the world and do things to other things. They act as functions on some set, or symmetries of some object. As such, we should be able to compose and invert with them. (You can't have both inversion and composition unless you also have a 'do nothing' transformation - the identity.)

There are two standard definitions of group actions. One is as a homomorphism $G\to{\rm Aut}(X)$, where in the category of sets ${\rm Aut}(X)$ just means the permutations of $X$. This encodes the fact that each element of $G$ acts as a permutation of $X$, and that the operation in $G$ corresponds to composition of functions in ${\rm Aut}(X)$. One might suspect we could just say $G\subseteq{\rm Aut}(X)$ is a subgroup of permutations, but this doesn't capture the idea fully: group actions needn't be faithful and so different group elements can behave as the same function on $X$ (i.e. $G\to{\rm Aut}(X)$ is not an injective homomorphism).

The other definition of a group action is as a map $G\times X\to X$ satisfying a particular set of properties. The map $G\times X\to X$ should be thought of as the "evaluation" map wherein $(g,x)$ is sent to $g(x)$ (here we treat $g\in G$ as a function of $X$). And so we want our functions to satisfy "associativity" viz. $f(g(x))=(f\circ g)(x)$; this is what the condition means.

...do you know what currying is in computer science?

Currying. For any function $$f$$ of the form $$A \times B \rightarrow C$$, there is an equivalent function that returns a function $$f'$$ of the form $$A \rightarrow (B \rightarrow C)$$. The reverse is also true: for any function $$g$$ of the form $$D \rightarrow (E \rightarrow F)$$, there is a function $$g'$$ of the form $$D \times E \rightarrow F$$.

For a group $$G$$, and the set of bijections of a set $$X$$ written as $$\mathrm{Sym}(X)$$, a homomorphism $$\varphi : G \rightarrow \mathrm{Sym}(X)$$ is really a function of the form $$\varphi : G \rightarrow (X \rightarrow X)$$. Because $$\varphi$$ returns a function, $$\varphi(g)$$ is not a complete expression! You need to write $$\varphi(g)(x)$$ to get a complete expression. Because mathematicians don't really like thinking in terms of a sequence of functions, they decided to decurry the homomorphism of a group action into an operator.

A group action is an operator $$\diamondsuit : G \times X \rightarrow X$$ that acts like a homomorphism.

But what axioms do we need to specify "acts like a homomorphism"?

If $$\diamondsuit$$ is going to act like a homomorphism, then for any element $$g \in G$$, we better make sure $$x \mapsto g \diamondsuit x$$ acts like a homomorphism! For convenience sake, we're going to call the map $$x \mapsto g \diamondsuit x$$ just $$f_g$$... to signify that it's the map generated by the element $$g$$. A different $$g'$$ is going to produce a different map $$f_{g'}$$ which would look like $$x \mapsto g'\diamondsuit x$$.

First obvious property: a homomorphism better take the identity element to the identity function. So the map $$(x \mapsto e \diamondsuit x) = x$$. So we have our first axiom.

A group action is an operator $$\diamondsuit : G \times X \rightarrow X$$ such that

1. $$e \diamondsuit x = x$$

2. ...and something else.

What's the second homomorphism property? That the group product transfers to the composition of function. So $$f_g \circ f_h = f_{gh}$$ Because we previously defined our notation such that $$f_g = g \diamondsuit x$$, $$f_h = h \diamondsuit x$$, $$f_{gh} = gh \diamondsuit x$$...

A group action is an operator $$\diamondsuit : G \times X \rightarrow X$$ such that

1. $$e \diamondsuit x = x$$

2. $$g \diamondsuit (h \diamondsuit x) = gh \diamondsuit x$$

Tadaaah! That's why we define a group action the way we do. The end.