3.12 Introduction to group actions
Suppose \(f:G\to M\) is a monoid homorphism from \(G\) to a monoid \(M\). Then \(f(G)\subseteq M^{\times }\) and the resulting map \(G\to M^{\times }\) is a group homomorphism.
Write \(1\) for the identity element of \(M\). Then, for \(g\in G\), we have \(f(g)f(g^{-1}) = f(gg^{-1}) = f(1) = 1\), and, similarly, \(f(g^{-1})f(g) = 1\). So \(f(g)\) is invertible with inverse \(f(g^{-1})\). Therefore \(f(G)\subseteq M^{\times }\).
In this section \(G\) is a group with identity element \(1\) and \(X\) is a set. We write \(A(X)\) for the group of permuations on \(X\).
An action of \(G\) on \(X\) is a homomorphism \(\rho :G\to A(X)\).
I think the prevailing belief is that group actions are the whole point of groups. In other words, what groups really do is act on things.
We could also define a monoid action in a similar way. So let me do it. Here \(E(X)\) denotes the endomorphism monoid of \(X\), that is the monoid of all maps \(f:X\to X\).
The action of a monoid \(M\) on \(X\) is a monoid homomorphism \(M\to E(X)\).
I’ll mostly talk about group actions, not monoid actions, but it’s interesting to have the definition of a monoid action out there to compare how group actions and monoid actions differ.
Often, when we have a fixed group action \(\rho :G\to A(X)\) in mind, we just say “\(G\) acts on \(X\)." If we want to be precise about what \(\rho \) is, we could say “\(G\) acts on \(X\) via \(\rho \)." Here’s a (somewhat trivial) example.
Suppose \(n\in \mathbb {P}\). The symmetric group \(S_n\) is, by definition, \(A([n])\) where \([n] = \{ 1,\ldots , n\} \). So \(S_n\) acts on \([n]\).
If \(G\) acts on \(X\) via \(\rho \) and \(H\leq G\), then \(H\) also acts on \(X\) via \(\rho \circ i_H\), where \(i_H:H\to G\) is the inclusion homomorphism. So any subgroup \(H\leq S_n\) acts on \([n]\).
There’s another way to think about group actions that is important.
Suppose \(\rho :G\to A(X)\) is a group action. Define a map \(a:G\times X\to X\) by setting \(a(g,x) = \rho (g)(x)\). To save space, we write \(a:G\times X\to X\) as an infix binary operation. In other words, we write \(gx := a(g,x)\) unless we need to be absolutely precise about what \(a\) is. Then \(a\) satisfies the following properties.
For all \(x\in X\), \(1x = x\).
For all \(g,h\in G\) and \(x\in X\), \(g(hx) = (gh)x\).
Conversely, \(a:G\times X\to X\) is a map satisfying 1 and 2. For every \(g\in G\), define a map \(\rho (g):X\to X\) by setting \(\rho (g)(x) = gx\). Then,
For all \(g\in G\), \(\rho (g)\in A(X)\).
The map \(\rho :G\to A(X)\) is an action of \(G\) on \(X\).
(\(\Rightarrow \)) Since \(\rho :G\to A(X)\) is a group homomorphism, \(\rho (1) = \operatorname{\mathrm{id}}_X\). So \(1x = \rho (1) x = \operatorname{\mathrm{id}}_X (x) = x\) That proves 1. To prove 2, suppose \(g,h\in G\) and \(x\in X\). Then \(g(hx) = \rho (g)(\rho (h)x) = (\rho (g)\circ \rho (h))(x) = \rho (gh)(x) = (gh)x\).
(\(\Leftarrow \)) Suppose \(a\) satisfies 1 and 2, and let \(g,h\in G\). Then \(\rho (g)\in E(X)\) for all \(X\).
Suppose \(g,h\in G\), and pick \(x\in X\). Then \(\rho (gh)(x) = (gh)(x) = g(hx) = \rho (g)(hx) = \rho (g)(\rho (h) x) = (\rho (g)\circ \rho (h))(x)\). So \(\rho (gh) =\rho (g)\rho (h)\). This shows that \(\rho :G\to E(X)\) is a magma homomorphism. And, for every \(x\in X\), \(\rho (1)x = 1x = x\). So \(\rho (1) = \operatorname{\mathrm{id}}_X\), and, therefore, \(\rho :G\to E(X)\) is a monoid homomorphism. Therefore, by Lemma 3.78 \(\rho (G)\subset A(X) = E(X)^{\times }\) and the map \(G\to A(X)\) given by \(g\mapsto \rho (g)\) is a group homomorphism.
The map \(a:G\times X\to X\) is sometimes called the action map. Because of Theorem 3.82 giving an action map satifying 1 and 2 is equivalent to giving a group homomorphism \(\rho :G\to A(X)\). And we basically think about these two pieces of data as being the same thing. Here’s an example where we really think about the action map more than the homomorphism \(\rho :G\to A(X)\).
Let \(G = \operatorname{\mathbf{GL}}_2(\mathbb {R})\) and let \(X = \mathbb {R}^2\). Then \(G\) acts on \(X\) via the map \(a(T,v) = Tv\), where \(T\in \operatorname{\mathbf{GL}}_2\) and \(v\in \mathbb {R}^2\). This action is called the natural action of \(G\) on \(\mathbb {R}^2\). (It works the same for \(\operatorname{\mathbf{GL}}_n(\mathbb {R})\) and \(\mathbb {R}^n\) with \(n\) any nonnegative integer.)
Suppose \(G\) acts on \(X\) and \(x\in X\). The orbit of \(x\) is the set \(\operatorname{\mathrm{Orb}}_G(x) = \{ gx:g\in G\} \). The stabilizer of \(x\) is the set \(\operatorname{\mathrm{Stab}}_G(x) = \{ g\in G: gx = x\} \).
The set of orbits of \(G\) is \(G\backslash X := \{ \operatorname{\mathrm{Orb}}_G(x): x\in X\} \).
So the orbit of \(x\) is a subset of \(X\) and the stabilizer of \(x\) is a subset of \(G\). Sometimes I write \(Gx\) for the orbit of \(x\). Confusingly, some people write \(G_x\) for the stabilizer of \(x\). (I won’t do that.) People usually say “\(G\)-orbit" instead of “orbit of \(G\) on \(X\)" if they want to be precise about what the group \(G\) is.
Suppose \(G\) acts on \(X\) and \(x\in X\). Then \(\operatorname{\mathrm{Stab}}_G x \leq G\).
We have \(1x = x\). So \(1\in \operatorname{\mathrm{Stab}}_G x\).
Suppose \(g,h\in \operatorname{\mathrm{Stab}}_G x\). Then \(gh^{-1}x = gh^{-1}x = gh^{-1}hx = g1x = gx = x\). So, by the one-step subgroup test, \(\operatorname{\mathrm{Stab}}_G x\leq G\).
Suppose \(G\) acts on \(X\). The set \(G\backslash X\) of \(G\)-orbits is a partition of \(X\).
Clearly \(x = 1x\in \operatorname{\mathrm{Orb}}_G x\). So the union of the \(G\)-orbits is \(X\). I claim that any two \(G\)-orbits, \(\operatorname{\mathrm{Orb}}_G(x)\) and \(\operatorname{\mathrm{Orb}}_G(y)\), are either equal or disjoint.
To see this, suppose \(z\in \operatorname{\mathrm{Orb}}_G(x)\cap \operatorname{\mathrm{Orb}}_G(y)\). Then we can find group elements \(g_1,g_2\in G\) such that \(z = g_1x = g_2 y\). So \(x = g_1^{-1}z = g_1^{-1} g_2 y\).
Pick \(w = g_3x\in \operatorname{\mathrm{Orb}}_G(x)\). Then \(w = g_3 x = g_3g_1^{-1}g_2 y \in Orb_G(y)\). This shows that \(\operatorname{\mathrm{Orb}}_G(x)\subseteq \operatorname{\mathrm{Orb}}_G(y)\). But the same argument with the rolls of \(x\) and \(y\) switched shows that \(\operatorname{\mathrm{Orb}}_G(y)\subseteq \operatorname{\mathrm{Orb}}_G(x)\). So \(\operatorname{\mathrm{Orb}}_G(x) = \operatorname{\mathrm{Orb}}_G(y)\).