Very likely you already know the basic properties of groups, rings and fields. Some of this Chapter is therefore review. We will be a little inconsistent in that we will develop group theory from scratch, but later when we begin the group representation theory, we will assume some properties of vector spaces and fields to be known to the reader. To accomplish a speedy development, we will leave many details to the reader in Exercises.

We will use the notation $X \subseteq Y$ to mean that $X$ is a subset of $Y$ that might equal $Y$, and $X \subsetneq Y$ if $X$ is not allowed to equal $Y$. Strictly speaking the notation $X \subset Y$ is equivalent to $X \subseteq Y$.

If $X$ is a set, we will denote by $|X|$ the cardinality of $X$. If $X$ is
finite, this is just the number of elements of $X$. This notation is commonly
used in group theory and combinatorics, but in other branches of mathematics
it is not universally standard. If $G$ is a group, its cardinality $|G|$ is
called its *order*, though this term is not usually used for sets that
are not groups, and the term "order'' has other common meanings in
mathematics.

A *group* is a set $G$ together with a map $\mu : G \times G
\longrightarrow G$ (called the *group law*) and a distinguished element
$1 \in G$ called the *unit* or *identity element* that satisfy
certain axioms that we will state momentarily. We regard $\mu$ as a
multiplication and denote $\mu (x, y)$ by $x \cdot y$ or $x y$. The axioms
are:

- We have $1 \cdot x = x \cdot 1 = x$ for $x \in G$.
- We have $x \cdot (y \cdot z) = (x \cdot y) \cdot z$ if $x, y, z \in G$.
- If $x \in G$ there is an element $x^{- 1}$ such that $x \cdot x^{- 1}
= x^{- 1} \cdot x = 1$. The map $x \longmapsto x^{- 1}$ from $G
\longrightarrow G$ is called the
*inverse map*.

We may denote $1$ by $1_G$ if there are several groups floating around and it is necessary to distinguish them. This notation has a drawback since occasionally if $X$ is a set we will use $1_X$ to denote the identity map $X \longrightarrow X$.

**Exercise 1.1.1:***
Prove that if $x, y$ are elements of a group, then $(x y)^{- 1} = y^{- 1}
x^{- 1}$.
*

If $x y = y x$ for $x, y \in G$ then the group $G$ is called *Abelian*.
For Abelian groups it is sometimes useful to use additive notation: thus we
denote the identity element $0 \in G$ and denote the group law $m (x, y) = x +
y$ instead of $x \cdot y$. However we will use multiplicative notation most of
the time.

If $G$ is a group and $H$ is a subset of $G$ such that $1_G \in H$, and
whenever $x, y \in H$ we have $x \cdot y, x^{- 1} \in H$, then $H$ is called a
*subgroup* of $G$. We express this assumption by saying that $H$ is
*closed under multiplication* and under the inverse map. We call $H$
*a proper subgroup* if it is subgroup that is a proper subset, that is,
$H \neq G$.

If $G$ and $H$ are groups, and $\phi : G \longrightarrow H$ is a map, we call
$\phi$ a *homomorphism* if $\phi (x \cdot y) = \phi (x) \cdot \phi
(y)$. We express this identity by saying that $\phi$ "preserves'' the group
law. It is easily deduced that $\phi (1_G) = 1_H$ and $\phi (x^{- 1}) = \phi
(x)^{- 1}$ for $x \in G$. The *image* of $\phi$ is $\phi (G) =\{\phi
(g) |g \in G\}$, and the *kernel* $\ker (\phi) =\{g \in G| \phi (g) =
1\}$. We will also sometimes denote the image $\phi (G) = \operatorname{im} (\phi)$.

**Exercise 1.1.2:***
Prove that if $\phi : G \longrightarrow H$ is a homomorphism, then
$\operatorname{im} (\phi)$ is a subgroup of $H$, and $\ker (\phi)$ is a subgroup of
$G$. Prove that $\phi$ is surjective if and only if $\operatorname{im} (\phi) = H$
and injective if and only if $\ker (\phi) = 1$. (Here $1$ means the subgroup
$\{1\}$ of $G$. Of course the surjectivity statement is obvious and is
included only to point up a certain duality between image and kernel; but
give details for the injectivity claim.)
*

If $X$ is a set, an *equivalence relation* on $X$ is a relation $\sim$
such that $x \sim y$ is defined to be true or false for $x$ and $y$ in $X$;
the relation is assumed to satisfy the following axioms.

- Reflexive: $x \sim x$ for $x \in X$;
- Symmetric: $x \sim y$ if and only if $y \sim x$;
- Transitive: $x \sim y$ and $y \sim z$ implies $x \sim z$.

**Exercise 1.1.3:***
If $x, y \in X$, prove that $c (x)$ and $c (y)$ are either disjoint or
equal. Moreover $x \in c (x)$ so $X = \bigcup_{x \in X} c (x)$. Thus $X$ is
the disjoint union of the equivalence classes.
*

**Exercise 1.1.4:***
Let $G$ be a group and $H$ a subgroup. Define a
relation $\sim$ on $G$ by $x \sim y$ if $x^{- 1} y \in H$. Prove that $\sim$
is an equivalence relation.
*

If $X$ and $Y$ are subsets of a group $G$, we will use the notation $X Y$ to denote the set $\{x y| x \in X, y \in Y\}$, and $X^{- 1} =\{x^{- 1} |x \in X\}$. The identity $(X Y) Z = X (Y Z)$ for three subsets $X, Y, Z$ is an immediate consequence of the associative law, and we will use the notation $X Y Z =\{x y z| x \in X, y \in Y, z \in Z\}$ for this triple product, and similarly for products of several subsets. We will also write (for an element $x$ and a subset $Y$) $x Y =\{x y| y \in Y\}$ and other obvious notations.

**Exercise 1.1.5:***
Let $G$ be a group and $H$ a subgroup. Let $\sim$ denote the equivalence
relation of Exercise 1.1.4. Show that the equivalence class
of $x$ is the left coset $x H$.
*

**Proposition 1.1.1:***
Let $G$ be a group and let $H$ be a subgroup. The cardinality of the set of
left cosets $x H$ equals the cardinality of the set of right cosets $H x$.
*

**Proof. **(Click to Expand/Collapse)

We have $(x H)^{- 1} = H^{- 1} x^{- 1} = H x^{- 1}$, so $x H \longmapsto H
x^{- 1}$ is a bijection between the set of left cosets and the set of right
cosets.

**Exercise 1.1.6:***
In the last proof, if we said that $x H \longmapsto H x$ is a bijection
between the set of left cosets and the set of right cosets, we would be
lying. What is wrong with this statement?
*

We will denote by $[G : H]$ the number of left (or right) cosets of a subgroup $H$ of a group $G$. It can be finite even if $G$ and $H$ are infinite – for example, the group $2\mathbb{Z}$ of even integers is a subgroup of the additive group $\mathbb{Z}$ of integers, and its index is $2$.

**Proposition 1.1.2:***
If $G$ and $H$ are finite groups, then $|H|$ divides $|G|$, and
\[ [G : H] = \frac{|G|}{|H|} . \]
*

**Proof. **(Click to Expand/Collapse)

We observe that $|x H| = |H|$ since $h \longmapsto x h$ is a bijection $H
\longrightarrow x H$. Thus each coset has cardinality $|H|$ and $G$ is the
disjoint union of the $[G : H]$ disjoint cosets. Therefore $|G| = [G : H]
\cdot |H|$.

**Exercise 1.1.7:***
Let $G$ be a group. Define a relation $\sim$ on $G$ by $x \sim y$ if and
only if there exists $z \in G$ such that $y = z x z^{- 1}$. Prove that this
relation, called conjugacy, is an equivalence relation. If $x \sim
y$ we say that $x$ is conjugate to $y$, or (equivalently) that $x$
is a conjugate of $y$ or that $x$ and $y$ are conjugates
(of each other).
*

If $G$ is a group and $K$ is a subgroup, then $K$ is called *normal* if
whenever $x \in K$ and $y$ is conjugate to $x$, we have $y \in K$. If $K$ is
normal, we write $K \trianglelefteq G$ (or $K \vartriangleleft G$ if $K$
is a proper subgroup).

**Proposition 1.1.3:***
If $K$ is the kernel of a homomorphism $\phi : G
\longrightarrow H$ between groups $G$ and $H$, then $K$ is a normal subgroup
of $G$.
*

**Proof. **(Click to Expand/Collapse)

If $y$ is conjugate to $x$ we may write $y = g x g^{- 1}$ for some $g \in
G$. If $x \in K$ then $\phi (x) = 1$. Now $\phi (y) = \phi (g) \cdot \phi
(x) \cdot \phi (g^{- 1}) = \phi (g) \cdot 1 \cdot \phi (g)^{- 1} = 1$, so $y
\in K$. This shows that $K$ is closed under conjugation, hence normal.

**Exercise 1.1.8:***
Let $\phi : G \longrightarrow H$ be a group homomorphism. If $\phi$ is
bijective, prove that the inverse map $\phi^{- 1} : H \longrightarrow G$ is
also a homomorphism. If this is true, $\phi$ is called an
isomorphism, and we say that $G$ is isomorphic to $H$ and
we write $G \cong H$. Explain why isomorphism has the three properties of an
equivalence relation: (reflexive) $G \cong G$, (symmetric) $G \cong H$
implies $H \cong G$ and (transitive) if $G \cong H$, $H \cong K$ then $G
\cong K$. Thus the class of groups is divided into equivalence classes,
which are called isomorphism classes.
*

If $G$ is a group and $g \in G$, let $\left\langle g \right\rangle =\{g^n |n
\in \mathbb{Z}\}$. This is a group, obviously. It is called the *cyclic
subgroup of $G$ generated by $g$*. The terms "cyclic'' and "generated by''
will be given more formal meanings later.

**Exercise 1.1.9:***
Prove that if $\left\langle g \right\rangle$ is infinite then it is
isomorphic to the additive
group $\mathbb{Z}$.
*

**Exercise 1.1.10:***
Prove that if $\left\langle g \right\rangle$ has finite
order $d$, then $d$ is the smallest positive integer such that $g^d = 1$.
Moreover, prove that $g^n = 1$ if and only if $d|n$. ( Hint:
Reverse the order of the two statements by first proving that there is an
integer $d$ such that $g^n = 1$ if and only if $d|n$; then show $d = |G|$.
Use the division algorithm for $\mathbb{Z}$, which is the
statement that given any $n \in \mathbb{Z}$ we can write $n = q r + d$ with
$0 \le d < r$.)
*

If $d$ is the integer in Exercise 1.1.10, then $d$ is called the
*order* of $g$. Thus the order of the element $g$ equals the order of
the cyclic subgroup it generates.

**Exercise 1.1.11:***
Prove that if $p$ is a prime, and if $|G| = |H| = p$, then $G \cong H$.
*

Suppose that $G$ is a group and $H$ a subgroup. We will denote by $G / H$ the set of left cosets $x H$. It is in general not a group, but there is a special case where it is.

**Proposition 1.1.4:***
Let $G$ be a group and $K$ a normal subgroup. If $x K$
and $y K$ are cosets, then so is $(x K) (y K)$; more precisely, $x K y K = x
y K$.
*

**Proof. **(Click to Expand/Collapse)

We have $x K y K = x y (y^{- 1} K y) K.$ Since $K$ is normal, $y^{- 1} K y =
K$ and so $(y^{- 1} K y) K = K$. Thus $x K y K = x y K$.

Because of Proposition 1.1.4, the set $G / K$ has a
multiplicative structure, and is in fact a group. The identity element in $G /
K$ is the coset $K = 1 \cdot K$. We call $G / K$ the *quotient group*.

In Proposition 1.1.3 we saw that the kernel of a homomorphism is normal; now let us observe that a normal subgroup is the kernel of a homomorphism.

**Proposition 1.1.5:***
Let $K \trianglelefteq G$. Then $K$ is the kernel of the map $\pi : G
\longrightarrow G / K$ defined by $\phi (g) = g K$, which is a surjective
homomorphism.
*

We will sometimes call the map $\pi$ the *projection* onto the quotient
group $G / K$.

**Proof. **(Click to Expand/Collapse)

The identity $x y K = x K y K$ in Proposition 1.1.4 can be
written $\pi (x y) = \pi (x) \pi (y)$, so $\pi$ is a homomorphism. It is
obviously surjective. If $x \in G$ then $x \in \ker (\pi)$ if and only if $x
K = K$, that is, if $x \in K$, so the kernel is $K$.

**Theorem 1.1.1:***
Let $\phi : G \longrightarrow H$ be any surjective group
homomorphism, and let $K = \ker (\phi)$. Then $H \cong G / K$. More
precisely, the rule $\lambda (g K) = \phi (g)$ gives a well-defined
bijection $\lambda : G / K \longrightarrow H$, which is a group isomorphism.
*

**Proof. **(Click to Expand/Collapse)

We first prove that if $g, g' \in G$, then

Indeed, $g K = g' K$ if and only if $K = g^{- 1} g' K$, which is if and only
if $g^{- 1} g' \in K$, which is if and only if $\phi (g^{- 1} g') = 1$, by
definition of $K$ as the kernel of $\phi$. And since $\phi$ is a
homomorphism, this is true if and only if $\phi (g)^{- 1} \phi (g') = 1$,
that is, $\phi (g) = \phi (g')$. This proves (1.1.1).

\[ \text{$g K = g' K$ if and only if $\phi (g) = \phi (g') .$} \] | (1.1.1) |

Now (1.1.1) may be interpreted as the statement that $\lambda (g K) = \phi (g)$ is a well defined map $G / K \longrightarrow H$, and furthermore, it is injective. It is surjective since $\phi$ was assumed surjective. The fact that it is a group homomorphism is a consequence of the identity $x y K = x K y K$ in Proposition 1.1.4.

**Proposition 1.1.6:***
Let $G$ be a cyclic group of order $n$. Then $G \cong \mathbb{Z}/
n\mathbb{Z}$ as an abelian group.
*

Note that $\mathbb{Z}$ is a group customarily written additively, and $n\mathbb{Z}$ is the subgroup of integer multiples of $n$. Since a cyclic group of order $n$ is determined by its order, it is good to have a consistent notation for it. The only completely standard notation for a cyclic group of order $n$ is $\mathbb{Z}/ n\mathbb{Z}$, but this has a connotation of being written additively, and sometimes the group would be written multiplicatively. We will denote a cyclic group of order $n$ by $Z_n$, though other authors might use the notation $C_n$.

**Proof. **(Click to Expand/Collapse)

Let $G = \left\langle g \right\rangle$, where $g$ has order $n$. There is a
homomorphism $c : \mathbb{Z} \longrightarrow G$ given by $c (k) = g^k$.
(Since we are writing $\mathbb{Z}$ additively and $G$ multiplicatively,
"homomorphism'' means $c (n + m) = c (n) c (m)$.) The kernel of this
homomorphism is the set of multiples of $n$ by Exercise 1.1.10,
that is, $n\mathbb{Z}$. The statement now follows from
Theorem 1.1.1.

Here is another application.

**Proposition 1.1.7:***
Let $G$ be a group, $H$ and $K$ subgroups, and assume
that $K \vartriangleleft G$. Then $H K$ is a group and $K$ is a normal
subgroup of $H K$. Moreover $K \cap H$ is a normal subgroup of $H$, and
*

**Proof. **(Click to Expand/Collapse)

We leave it to the reader to show that $H K$ is a
group (Exercise 1.1.12). Obvious $K$ is normal in $H K$ since it
is normal in the larger group $G$. Consider the homomorphism $\phi : H
\longrightarrow H K / K$, which is the composite of the inclusion $H
\longrightarrow H K$ with the projection $H K \longrightarrow H K / K$. This
composite is surjective and the kernel is $H \cap K$
(Exercise 1.1.12). Now (1.1.2) follows from
Theorem 1.1.1.

**Exercise 1.1.12:***
Complete the proof of Proposition 1.1.7 by
showng that $H K$ is a group, and that the homomorphism $H \longrightarrow H
K / K$ is surjective, with kernel $H \cap K$.
*

**Proposition 1.1.8:***
If $G$ is a group, and $S \subset G$ a subset of $G$, there is a unique
smallest subgroup $H$ of $G$ containing $S$. Specifically, $H$ is a subgroup
of $G$ satisfies the two conditions
*

- $H \supseteq S$;
- If $H'$ is any subgroup of $G$ containing $S$ then $H \subseteq H'$.

The group $H$ is called the group *generated* by $S$, and is denoted
$\left\langle S \right\rangle$. In particular if $S =\{x\}$ consists of a
single element, we write $H = \left\langle x \right\rangle$ and call it a
*cyclic group* with generator $x$. If $S =\{x_1, x_2, \cdots\}$ we may
also write $\left\langle S \right\rangle = \left\langle x_1, x_2, \cdots
\right\rangle$.

In the proof we will first prove the uniqueness assertion: that there cannot be more than one smallest group containing $S$. Then we will give two proofs of existence, one abstract, one constructive, illustrating two different ways of thinking about this. Either proof can be adapted to many other situations: if we are give a subset $S$ of a ring, field, vector space, etc. we may similarly consider the subring (or ideal), subfield, vector subspace, etc. generated by $S$.

**Proof. **(Click to Expand/Collapse)

First we prove uniqueness, then existence. Suppose $H_1$ and $H_2$ are both
satisfy the two bulleted conditions. Then applying the two conditions first
with $H = H_1$ and $H' = H_2$ gives $H_1 \subseteq H_2$, and symmetrically,
$H_2 \subseteq H_1$. Hence $H_1 = H_2$.

We now give two constructions of a group $H$ satisfying the two properties. The first proof is to define \[ H = \bigcap_{\begin{array}{c} H' \supseteq S\\ \text{$H'$ is a subgroup of $G$} \end{array}} H' . \] Note that the intersection is nonempty since $H' = G$ will be one of the subgroups in the intersection. Clearly $H$ is a subgroup of $G$, containing $S$, since all the $H'$ in the intersection contain it. If $H'$ is a subgroup of $G$ containing $S$, then $H'$ is among the groups in the intersection, so $H \subseteq H'$.

We give an alternative, more constructive proof based on different concepts. Define $H$ to be the set consisting of $1$ (just in case $S$ is empty) and all products $x_1 x_2 \cdots x_n$ such that either $x_i \in S$ or $x_i^{- 1} \in S$. It is clear that $H$ is closed under multiplication and the inverse map, so it is a subgroup. On the other hand if $H'$ is any group containing $S$ then since it is closed under multiplication and the inverse map, it obviously contains all the elements we have mandated to be in $H$, so $H \subseteq H'$.

We say that a subset $S$ of $G$ is a *generating set* if $G =
\left\langle S \right\rangle$.

**Exercise 1.1.13:***
Let $f, f' : G \longrightarrow H$ be a homomorphism, and
let $S$ be a generating set for $G$. That is, assume that $f (s) =
f' (s)$ for all $s \in S$. Prove that $f = f'$.
*