# Notes on Group Theory

Last Modified 2024-01-02

## Introduction

Group theory is the first subject in a field of mathematics known as * abstract algebra* or

*. Abstract algebra is one of the main branches of modern pure mathematics, and every mathematician, scientist, or engineer who wants a firm understanding of theoretical foundations should know a thing or two about group theory.*

**modern algebra**The goal of abstract algebra is to study the properties of algebraic systems and the interrelations between them. For example, you probably know about a handful of different number systems, including the integers $\Z$, the rational numbers $\Q$, the real numbers $\R$, and possibly the complex numbers $\C$. You may have also learned about their * algebraic properties*, such as the

*of addition, $x + y = y + x$, and the*

**commutative property***of multiplication, $x \cdot (y \cdot z) = (x \cdot y) \cdot z$.*

**associative property**In abstract algebra, we will go beyond these familiar number systems to study the properties of algebraic systems in general. We will learn how these properties can inform us about the structure of an unknown algebraic system, and we will see how this knowledge can be used to quickly and easily solve problems that would otherwise be difficult or tedious.

To begin our study of abstract algebra, we will focus on one particular type of algebraic system, known as a * group*. Groups are an ideal starting point because they strike an excellent balance between simplicity and complexity. They are straightforward to define, involving only a single binary operation, and yet they capture most of the key ideas used to study more complicated algebraic structures. In addition, groups are widely used in both pure and applied contexts. Group theory plays a foundational role not only in abstract algebra itself, but also in mathematical analysis, combinatorics, cryptography, physics, chemistry, and materials science.

## Groups

**Definition: Group**

A * group* is an algebraic structure $\alg{G; 1, {}^{-1}, \cdot}$ consisting of:

- a set $G$, called the
;**underlying set** - a distinguished element $1 \in G$, called the
;**identity element** - a unary operation ${}^{-1}: G \to G$, written as $x \mapsto x^{-1}$, called
;**inversion** - a binary operation $\cdot : G \times G \to G$, written as $(x, y) \mapsto x \cdot y$, called the
or**group operation**;**group product**

satisfying the following requirements:

: $(x \cdot y) \cdot z = x \cdot (y \cdot z)$ for all $x, y, z \in G$.**Associative property**: $1 \cdot x = x$ for all $x \in G$.**Identity property**: $x^{-1} \cdot x = 1$ for all $x \in G$.**Inverse property**

The * commutative property* $(x \cdot y = y \cdot x)$ is conspicuously absent from the definition of a group.

**Identity is the Only Idempotent**

**Theorem:** Let $\alg{G; 1, {}^{-1}, \cdot}$ be a group. The only element $x \in G$ having the property that $x \cdot x = x$ is the identity element.

*Proof:* First, to see that the identity element $1 \in G$ actually has the claimed property, observe that $1 \cdot 1 = 1$ follows from the identity property.

Next, to see that no other element of $G$ has this property, let $x \in G$ be given, and suppose $x \cdot x = x$. By multiplying both sides by $x^{-1}$ on the left, we have $x^{-1} \cdot (x \cdot x) = x^{-1} \cdot x$, from which it follows that:

$x = 1 \cdot x = (x^{-1} \cdot x) \cdot x = x^{-1} \cdot (x \cdot x) = x^{-1} \cdot x = 1$Thus, we have proven that $x \cdot x = x$ implies $x = 1$. □

**Right Inverse Property**

**Theorem:** Let $\alg{G; 1, {}^{-1}, \cdot}$ be a group. For all $x \in G$, we have $x \cdot x^{-1} = 1$.

*Proof:* Let $x \in G$ be given. We know, from the definition of a group, that $x^{-1} \cdot x = 1$ and $1 \cdot x^{-1} = x^{-1}$. It follows that:

This shows that the element $x \cdot x^{-1} \in G$ remains unchanged when it is multiplied by itself. Using the previous result, this proves that $x \cdot x^{-1} = 1$. □

**Right Identity Property**

**Theorem:** Let $\alg{G; 1, {}^{-1}, \cdot}$ be a group. For all $x \in G$, we have $x \cdot 1 = x$.

*Proof:* Let $x \in G$ be given. We know, from the definition of a group, that $x^{-1} \cdot x = 1$ and $1 \cdot x = x$. It follows that:

This proves that $x \cdot 1 = x$. □

## Subgroups

**Definition: Subgroup, $H \le G$**

Let $\alg{G; 1, {}^{-1}, \cdot}$ be a group. A * subgroup* of $\alg{G; 1, {}^{-1}, \cdot}$ is a subset $H \subseteq G$ of its underlying set that satisfies the following requirements:

- $1 \in H$.
- If $x \in H$, then $x^{-1} \in H$.
- If $x, y \in H$, then $x \cdot y \in H$.

We write $H \le G$ to denote that $H$ is a subgroup of $\alg{G; 1, {}^{-1}, \cdot}$.

In other words, $H$ is a subgroup of $G$ if and only if the structure $\alg{H; 1, {}^{-1}|_H, \cdot|_H}$ is a group in its own right, where ${}^{-1}|_H$ and $\cdot|_H$ denote the inverse operation ${}^{-1}$ and the group operation $\cdot$ restricted to $H$ and $H \times H$, respectively.

This definition introduces an important notational convention. We will often refer to a group $\alg{G; 1, {}^{-1}, \cdot}$ simply by the name of its underlying set $G$, omitting explicit mention of the identity element, the inversion operation, and the group operation.

It is also common in the mathematics literature to not use any symbol, such as $x \times y$ or $x \cdot y$, to denote the binary operation in a group. Instead, most mathematicians simply write $xy$, using *juxtaposition* to indicate application of the group operation. Note that the associative property allows us to write expressions like $xyz$ without needing to specify which product should be evaluated first, since both $(xy)z$ and $x(yz)$ are guaranteed to produce the same result.

**Definition: Commute, Commutes**

Let $G$ be a group. We say that two elements $x, y \in G$ * commute*, or we say that $x$

*with $y$, if $xy = yx$.*

**commutes****Definition: Central Element, Center, $Z(G)$**

Let $G$ be a group. An element $x \in G$ is * central* if $x$ commutes with every element of $G$. The set of all central elements of $G$ is called the

*of $G$, and is denoted by $Z(G)$.*

**center****Definition: Abelian Group**

An * abelian group* is a group $\alg{G; 1, {}^{-1}, \cdot}$ that satisfies the following additional requirement:

: $x \cdot y = y \cdot x$ for all $x, y \in G$.**Commutative property**

**Definition: Left Coset, Right Coset, $G/H$, $H \backslash G$**

Let $G$ be a group, and let $H \le G$ be a subgroup. A * left coset* of $H$ is a set of the form

for some fixed element $g \in G$. Similarly, a * right coset* of $H$ is a set of the form

for some fixed element $g \in G$. The collection of all left cosets of $H$ is denoted by $G/H$, while the collection of all right cosets of $H$ is denoted by $H \backslash G$.

This definition introduces a new notational convention. Whenever we apply a group operation, such as $gH$ or $H^{-1}$, to a *subset* of a group, we mean the set formed by applying that operation to each element of the subset. For example:

Note that we can write $g_1 H g_2$ without ambiguity, since the associative property guarantees that $(g_1 H) g_2 = g_1 (H g_2)$.

**Definition: Index, $\abs{G:H}$**

Let $G$ be a group, and let $H \le G$ be a subgroup. The * index* of $H$ in $G$, denoted by $\abs{G:H}$, is the cardinality of the

**Cosets Partition a Group**

**Theorem:** Let $G$ be a group, and let $H \le G$ be a subgroup.

*Proof:*

Suppose two cosets $xH$ and $yH$ of $H$ intersect, i.e., there exists an element $z \in xH \cap yH$

**Definition: Normal Subgroup**

Let $G$ be a group. We say that a subgroup $H \le G$ is * normal* if for all $g \in G$ and $h \in H$, we have $g^{-1}hg \in H$.

* left conjugation* — this is a left action

* right conjugation* — this is a right action

**Center is a Subgroup**

**Theorem:** Let $G$ be a group. Its center $Z(G)$ is a subgroup of $G$.

*Proof:* We must show that $Z(G)$ contains the identity element $1$ and is closed under taking inverses and products.

- Clearly, $1 \in Z(G)$, since the identity element $1$ commutes with everything.
- Let $a \in Z(G)$ and $g \in G$. By definition, we have $ag = ga$. By multiplying on the left and right by $a^{-1}$, we obtain $ga^{-1} = a^{-1}g$. Hence, $a^{-1}$ commutes with $g$. Since $g \in G$ was arbitrary, this proves that $a^{-1} \in Z(G)$.
- Let $a, b \in Z(G)$ and $g \in G$. Because $a$ and $b$ are central, we can write $abg = agb = gab$. Hence, $ab$ commutes with $g$. Again, because $g \in G$ was arbitrary, this proves that $ab \in Z(G)$. □

**Center is a Normal Subgroup**

**Theorem:** Let $G$ be a group. Its center $Z(G)$ is a normal subgroup of $G$.

*Proof:* Let $g \in G$ and $z \in Z(G)$. Because $z$ is central, we have $g^{-1}zg = zg^{-1}g = z \in Z(G)$. □