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 $(g h) \cdot x = g \cdot (h \cdot x)$.

**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 $x \cdot (g h) = (x \cdot g) \cdot h$.

**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 $[G : H]$.

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 $\phi (\gamma g H) = \gamma g \cdot x =
\gamma \cdot (g \cdot x) = \gamma \phi (g H)$.

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) |

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) |

\[ \text{The $C_G (g_i)$ are all proper subgroups of $G$} . \] | (1.5.3) |

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

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
\[ \mathbf{e}_i = \left( \begin{array}{l}
0\\
\vdots\\
1\\
\vdots\\
0
\end{array} \right) \begin{array}{l}
\\
\\
\longleftarrow \text{$i$-th row}\\
\\
\end{array} \]

\[ \label{xvectorn} \mathbf{x}= \left( \begin{array}{l} x_1\\ \vdots\\ x_k \end{array} \right), \hspace{2em} x_k \in \mathbb{Z}/ n\mathbb{Z}. \] | (1.5.4) |

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.