Groups
Table of Contents

Groups

Recall from the Associativity and Commutativity of Binary Operations that an operation $*$ on $S$ is said to be associative if for all $a, b, c \in S$ we have that $a * (b * c) = (a * b) * c$ and $*$ is said to be commutative if for all $a, b \in S$ we have that $a * b = b * a$.

On the Identity and Inverse Elements of Binary Operations page we said that an element $e \in S$ is the identity element of $S$ under $*$ if for all $a \in S$ we have that $a * e = a$ and $e * a = a$. We said that $a^{-1} \in S$ is said to be the Inverse of $a \in S$ if $a * a^{-1} = e$ and $a^{-1} * a = e$.

We can now begin to describe our first type of algebraic structures known as groups, which are a set $G$ equipped with a binary operation $*$ that is associative, contains an identity element, and contains inverse elements under $*$ for each element in $G$.

Definition: If $*$ is a binary operation on the set $G$, then $G$ is called a Group under $*$ denoted $(G, *)$ if $G$ under $*$ satisfies the following properties:
1. For all $a, b \in G$ we have that $(a * b) \in G$ (Closure under $*$).
2. For all $a, b, c \in G$, $a * (b * c) = (a * b) * c$ (Associativity of elements in $G$ under $*$).
3. There exists an $e \in G$ such that for all $a \in G$, $a * e = a$ and $e * a = a$ (The existence of an Identity Element of $G$ under $*$).
4. For all $a \in G$ there exists an $a^{-1} \in G$ such that $a * a^{-1} = e$ and $a^{-1} * a = e$ (The existence of inverses for each element in $G$ under $*$).
Furthermore, if $G$ is a finite set then the group $(G, *)$ is said to be a Finite Group and if $G$ is an infinite set then the group $(G, *)$ is said to be an Infinite Group. The Order of $(G, *)$ is defined as the number of elements in $G$ and is denoted $\mid G \mid$.

Sometimes we will denote the inverse of $a \in G$ with respect to $*$ as $-a \in G$.

Some of the sets and binary operations we have already seen can be considered groups. For example, $(\mathbb{R}, +)$ is a group under standard addition $+$ since the sum of any two real numbers is a real number, $+$, is associative, an additive identity $0 \in \mathbb{R}$ exists and inverse elements exist for every $a \in \mathbb{R}$ (namely $-a \in \mathbb{R}$).

Furthermore, $(\mathbb{Z}, +)$ is also a group under the operation of standard addition since the sum of any two integers is an integer, addition is associative, the additivity identity is $0 \in \mathbb{Z}$, and for all $a \in \mathbb{Z}$ we have $-a \in \mathbb{Z}$ as additive inverses.

Let's now look at an example of a set and a binary operation that is NOT a group. Consider the set of integers $\mathbb{Z}$ and define $*$ for all $a, b \in \mathbb{Z}$ by:

(1)
\begin{align} \quad a * b = a + 2b \end{align}

We will show that $(\mathbb{Z}, *)$ is not a group by showing that $*$ is not associative. Let $a, b, c \in \mathbb{Z}$. Then $*$ is not associative since:

(2)
\begin{align} \quad a * (b * c) = a * (b + 2c) = a + 2(b + 2c) = a + 2b + 4c \end{align}
(3)
\begin{align} \quad (a * b) * c = (a + 2b) * c = (a + 2b) + 2c = a + 2b + 2c \end{align}

Clear $a * (b * c) \neq (a * b) * c$ so $\mathbb{Z}$ does not form a group under the operation $*$.

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