# Group Actions

Let $G$ be a group, $X$ a set. By a (left) action of $G$ on $X$, we mean a map $\mu : G \times X \longrightarrow X$ that satisfies certain axioms, to be stated momentarily. We think of $\mu$ as a sort of "multiplication'' even though $X$ is just a set with no algebraic structure. Thus we may write $\mu (g, x) = g \cdot x$ or $g x$. Here are the axioms:

• If $x \in X$ then the unit element $1 \in G$ acts trivially; that is, $1 \cdot x = x$.
• If $x \in X$ and $g, h \in G$, then $(g h) \cdot x = g \cdot (h \cdot x)$.
Intuitively we think of the group as moving things around inside of $X$, and our language will reflect this intuition; we will say that $G$ acts on $X$, when what we mean is that we are given a group action.

Remark 1.5.1: Sometimes the notation is bad for one reason or another; then we will choose a similar but different notation, such as $g \star x$ for $\mu (g, x)$.

Remark 1.5.2: There is a certain metaphor in talking or thinking about group actions: we think of the group moving the points (elements) of $X$ around. In short, it "acts'' on $X$.

There are also right actions. A right action of $G$ on $X$ is a map $\rho : X \times G \longrightarrow X$ that satisfies (denoting $\rho (x, g)$ by $x \cdot g$)

• If $x \in X$ then the unit element $1 \in G$ acts trivially; that is, $x \cdot 1 = x$.
• If $x \in X$ and $g, h \in G$, then $x \cdot (g h) = (x \cdot g) \cdot h$.
The notions of left and right actions are actually equivalent:

Exercise 1.5.1: Let there be given a right action of $G$ on $X$. Define $\mu (g, x) = \rho (x, g^{- 1})$, that is, $g \cdot x = x \cdot g^{- 1}$. Show that this is a left action.

Exercise 1.5.2: Suppose that $X$ is a set with left and right actions that are compatible in the sense that if $g, h \in G$ and $x \in X$, then $g \cdot (x \cdot h) = (g \cdot x) \cdot h$. Show that we can define a left action of $G \times H$ on $X$ by $(g, h) \cdot x = g \cdot x \cdot h^{- 1}$.

If we speak of a group action with no further explanation, we will always mean a left action. For brevity, we will sometimes call a set with a left $G$-action a $G$-set. Exercise 1.5.1 explains why we usually do not have to consider right $G$-actions. However, sometimes they do come up naturally. Exercise 1.5.2 shows one context in which left and right can work together.

Proposition 1.5.1: If $G$ is a group and $X$ a set on which $G$ acts. Define $\phi : G \longrightarrow \operatorname{Bij} (X)$ by letting $\phi (g) : X \longrightarrow X$ be the map $\phi (g) (x) = g \cdot x$. Then $\phi$ is a group homomorphism. Conversely, let there be given a homomorphism $\phi : G \longrightarrow \operatorname{Bij} (X)$. Then $g \cdot x = \phi (g) (x)$ defines a group action.

Proof. (Click to Expand/Collapse)

Since the group law in $\operatorname{Bij} (X)$ is composition of functions, for $g, h$ in $G$ we have $(\phi (g) \phi (h)) x = \phi (g) (\phi (h) x) = \phi (g) (h \cdot x) = g \cdot (h \cdot x),$ while $\phi (g h) (x) = (g h) \cdot x.$ So the condition that $\phi$ is a homomorphism expresses exactly the associative property of a group action. We leave discussion of the other axiom $1 \cdot x = x$ to the reader.

We see that giving an action of $G$ on $X$ is the same as giving a homomorphism $G \longrightarrow \operatorname{Bij} (X)$. If $X$ is finite, with $n$ elements, this is the same as giving a homomorphism $G \longrightarrow S_n$, by Exercise 1.4.1.

Let $X$ be a set on which $G$ acts. If $x, y \in X$, define $x \sim y$ to mean that $y = g \cdot x$ for some $g \in G$. This is an equivalence relation, and the equivalence classes are called the orbits of the $G$-action. If $x \in X$, we call the unique equivalence class containing $x$ the orbit of $X$. The action is called transitive if there is only one orbit. We may also say that a $G$-set is transitive if the action is transitive.

Example 1.5.1: Let $G$ be a group. Then $G$ acts on itself by left translation: the action is $\mu (g, x) = g x$. This action is transitive.

Example 1.5.2: Let $G$ be a group. Then $G$ acts on itself by right translation: the action is $\mu (g, x) = x g^{- 1}$. This action is also transitive.

Exercise 1.5.3: Explain why we need $g^{- 1}$ in this definition.

Example 1.5.3: Let $G$ be a group. $G$ acts on itself by conjugation: the action is $\mu (g, x) = g x g^{- 1}$, $g, x \in G$. This is an example where the notation $g \cdot x$ would be bad. (See Remark 1.5.1.) Thus we might prefer to write this action by $g \star x = g x g^{- 1}$, though this is generally not done in the literature. The action is not transitive, unless $G$ has only one element. The orbits are just the conjugacy classes.

Example 1.5.4: Let $G$ be a group, let $X$ be the set of subgroups of $G$. Then $G$ acts on $X$ by conjugation. This is an example where the notation $g \cdot x$ would be bad. (See Remark 1.5.1.) So if $H$ is a subgroup of $G$ we will denote the action of $g$ with the notation $g \star H = g H g^{- 1}$. If $G = S_3$, then $X$ has seven elements, and four orbits, $\mathcal{O}_1$, $\mathcal{O}_2$, $\mathcal{O}_3$, $\mathcal{O}_4$: \begin{eqnarray*} \mathcal{O}_1 & = & \{ \left\langle 1 \right\rangle \}\\ \mathcal{O}_2 & = & \{ \left\langle (12) \right\rangle, \left\langle (13) \right\rangle, \left\langle (23) \right\rangle \}\\ \mathcal{O}_3 & = & \{ \left\langle (123) \right\rangle \},\\ \mathcal{O}_4 & = & \{S_3 \} \end{eqnarray*} We can read off the normal subgroups from this data: they are $1$, $S_3$ itself and $\left\langle (123) \right\rangle$.

Exercise 1.5.4: Find the orbits of subgroups of $S_4$ in the action by conjugation.

Example 1.5.5: Let $G$ be a group and $H$ a subgroup. Then $H$ acts on $G$ by left translation: this is the action $h \star g = h g$ ($h \in H$, $g \in G$). The orbits are the right-cosets $H g$. Similarly, there is a right action of $H$ on $G$ by right translation: $g \ast h = g h$ ($g \in G, h \in H$). The orbits in this right action are the left cosets $g H$.

There is another, more important connection between cosets and group actions, as we will now explain. In a nutshell we will now construct all transitive group actions. But when are two actions the same?

If $X$ and $Y$ are $G$-sets, we say that a map $\phi : X \longrightarrow Y$ is a $G$-set homomorphism or is equivariant if $\phi (g \cdot x) = g \cdot \phi (x)$ for $g \in G$, $x \in X$. If it is a bijection, we say that it is an isomorphism of $G$-sets, or that the $G$-actions are equivalent. Our goal is to classify all transitive $G$-actions up to equivalence.

First, some notation. If $X$ is a set with a left $G$-action, we will denote by $G \backslash X$ the set of orbits; similarly, if $X$ is a set with a right action, we will denote the orbits by $X / G$. For example, if $H$ is a subgroup of $G$, then $H$ acts on $G$ by either left or right translation, so in accordance with Example 1.5.5, $H \backslash G = \left\{ H x|x \in G\}, \hspace{2em} G / H = \left\{ x H|x \in G\}. \right. \right.$ Note that this is consistent with the previous notation, introduced above Proposition 1.1.4.

We have a group action of $G$ on $G / H$ by left translation. Thus if $x H \in G / H$ and $g \in G$, then $g x H$ is another coset, and $(g, x H) \longmapsto g x H$ is a group action. This action is obviously transitive.

If $X$ is a $G$-set and $x \in X$, the isotropy subgroup or stabilizer of $x$ is $G_x =\{g \in G|g \cdot x = x\}.$ It is the set of group elements that don't move $x$. The following result is of fundamental importance in mathematics.

Theorem 1.5.1: Let $X$ be a transitive $G$-set, let $x \in X$, and let $H = G_x$.

(a) The cardinality of the orbit $\mathcal{O}$ of $x$ is $[G : H]$.

(b) Suppose that the action is transitive. The conjugacy class of $H$ is independent of the choice of $x$; that is, if $y \in X$ then $G_y$ is conjugate to $H$. The set $X$ is equivalent to the $G$-set $G / H$.

Proof. (Click to Expand/Collapse)

Let us first see that (b) implies (a). Even if the action of $G$ on $X$ is not transitive, it induces a transitive action on $\mathcal{O}$. Applying (b) shows that $\mathcal{O}$ is in bijection with $G / H$, so its cardinality is $[G : H]$.

We turn to the proof of (b). First, if $y \in X$, since $G$ acts transitively on $X$, we have $\gamma x = y$ for some $\gamma \in G$. Now \begin{eqnarray*} g \in G_x \hspace{1em} \Longleftrightarrow \hspace{1em} g x = x & \Longleftrightarrow & \\ g \gamma^{- 1} y = \gamma^{- 1} y & \Longleftrightarrow & \\ \gamma g \gamma^{- 1} \cdot y = y & \Longleftrightarrow & \\ \gamma g \gamma^{- 1} \in G_y & \Longleftrightarrow & g \in \gamma^{- 1} G_y \gamma . \end{eqnarray*} We see that $G_x = \gamma^{- 1} G_y \gamma$. This proves that the conjugacy class of the stabilizer is independent of the choice of $x$.

Now with $H = G_x$ we exhibit an equivalence of $G$-actions on the $G$-sets $X$ and $G / H$. We show that there is a well-defined map $\phi : G / H \longrightarrow X$ such that $\phi (g H) = g \cdot x$. This will depend on checking that
 $g H = g' H \hspace{1em} \Longleftrightarrow \hspace{1em} g \cdot x = g' \cdot x.$ (1.5.1)

Assume for the moment that (1.5.1) is known. Then $\Longrightarrow$ shows that we can define $\phi (g H) = g \cdot x$, and that this map is well-defined. Moreover $\Longleftarrow$ shows that the map $\phi : G / H \longrightarrow X$ is injective. The fact that it is surjective is just a paraphrase of the assumption that $X$ is transitive. And $\phi$ is equivariant since $\phi (\gamma g H) = \gamma g \cdot x = \gamma \cdot (g \cdot x) = \gamma \phi (g H)$.

Thus (1.5.1) is sufficient to complete the proof. Here is the proof of (1.5.1): \begin{eqnarray*} g H = g' H \hspace{1em} \Longleftrightarrow \hspace{1em} H = g^{- 1} g' H & \Longleftrightarrow & \\ g^{- 1} g' \in H & \Longleftrightarrow & \\ g^{- 1} g' \cdot x = x & \Longleftrightarrow & g' \cdot x = g \cdot x. \end{eqnarray*}

We may summarize the situation as follows: equivalence classes of transitive $G$-sets are in bijection with conjugacy classes of subgroups of $G$.

Let us give some examples. The group $G$ acts on itself by conjugation. Let $x \in G$, and let $\mathcal{C}_x$ be the conjugacy class of $x$. Let $C (x) = C_G (x)$ be the group of elements of $G$ that commute with $x$.

Proposition 1.5.2: The cardinality of $\mathcal{C}_x$ is $[G : C (x)]$. In particular, if $G$ is finite, $|\mathcal{C}_x |$ divides $|G|$.

Proof. (Click to Expand/Collapse)

The group $G$ acts on $\mathcal{C}_x$ by conjugation. Since $\mathcal{C}_x$ is a single conjugacy class, this action is transitive. The stabilizer of $x$ is $\{g \in G| g x g^{- 1} = x\}=\{g \in G| g x = x g\}= C (x) .$ Thus by Theorem 1.5.1 $\mathcal{C}_x$ is in bijection with the cosets of $C (x)$, and the statement follows.

Similarly, $G$ acts on its set of subgroups by conjugation. If $H$ is a subgroup, its isotropy subgroup under this action is $N_G (H) =\{g \in G| g H g^{- 1} = H\},$ the normalizer of $H$ in $G$. Thus the number of conjugates of $G$ is $[G : N_G (H)]$.

Let us return to the action of $G$ on itself by conjugation. Every element of the center $Z (G) =\{g \in G| \text{g x = x g for all x \in G} \}$ constitutes a single conjugacy class. Such a conjugacy class $\{g\}$ is called central. Any noncentral conjugacy class has more than one element. Let $g_1, \cdots, g_h$ be representatives of these noncentral conjugacy classes, so that $G = Z (G) \cup \mathcal{C}_{g_1} \cup \cdots \cup \mathcal{C}_{g_h} .$ This is a disjoint union, so $|G| = |Z (G) | + \sum_{i = 1}^h |\mathcal{C}_{g_i} |$. By Proposition 1.5.2, we may write this
 $|G| = |Z (G) | + \sum_{i = 1}^h [G : C_G (g_i) |.$ (1.5.2)

This is called the class equation. We observe that in the class equation:
 $\text{The C_G (g_i) are all proper subgroups of G} .$ (1.5.3)

One way to see this is to note that that for the noncentral conjugacy classes, $\mathcal{C}_{g_i}$ consists of more than one lement, so by Proposition 1.5.2, we have $[G : C_G (g_i)] > 1$. Alternatively, it is clear from the definitions that if $C_G (g_i) = G$ then $g_i \in Z (G)$, and we've excluded the elements of $Z (G)$ by splitting them off separately.

The class equation, together with (1.5.3) can be used as the basis for induction arguments. Typically, one proves something about a group $G$ to be a minimal counterexample, so that the property can be assumed for proper subgroups – like the $C_G (g_i)$. Here is an example.

Proposition 1.5.3: If $p$ is a prime dividing the order of the finite group $G$, then $G$ has an element of order $p$.

Proof. (Click to Expand/Collapse)

This is already proved for abelian groups; for example in Proposition 1.3.9 if $p = p_i$ then $|P_i | > 1$, and if $x$ is any nonidentity element of $P_i$, its order is $p^k$ for some $k > 1$, and so the order of $x^{p^{k - 1}}$ is exactly $p$.

Since $Z (G)$ is abelian, if $p| |Z (G) |$, then the above remark shows that the abelian group $Z (G)$, and therefore $G$ itself, has an element of order $p$. We may therefore assume that $p \nmid |Z (G) |$. Since $|G|$ is a multiple of $p$, it follows from the class equation that for some noncentral conjugacy class with representative $g_i$, we have $p \nmid [G : C_G (g_i)]$. Since $p| |G|$ it follows that $p| |C_G (g_i) |$, and because $C_G (g_i)$ is a proper subgroup of $G$, we may assume inductively that it contains an element of order $p$; hence so does $G$.

Exercise 1.5.5: Show that if $G$ has order $p^k$ with $k > 1$, then $Z (G)$ is nontrivial. (Hint: use the class equation.)

Proposition 1.5.4: Any group of order $p^2$ is abelian.

Proof. (Click to Expand/Collapse)

Suppose $|G| = p^2$. By Exercise 1.5.5, the center of $G$ is nontrivial, it contains a cyclic subgroup order $p$; let $x$ be a generator of this subgroup of $Z (G)$, and let $a \in G - \left\langle x \right\rangle$. Then $\left\langle a, x \right\rangle$ is a subgroup of $G$ properly containing $Z$, so $\left\langle a, x \right\rangle = G$. Now $a$ commutes with $x$ since $x \in Z (G)$, and $a$ commutes with itself obviously. Since $a$ commutes with a set of generators, it is also in the center, so $G = \left\langle a, x \right\rangle \subseteq Z (G)$.

The case where a group acts on another group by automorphisms is a very special kind of group action. It will be useful to know something about the automorphism groups of some finite groups. First, let us consider the case of the automorphism group of a cyclic group $Z_n$. We recall that if $G$ is any abelian group then $R = \operatorname{End} (G)$ is a ring and $\operatorname{Aut} (G)$ is its multiplicative group $R^{\times}$.

Proposition 1.5.5: We have $\operatorname{End} (Z_n) \cong \mathbb{Z}/ n\mathbb{Z}$ as rings, and so $\operatorname{Aut} (Z_n) \cong (\mathbb{Z}/ n\mathbb{Z})^{\times}$.

Proof. (Click to Expand/Collapse)

If $a \in \mathbb{Z}$ then multiplication by $a$ gives an endomorphism $\lambda (a)$ of the underlying additive group of the ring $\mathbb{Z}/ n\mathbb{Z}$; this group is $Z_n$. Thus we obtain a ring homomorphism $\lambda : \mathbb{Z} \longrightarrow \operatorname{End} (Z_n)$, which is easily seen to be surjective and to have kernel $n\mathbb{Z}$, and the statement follows from Exercise 1.2.2.

Thus $\operatorname{Aut} (Z_n)$ is the multiplicative group modulo $n$ that are prime to $n$. This group is one of the first topics in any course on number theory; its cardinality is $\phi (n)$, where $\phi$ is Euler's totient function.

Proposition 1.5.6: If $m$ and $n$ are coprime, then $\operatorname{Aut} (Z_{m n}) \cong \operatorname{Aut} (Z_m) \times \operatorname{Aut} (Z_n)$.

Proof. (Click to Expand/Collapse)

This follows from the Chinese Remainder Theorem and Exercise 1.3.7.

Thus to determine the structure of $\operatorname{Aut} (Z_n)$ we are reduced to the case where $n$ is a prime power. The most important case is when $n = p$ is a prime. In this case, $\mathbb{Z}/ p\mathbb{Z}$ is a field, and the fact that $(\mathbb{Z}/ p\mathbb{Z})^{\times} \cong \operatorname{Aut} (Z_p)$ is cyclic is a special case of a more general theorem.

Theorem 1.5.2: If $F$ is a finite field, then $F^{\times}$ is cyclic.

Proof. (Click to Expand/Collapse)

If $F$ has $q$ elements, then $F^{\times}$ has $q - 1$. Let $m$ be the exponent of $F^{\times}$. Thus $m|q - 1$, and if $F^{\times}$ is not cyclic, $m$ is strictly less than $q - 1$ (Exercise 1.3.13). Now by definition of the exponent, every element of $F^{\times}$ is a root of the polynomial equation $x^m - 1 = 0$. Since $F$ is a field, a polynomial cannot have more roots than its degree, so $m = q - 1$ and $F^{\times}$ is cyclic.

Let $A$ is a ring, which we will assume commutative. Let $\operatorname{Mat}_k (A)$ be the ring of $k \times k$ matrices over $A$.

Proposition 1.5.7: If $\alpha \in \operatorname{Mat}_k (A)$, then $\alpha$ is a unit in $\operatorname{Mat}_k (A)$ if and only if $\det (\alpha)$ is a unit in $A$.

Proof. (Click to Expand/Collapse)

If $\alpha$ is a unit, then $\alpha \beta = \beta \alpha = 1$ for some $\beta \in \operatorname{Mat}_k (A)$, and taking determinants gives $\det (\alpha) \det (\beta) = \det (\beta) \det (\alpha) = 1$, so $\det (\alpha)$ is a unit.

Conversely, if $\det (\alpha)$ is a unit, then $\alpha$ has an inverse, given by the familiar formula expressing the inverse of a matrix by its matrix of cofactors, divided by the determinant. For example, if $k = 2$ ${\left(\begin{array}{cc} a & b\\ c & d \end{array}\right)}^{- 1} = \frac{1}{a d - b c} \left(\begin{array}{cc} d & - b\\ - c & a \end{array}\right) .$

Thus the multiplicative group of $\operatorname{Mat}_k (A)$ is the general linear group $\operatorname{GL}(k,A) =\{\alpha \in \operatorname{Mat}_k (\alpha) | \det (\alpha) \in A^{\times} \}.$ Let $\operatorname{SL}_k (A)$ be the special linear group, which is the subgroup of matrices of determinant 1. The center of $\operatorname{GL}(k,A)$ consists of the subgroup of scalar matrices $Z = \left\{ \left(\begin{array}{ccc} z & & \\ & \ddots & \\ & & z \end{array}\right) {\LARGE{|}} z \in A^{\times} \right\} .$ The projective linear group $\operatorname{PGL}(k,A)$ is $\operatorname{GL}(k,A) / Z$. Finally, the projective special linear group $\operatorname{PSL}(k,A)$ is $\operatorname{SL}(k,A) / Z_1$, where $Z_1 = Z \cap \operatorname{SL}(k,A)$.

Exercise 1.5.6: Prove that if $F =\mathbb{F}_q$ is a finite field with $q$ elements, then the order of $\operatorname{GL}(k,\mathbb{F}_q)$ is $(q^k - 1) (q^k - q) \cdots (q^k - q^{k - 1})$.

Exercise 1.5.7: If $F =\mathbb{F}_q$ is a finite field with $q$ elements, compute the orders of $\operatorname{SL}_k (\mathbb{F}_q)$ and $\operatorname{PGL}(k,(\mathbb{F}_q)$. They have the same order – but prove they are not isomorphic if $k = 2$ and $q = 3$.

Exercise 1.5.8: Show that $\operatorname{PSL}_k (\mathbb{F}_q)$ is isomorphic to a subgroup of $\operatorname{PGL}(k,\mathbb{F}_q)$ and compute its order. Your answer should depend on $\gcd (n, q - 1)$.

Proposition 1.5.8: (a) The endomorphism ring $\operatorname{End} (Z_n^k) \cong \operatorname{Mat}_k (\mathbb{Z}/ n\mathbb{Z})$ is the ring of $k \times k$ matrices over the finite ring $Z_n$.

(b) The automorphism group $\operatorname{Aut} (Z_n^k) \cong \operatorname{GL}(k,\mathbb{Z}/ n\mathbb{Z})$.

Proof. (Click to Expand/Collapse)

We can identify $Z_n^k$ as the module of column vectors
 $\label{xvectorn} \mathbf{x}= \left( \begin{array}{l} x_1\\ \vdots\\ x_k \end{array} \right), \hspace{2em} x_k \in \mathbb{Z}/ n\mathbb{Z}.$ (1.5.4)

We have a homomorphism $\lambda : \operatorname{Mat}_k (\mathbb{Z}/ n\mathbb{Z}) \longrightarrow \operatorname{End} (Z_n^k)$ in which $\lambda (M)\mathbf{x}= M\mathbf{x}$, where $M\mathbf{x}$ is the usual matrix multiplication. To see that this homomorphism is injective, let $\mathbf{e}_i = \left( \begin{array}{l} 0\\ \vdots\\ 1\\ \vdots\\ 0 \end{array} \right) \begin{array}{l} \\ \\ \longleftarrow \text{i-th row}\\ \\ \end{array}$

be the standard basis of $Z_n^k$, and let $\mu_i : Z_n^k \longrightarrow \mathbb{Z}/ n\mathbb{Z}$ be the standard linear functionals mapping the vector $\mathbf{x}$ as in (1.5.4) to $\mu_i (\mathbf{x}) = x_k$. Then if $M = (m_{i j})$ then $\mu_i (\lambda (M)\mathbf{e}_j) = m_{i j}$, so the matrix $M$ can be reconstructed from the endomorphism $\lambda (M)$. Thus $\lambda$ is injective. Also if $L \in \operatorname{End} (Z_n^k)$ is an arbitrary endomorphism and if we define $M = (m_{i j})$ by the formula $m_{i j} = \mu_i (L (\mathbf{e}_j))$, then the two endomorphisms $\lambda (M)$ and $L$ are easily seen to have the same effect, so $\lambda$ is also surjective. The fact that $\lambda$ is a ring homomorphism is easily deduced from the fact that matrix multiplication is linear and associative.