 # Introduction to Groups and Geometries

## Section2.5Group actions

### Subsection2.5.1

#### Definition2.5.1.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
\begin{equation*} \phi \colon G\to \Perm(X). \end{equation*}
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
\begin{equation*} \Orb(x)=\{gx\colon g\in G\}, \end{equation*}
called the orbit of $$x\text{,}$$ and we write $$\Stab(x)$$ to denote the set
\begin{equation*} \Stab(x) = \{g\in G\colon gx=x\}, \end{equation*}
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.
Find the indicated orbits and stabilizers for each of the following group actions.
1. $$D_4$$ acts on the square $$X=\{(x,y)\in \R^2\colon -1\leq x,y\leq 1\}$$ by rotations 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 $$(1,0)\text{?}$$
2. The circle group $$S^1$$ (see Subsection 2.1.3) acts on the two-sphere $$S^2$$ by rotation about the $$z$$-axis. Given an element $$e^{i\alpha}$$ in $$S^1$$ a point $$(\theta,\phi)$$ in $$S^2$$ (in spherical coordinates), the action is given by
\begin{equation*} e^{i\alpha}\cdot (\theta,\phi)=(\theta,\phi+\alpha). \end{equation*}
What is the orbit of $$(\pi/4,\pi/6)\text{?}$$ What is the orbit of the north pole $$(0,0)\text{?}$$ What is the stabilizer of $$(\pi/4,\pi/6)\text{?}$$ What is the stabilizer of the north pole?
3. Any group $$G$$ acts on itself by conjugation, that is, by $$(\phi(g))(x)=gxg^{-1}=C_g(x)$$ (see Exercise 2.4.2.10). Describe the orbit and stabilizer of a group element $$x\text{.}$$
1. $$\Orb((1,1))=\{(1,1),(1,-1),(-1,1),(-1,-1)\}\text{,}$$ $$\Orb((1,0))=\{(1,0),(-1,0),(0,1),(0,-1)\}\text{,}$$ $$\Stab((1,1))= \{R_0,F_{D'}\}\text{,}$$ $$\Stab((1,0))=\{R_0,F_H\}$$
2. $$\Orb(\pi/4,\pi/6)$$ is the horizontal circle on $$S^2$$ with "latitude" $$\pi/4\text{,}$$ $$\Orb(0,0)=\{(0,0)\}\text{,}$$ $$\Stab(\pi/4,\pi/6)=\{1\}\text{,}$$ $$\Stab{(0,0)}=S^1$$
3. $$\Orb(x)=\{gxg^{-1}\colon g\in G\}\text{,}$$ $$\Stab(x)=C(x)$$ (the centralizer of $$x$$)
Show that the stabilizer of an element $$x$$ in a $$G$$-space $$X$$ is a subgroup of $$G\text{.}$$

#### Instructor’s solution for Checkpoint 2.5.3.

We use the 2-step subgroup test. First, note that $$\Stab(x)$$ is not empty, because $$\Stab(x)$$ contains $$e_G$$ (the action $$G\to \Perm(X)$$ is a group homomorphism, so $$e_G$$ must go to the identity transformation on $$X$$ by Property 1 of Proposition 2.4.5). Now suppose $$g,h\in \Stab(x)\text{.}$$ Then we have
\begin{equation*} (gh)(x)=g(hx)=gx=x \end{equation*}
so $$gh\in \Stab(x)\text{.}$$ Also, we have
\begin{equation*} x=(g^{-1}g)x =g^{-1}(gx)=g^{-1}x\text{,} \end{equation*}
so $$g^{-1}\in \Stab(x)\text{.}$$ This completes the proof.

#### Instructor’s solution for Checkpoint 2.5.5.

1. There are 3 types of orbits for the action of $$D_4$$ on the square.
• the 1-point set $$\{(0,0)\}$$ (a single orbit)
• a 4-point set, the vertices of a square, of the form $$(\pm a,\pm a)$$ or $$\{(\pm a,0),(0\pm a)\}$$ (infinitely many orbits, one for each $$a$$ in the range $$0\lt a\leq 1)$$
• an 8-point set of the form $$\{(\pm a,\pm b),(\pm b,\pm a)\}$$ (infinitely many orbits, one for each $$0\leq a\lt b$$)
2. The orbits are all the latitude circles with latitude $$\theta\text{,}$$ for $$0\lt \theta\lt \pi/2\text{,}$$ and two 1-point orbits, the north and south poles.
3. For the action of a group on itself by conjugation, the orbits are conjugacy classes. How many orbits and how they look depends on the group.

### Exercises2.5.3Exercises

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

Let $$G$$ be a group. Here are three actions $$G\to \Perm(G)$$ of $$G$$ on itself. Left multiplication is given by
\begin{equation*} g\to L_g \end{equation*}
where $$L_g$$ is given by $$L_g(h)=gh\text{.}$$ Right inverse multiplication is given by
\begin{equation*} g\to R_g \end{equation*}
where $$R_g$$ is given by $$R_g(h)=hg^{-1}\text{.}$$ Conjugation is given by
\begin{equation*} g\to C_g \end{equation*}
where $$C_g$$ is given by $$C_g(h)=ghg^{-1}\text{.}$$
1. Show that, for $$g\in G\text{,}$$ the maps $$L_g,R_g,C_g$$ are elements of $$\Perm(G)\text{.}$$
2. Show that each of these maps $$L,R,C$$ is indeed a group action.
3. 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.

#### Instructor’s solution for Exercise 2.5.3.1.

1. We will show that $$L_g$$ is invertible. Similar arguments show that $$R_g,C_g$$ are invertible. We have
\begin{equation*} L_{g^{-1}}(L_g(x))=g^{-1}gx=x \end{equation*}
for all $$x\in G\text{.}$$ A similar argument shows that $$L_g(L_{g^{-1}}(x))=x$$ for all $$x\text{.}$$ We conclude that $$L_g$$ is invertible, with inverse $$(L_g)^{-1}=L_{g^{-1}}\text{.}$$ Thus $$L_g$$ is an element of $$\Perm(G)\text{,}$$ as desired.
2. You just have to check that $$L_{gh}=L_g\circ L_h\text{,}$$ $$R_{gh}=R_g\circ R_h$$ (here you see why the inverse is needed!), and $$C_{gh}=C_g\circ C_h\text{.}$$ Here is the argument for $$L\text{.}$$ Similar arguments work for $$R,C\text{.}$$ Let $$g,h\in G\text{.}$$ We have $$L_g(L_h(x))=g(hx) = (gh)x = L_{gh}(x)$$ for all $$x\in G\text{,}$$ so we have $$L_g\circ L_h = L_{gh}\text{.}$$
3. For injectivity, if $$L_g=L_h\text{,}$$ then $$g=L_g(e)=L_h(e)=h\text{.}$$ By the First Isomorphism Theorem, we conclude that $$G\approx L(G)\text{,}$$ which is subgroup of $$\Perm(G)\text{.}$$

#### 2.Cosets, revisited.

Let $$H$$ be a subgroup of a group $$G\text{,}$$ and consider the map
\begin{equation*} R\colon H\to \Perm(G) \end{equation*}
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.

#### Instructor’s solution for Exercise 2.5.3.2.

An orbit of the group action $$H\to \Perm(G)$$ is a set of the form
\begin{equation*} \Orb(g)=\{gh^{-1}\colon h\in H\} \end{equation*}
for some $$g\in G\text{.}$$ Consider the map $$x\to gx^{-1}g\text{.}$$ For $$x=gh^{-1}\in \Orb(g)\text{,}$$ we have $$i(gh^{-1})=g(gh^{-1})g=gh\text{,}$$ so the image of $$i$$ is contained in the coset $$gH\text{.}$$ The map $$i$$ is its own inverse:
\begin{equation*} i(i(x))= i(gxg^{-1})= g(gxg^{-1})g=x\text{,} \end{equation*}
so we conclude that $$i\colon \Orb(g)\to gH$$ is bijective. Thus we have established that $$\Orb(g)=gH\text{,}$$ as desired.

#### 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
\begin{equation*} g\to [v\to g\cdot v], \end{equation*}
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.

#### Instructor’s solution for Exercise 2.5.3.3.

Let $$\phi$$ denote the map $$\phi\colon G\to \Perm(X)$$ given by
\begin{equation*} g\to [v\to g\cdot v], \end{equation*}
so that $$(\phi(g))(v)=gv$$ for $$g\in G\text{,}$$ $$v\in X\text{.}$$ To see that $$\phi$$ is a homomorphism, we have,
\begin{equation*} (\phi(gh))(v)=(gh)v = g(hv) =(\phi(g)\circ\phi(h))(v) \end{equation*}
for all $$v\in X\text{.}$$ [Note: the equality $$(gh)v=g(hv)$$ uses the fact that matrix multiplication is associative.] We conclude that $$\phi(gh)=\phi(g)\circ \phi(h)\text{,}$$ as required.

#### Instructor’s solution for Exercise 2.5.3.4 (proof of Proposition 2.5.4).

First, we check that $$\sim_G$$ is an equivalence relation. For symmetry, we have $$x=e_Gx$$ for all $$x\in X\text{.}$$ For reflexivity, if $$y=gx\text{,}$$ then $$x=g^{-1}y\text{.}$$ For transitivity, if $$y=gx$$ and $$z=hy\text{,}$$ then $$z=hy=h(gx)=(hg)x\text{.}$$ It follows that the set of equivalence classes $$X/\!\!\sim_G$$ forms a partition of $$X\text{.}$$ Since $$X/G=X/\!\!\sim_G$$ by definition, it follows that orbits form a partition.

#### 5.

Prove The Orbit-Stabilizer Theorem (Theorem 2.5.6).

#### Instructor’s solution for Exercise 2.5.3.5.

Let $$x_0\in X$$ and consider the function $$f_{x_0}\colon G\to X$$ given by $$f_{x_0}(g)= gx_0\text{.}$$ Applying Fact 1.4.5 to $$f_{x_0}\text{,}$$ we have $$G/\!\!\sim_{f_{x_0}}=G/\Stab(x_0)\text{,}$$ and $$f_{x_0}(G)=\Orb(x_0)\text{.}$$
Alternatively, one can check that the given map is well-defined, one-to-one, and onto. Let $$\pi$$ denote the map $$\pi(g\Stab(x))=gx\text{.}$$
• ($$\pi$$ is well-defined) Suppose $$g\Stab(x)=h\Stab(x)\text{.}$$ By Proposition 2.3.9, we have $$g^{-1}h\in \Stab(x)\text{.}$$ Thus we have $$g^{-1}hx=x\text{,}$$ so $$gx=hx\text{.}$$ We conclude that $$\pi$$ is onto.
• ($$\pi$$ is one-to-one) Suppose $$\pi(g\Stab(x))=gx=hx=\pi(h\Stab(x))\text{.}$$ Then $$g^{-1}hx=x\text{,}$$ so $$g^{-1}h\in \Stab(x)\text{,}$$ so $$g\Stab(x)=h\Stab(x)\text{.}$$ We conclude that $$\pi$$ is one-to-one.
• ($$\pi$$ is onto) Given $$gx\in \Orb(x)\text{,}$$ we have $$\pi(g\Stab(x))=gx\text{.}$$ We conclude that $$\pi$$ is onto.

#### 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 $$\Proj(V).$$
1. 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 $$\alpha\in\F^\ast\text{.}$$
2. 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})]\tag{2.5.1} \end{equation}
for $$g\in GL(V)$$ and $${v}\in V\setminus\!\{0\}\text{.}$$
3. Show that the kernel of the map $$GL(V)\to \Perm(\Proj(V))$$ given by (2.5.1) is the subgroup $$K=\{\alpha\Id\colon \alpha\in \F^\ast\}\text{.}$$
4. Conclude that the projective linear group $$PGL(V):=GL(V)/K$$ acts on $$\Proj(V)\text{.}$$
5. Show that $$\F^\ast$$ acts on $$GL(V)$$ by $$\alpha\cdot T=\alpha T\text{,}$$ and that $$PGL(V)\approx GL(V)/\F^\ast\text{.}$$
6. Let $$s\colon S^2 \to \extC$$ denote the stereographic projection (see (1.3.6)). 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.

#### Instructor’s solution for Exercise 2.5.3.6.

1. Apply Fact 1.4.4 (check that the action of $$g$$ is constant on projective equivalence classes of vectors).
2. Let $$\phi\colon GL(V)\to \Perm(\Proj(V))$$ be the action defined by Equation (2.5.1). It is easy to check that $$K\subseteq \ker(\phi)\text{.}$$ Containment in the other direction requires a short linear algebra argument. Suppose $$g\in \ker(\phi)\text{.}$$ This means that for every $$v\in V\text{,}$$ there is a nonzero scalar $$\lambda_v$$ such that $$gv=\lambda_v v$$ (that is, every vector in $$V$$ is an eigenvector of the operator $$g$$). If $$\dim(V)=1\text{,}$$ we are done. If $$\dim(V)\gt 1\text{,}$$ choose independent vectors $$v,w\in V\text{.}$$ We have $$\lambda_{v+w}(v+w)=g(v+w)=\lambda_v v+\lambda_w w\text{.}$$ Because $$v,w$$ are independent, it follows that $$\lambda_v=\lambda_w=\lambda_{v+w}\text{.}$$ Thus it must be that there is a single $$\lambda\in \F^\ast$$ such that $$gv=\lambda v$$ for all $$v\in V\text{,}$$ that is, $$g$$ is an element of $$K\text{.}$$
3. Apply the First Isomorphism Theorem (Corollary 2.4.7) to $$\phi\colon GL(V)\to \Perm(\Proj(V))$$ to obtain
\begin{equation*} GL(V)/K \approx \phi(G) \stackrel{\text{inclusion}}{\longrightarrow} \Perm(\Proj(V)). \end{equation*}
4. Given $$\alpha,\beta$$ in $$\F^\ast$$ and $$T$$ in $$GL(V)\text{,}$$ the fact that $$(\alpha \beta)T=\alpha(\beta T)$$ verifies that the map $$\F^\ast \to \Perm(GL(V))$$ given by $$\alpha\to [T\to \alpha T]$$ is a homomorphism. The equivalence relation that corresponds to this action is given by $$T\sim_{\F^\ast} S$$ if and only if $$S=\alpha T$$ for some $$\alpha\in \F^\ast\text{.}$$ Because $$S=\alpha T$$ if and only if $$S=(\alpha \Id)T\text{,}$$ the equivalence classes that define $$PGL(V)$$ in the previous part of this problem are the same as the equivalence classes $$GL(V)/\F^\ast\text{.}$$
5. Factor the given map as $$\Proj(\C^2)\to \extC\to S^2$$ given by
\begin{equation*} [(\alpha,\beta)] \to \alpha/\beta \to s^{-1}(\alpha/\beta). \end{equation*}
Let $$\mu$$ denote the map $$[(\alpha,\beta)] \to \alpha/\beta \text{.}$$ That $$\mu$$ is well-defined follows from the observation that $$(k\alpha)/(k\beta)=\alpha/\beta$$ for all $$k\in \F^\ast\text{.}$$ To see that $$\mu$$ is bijective, we will exhibit an inverse map. Let $$\nu\colon \extC\to \Proj(\C^2)$$ be given by $$\gamma \to [(\gamma,1)]$$ for $$\gamma \in \C$$ and $$\infty \to [(1,0)]\text{.}$$ To check that $$\nu\circ\mu=\Id\text{,}$$ for $$\beta\neq 0\text{,}$$ we have
\begin{equation*} (\nu\circ \mu)([(\alpha,\beta)])=\nu(\alpha/\beta) = [(\alpha/\beta,1)]=[(\alpha,\beta)] \end{equation*}
and for $$\beta=0\text{,}$$ we have
\begin{equation*} (\nu\circ \mu)([(\alpha,0)])=\nu(\infty) = [(1,0)]=[(\alpha,0)]. \end{equation*}
Reading the two displayed equations from right to left establishes that $$\mu\circ \nu=\Id\text{.}$$ We conclude that $$\nu=\mu^{-1}\text{.}$$ Using the fact that stereographic projection $$s$$ is bijective, we have $$s^{-1}\circ \mu$$ is bijective, as desired.
Other notations for $$(\phi(g))(x)$$ are $$g(x)\text{,}$$ $$g\cdot x\text{,}$$ and $$g.x\text{.}$$
It must be proved that $$\Stab(x)$$ is indeed a subgroup of $$G\text{.}$$ See Checkpoint 2.5.3 below.