# Groups

Learn about group axioms and finite groups in this lesson.

We'll cover the following

## Binary operators

Definition:

A binary operator on a nonempty set $G$ is a function $*$, which combines two given elements $a, b \in G$ and maps them to a new element $*(a, b)$, to say i.e.,

$*: G \times G \rightarrow G, \quad(a, b) \mapsto *(a, b)$

Instead of writing $*(a, b)$, we write $a * b$ from now on.

Example:

The addition $+$ is a binary operation defined on the integers $\mathbb{Z} .$ Instead of writing $+(5,7)=12$, we write $5+7=12$.

Note: For any given $a, b \in G$, it follows that $a * b \in G$, according to the definition given above. Thus, we say that the set $G$ is closed with respect to $(*)$.

Example:

We consider the following examples of closure:

1. The set of integers $\mathbb{Z}$ is closed with respect to the binary operation $+$ of addition, meaning that the sum of two integers is also an integer.
2. The set $2 \mathbb{Z}$ of even integers is closed under addition because the sum of two even integers gives an even integer since $2 a+2 b=2(a+b) \in 2 \mathbb{Z}$. Conversely, the set $\mathbb{Z} \backslash 2 \mathbb{Z}$ of odd integers is not closed under addition because the addition of two odd integers always yields an even integer.

## The group axioms

Definition:

A set of elements $G$ together with a binary operation $*$ is called a group if the following axioms are satisfied:

• Closure property: The group operation $*$ is closed, i.e., for all $a, b \in G$, it holds that $a * b=$ $c \in G$.

• Associativity: The group operation * is associative, i.e., $a *(b * c)=(a * b) * c$ for all $a, b, c \in G$.

1. Identity property: There’s an identity or neutral element $e \in G$, such that, $e * a=a * e=$ $a$, for all $a \in G$.

2. Inverse property: For each $a \in G$, there exists an inverse element of $a$, namely $a^{-1}$, such that $a^{-1} * a=a * a^{-1}=e$.

Furthermore, a group $G$ is said to be abelian (or commutative) if $a * b=b * a$, for all $a, b \in G$.

Note: We usually write $(G, *)$ to $G$ in order to make clear that the group $G$ takes the operation $*$ as a basis. If $(G, *)$ forms an additive group, we write $a * b=$ $a+b$, whereas the neutral element is written as $e=0$, and the inverse element to $a$ is given by $-a .$ In case the group is multiplicative, we write $a * b=a \cdot b=a b$ and denote the neutral element by $e=1$ and the inverse by $a^{-1}=\frac{1}{a}$.

Example:

1. $\left(\mathbb{N}_{0},+\right)$ doesn’t form a group since there’s no inverse element to $a$, such that $a+(-a)=0$.

2. $(\mathbb{Z},+)$ with the neutral element 0 and the inverse element $-a$ forms an abelian group because it holds that $0+a=a+0=a$ for all $a \in \mathbb{Z}$ and $(-a)+a=a+(-a)=0$ for all $a \in \mathbb{Z}$.

3. $(\mathbb{Z},+),(\mathbb{R},+)$, and $(\mathbb{C},+)$ form abelian groups with $e=0$ and $a^{-1}=$ $-a$.

4. $(\mathbb{Q}, \cdot)$ and $(\mathbb{R}, \cdot)$ are not groups because there’s no inverse element for $0$.

5. We denote $\mathbb{Q}^{*}=\mathbb{Q} \backslash\{0\}$ and $\mathbb{R}^{*}=\mathbb{R} \backslash\{0\}$. Then, $\left(\mathbb{Q}^{*}, \cdot\right)$ and $\left(\mathbb{R}^{*}, \cdot\right)$ are abelian groups with $e=1$ and $a^{-1}=\frac{1}{a} .$ Note that $\mathbb{Z}^{*}=\mathbb{Z} \backslash\{0\}$ is not a group, because there is no inverse element $\frac{1}{a}$ for any element $a .$

### Proposition 1:

Let $n \in \mathbb{N}$ with $n \geq 2 .$ Then, $\left(\mathbb{Z}_{n}, \oplus\right)($ with $\oplus$ of the definition: Addition and multiplication modulo :Addition_and_Multiplication_Modulo ) is an abelian group.

Proof:

Let $a, b, c \in \mathbb{Z}_{n} .$ We show that the group axioms are satisfied:

• G0: The group operation $\oplus$ is closed because of the operator’s definition: addition and multiplication modulo:Addition_and_Multiplication_Modulo

• G1: Because of the associativity of $(\mathbb{Z},+)$, we conclude that

\begin{aligned} \bar{a} \oplus(\bar{b} \oplus \bar{c}) &=\bar{a} \oplus \overline{b+c}=\overline{a+(b+c)}=\overline{(a+b)+c}=\overline{a+b} \oplus \bar{c} \\ &=(\bar{a} \oplus \bar{b}) \oplus \bar{c}, \end{aligned}

which shows that $\oplus$ is associative.

• G2:
1. The element $0 \in \mathbb{Z}_{n}$ is the identity element because

$\overline{0} \oplus \bar{a}=\overline{0+a}=\bar{a}$

and

$\bar{a} \oplus \overline{0}=\overline{a+0}=\bar{a}$

for every $\bar{a} \in \mathbb{Z}_{n}$.

1. There’s an inverse for each element $\bar{a} \in \mathbb{Z}_{n}$, because

$\bar{a} \oplus \overline{n-a}=\overline{a+(n-a)}=\bar{n}=\overline{0}$

for every $\bar{a} \in \mathbb{Z}_{n}$.

Furthermore, the group $\left(\mathbb{Z}_{n}, \oplus\right)$ is commutative because

$\bar{a} \oplus \bar{b}=\overline{a+b}=\overline{b+a}=\bar{b} \oplus \bar{a}$

for every $\bar{a}, \bar{b} \in \mathbb{Z}_{n}$ since $(\mathbb{Z},+)$ is also commutative.

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