## Section 2.5 Group actions

### Subsection 2.5.1

###### Definition 2.5.1. Group action, orbit, stabilizer.

Let \(G\) be a group and let \(X\) be a set. An action of the group \(G\) on the set \(X\) is a group homomorphism

We say that the group \(G\) acts on the set \(X\text{,}\) and we call \(X\) a \(G\)-space. For \(g\in G\) and \(x\in X\text{,}\) we write \(gx\) to denote \((\phi(g))(x)\text{.}\)^{ 1 } We write \(\Orb(x)\) to denote the set

called the orbit of \(x\text{,}\) and we write \(\Stab(x)\) to denote the set

called the stabilizer or isotropy subgroup^{ 2 } of \(x\text{.}\) A group action is transitive if there is only one orbit. A group action is faithful if the map \(G\to \Perm(X)\) has a trivial kernel.

###### Checkpoint 2.5.2.

Find the indicated orbits and stabilizers for each of the following group actions.

- \(D_4\) acts on the square \(X=\{(x,y)\in \R^2\colon -1\leq x,y\leq 1\}\) by rotations and and reflections. What is the orbit of \((1,1)\text{?}\) What is the orbit of \((1,0)\text{?}\) What is the stabilizer of \((1,1)\text{?}\) What is the stabilizer of \((1,0)\text{?}\)
- \(\Z\) acts on \(\R\) by
*translation*, that is, by \((\phi(n))(x)= x+n\text{.}\) What is the orbit of \(1\text{?}\) What is the orbit of \(\pi\text{?}\) What is the stabilizer of \(1\text{?}\) What is the stabilizer of \(\pi\text{?}\) - Any group \(G\) acts on itself by
*conjugation*, that is, by \((\phi(g))(x)=gxg^{-1}=C_g(x)\) (see Exercise 2.4.2.9). Describe the orbit and stabilizer of a group element \(x\text{.}\)

- \(\Orb((1,1))=\{(1,1),(1,-1),(-1,1),(-1,-1)\}\text{,}\) \(\Orb((1,0))=\{(1,0),(-1,0),(0,1),(0,-1)\}\text{,}\) \(\Stab((1,1))= \{R_0,F_{D'}\}\text{,}\) \(\Stab((1,0))=\{R_0,F_H\}\)
- \(\Orb(1)=\Z\text{,}\) \(\Orb(\pi)=\{\pi+n\colon n\in \Z\}\text{,}\) \(\Stab(1)=\{0\}=\Stab(\pi)\)
- \(\Orb(x)=\{gxg^-1\colon g\in G\}\text{,}\) \(\Stab(x)=C(x)\) (the centralizer of \(x\))

###### Checkpoint 2.5.3.

Show that the stabilizer of an element \(x\) in a \(G\)-space \(X\) is a subgroup of \(G\text{.}\)

###### Proposition 2.5.4. Orbits of a group action form a partition.

Let group \(G\) act on set \(X\text{.}\) The collection of orbits is a partition of \(X\text{.}\) The corresponding equivalence relation \(\sim_G\) on \(X\) is given by \(x\sim_G y\) if and only if \(y=gx\) for some \(g\in G\text{.}\) We write \(X/G\) to denote the set of orbits, which is the same as the set \(X/\!\!\sim_G\) of equivalence classes.

###### Checkpoint 2.5.5.

Describe \(X/G\) for each of the three group actions in Checkpoint 2.5.2.

### Subsection 2.5.2

###### Theorem 2.5.6. The Orbit-Stabilizer Theorem.

Let \(G\) be a group acting on a set \(X\text{,}\) and let \(x\) be an element of \(X\text{.}\) There is a one-to-one correspondence

given by

###### Proof.

### Exercises 2.5.3 Exercises

###### 1. Actions of a group on itself.

Let \(G\) be a group. Here are two actions \(G\to \Perm(G)\) of \(G\) on itself. Left multiplication is given by

where \(L_g\) is given by \(L_g(h)=gh\text{.}\) Right inverse multiplication is given by

where \(R_g\) is given by \(R_g(h)=hg^{-1}\text{.}\) Conjugation is given by

where \(C_g\) is given by \(C_g(h)=ghg^{-1}\text{.}\)

- Show that, for \(g\in G\text{,}\) the maps \(L_g,R_g,C_g\) are elements of \(\Perm(G)\text{.}\)
- Show that each of these maps \(L,R,C\) is indeed a group action.
- Show that the map \(L\) is injective, so that \(G\approx L(G)\text{.}\)

*Consequence of this exercise:* Every group is isomorphic to a subgroup of a permutation group.

###### 2. Cosets, revisited.

Let \(H\) be a subgroup of a group \(G\text{,}\) and consider the map

given by \(h\to R_h\text{,}\) where \(R_h(g)=gh^{-1}\) (this is the restriction of right inverse multiplication action in Exercise 2.5.3.1 to \(H\)). Show that the orbits of this action of \(H\) on \(G\) are the same as the cosets of \(H\text{.}\) This shows that the two potentially different meanings of \(G/H\) (one is the set of cosets, the other is the set of orbits of the action of \(H\) on \(G\) via \(R\)), are in fact in agreement.

###### 3. The natural action of a matrix group on a vector space.

Let \(G\) be a group whose elements are \(n\times n\) matrices with entries in a field \(\F\) and with the group operation of matrix multiplication. The natural action \(G\to \Perm(X)\) of \(G\) on the vector space \(X=\F^n\) is given by

where the "dot" in the expression \(g\cdot v\) is ordinary multiplication of a matrix times a column vector. Show that this is indeed a group action.

###### 4.

Prove Proposition 2.5.4.

###### 5.

Prove Theorem 2.5.6.

###### 6. The projective linear group action on projective space.

Let \(V\) be a vector space over a field \(\F\) (in this course, the base field \(\F\) is either the real numbers \(\R\) or the complex numbers \(\C\)). The group \(\F^\ast\) of nonzero elements in \(\F\) acts on the set \(V\setminus \!\{0\}\) of nonzero elements in \(V\) by scalar multiplication, that is, by the map \(\alpha \to [v\to \alpha v]\text{.}\) The set of orbits \((V\setminus\!\{0\})/\F^\ast\) is called the *projectivization* of \(V\text{,}\) or simply projective space, and is denoted \(\Proj(V).\)

- Let \(\sim_{\text{proj}}\) denote the equivalence relation that defines the orbits \((V\setminus \!\{0\})/\F^\ast\text{.}\) Verify that \(\sim_{\text{proj}}\) is given by \(x\sim_{\text{proj}} y\) if and only if \(x=\alpha y\) for some \(\alpha\in\F^\ast\text{.}\)
- Verify that the group \(GL(V)\) (the group of invertible linear transformations of \(V\)) acts on \(\Proj(V)\) by\begin{equation} g\cdot [{v}] = [g({v})]\label{glnprojaction}\tag{2.5.1} \end{equation}for \(g\in GL(V)\) and \({v}\in V\setminus\!\{0\}\text{.}\)
- Show that the kernel of the map \(GL(V)\to \Perm(\Proj(V))\) given by (2.5.1) is the subgroup \(K=\{\alpha\Id\colon \alpha\in \F^\ast\}\text{.}\)
- Conclude that the projective linear group \(PGL(V):=GL(V)/K\) acts on \(\Proj(V)\text{.}\)
- Show that \(\F^\ast\) acts on \(GL(V)\) by \(\alpha\cdot T=\alpha T\text{,}\) and that \(PGL(V)\approx GL(V)/\F^\ast\text{.}\)
- Show that the map \(\Proj(\C^2)\to S^2\) given by \([(\alpha,\beta)]\to s^{-1}(\alpha/\beta)\) if \(\beta\neq 0\) and given by \([(\alpha,\beta)]\to (0,0,1)\) if \(\beta=0\) is well-defined and is a bijection.