## Invertible elements

**Definition:** Let $(S, \cdot)$ be a groupoid and let* e *be its neutral element. If for $x \in S$ exists $y \in S$ such that

**$$xy = yx = e,$$**

we say that element *x* is **invertible**. In that case, element *y* is called an **inverse** of element *x*.

**Note 1:** If only $yx = e$ is valid, we say that *y* is an **left inverse**. Similarly, if only $xy = e$ is valid, we say that *y* is a **right inverse**.

Furthermore, if some element *x* has both left and right inverse, they are equal.

**Note 2:** If an element x is invertible, its **inverse is unique**. If a binary operation is of ”multiplicative type”, we denote an inverse with $x^{-1}$. Furthermore, if a binary operation is of ”additive type”, we denote an inverse with $-x$ and call it the opposite element.

## Examples of inverses

**Example 1:** Which elements of a groupoid $(\mathbf{N}, \cdot)$ are invertible? Explain your answer.

**Solution:** In a groupoid $(\mathbf{N}, \cdot)$ only number $1$ has an inverse. More precisely, other elements don’t have an inverses because their potential inverses aren’t elements of a set $\mathbf{N}$. For instance, $5^{-1} = \frac{1}{5} \notin \mathbf{N}$.

**Example 2:** In $(\mathbf{Z}, +), (\mathbf{Q}, +), (\mathbf{R}, +)$ all elements are invertible. In other words, every element x has an opposite element $-x$ from the same set.

**Example 3: **Is number $0$ invertible in $(\mathbf{Q}, \cdot)$ and $(\mathbf{R}, \cdot)$?

**Solution: **No. Moreover, every element except for $0$ is invertible in $(\mathbf{Q}, \cdot)$ and $(\mathbf{R}, \cdot)$.

**Example 4:** What are the invertible elements in $(\{f: S \rightarrow S\}, \circ)$?

**Solution:** The invertible elements of a given set are only bijective functions.

## Abelian group

**Definition:** A monoid in which every element is invertible is called a **group**. Moreover, if a binary operation is commutative, we call it a **commutative** or **Abelian group**.

More precisely, ordered pair $(G, \cdot)$ is a group if the following properties are valid:

$$(1) \ x \cdot y \in G, \ \forall x, y \in G \ \ (closure)$$

$$(2) \ x(yz) = (xy)z \ \forall x, y, z \in G \ \ (associativity)$$

$$(3) \ There \ exists \ e \in G \ such \ that \ ex = xe = x \ \forall x \in G \ \ (neutral \ element)$$

$$(4) \ For \ every \ x \in G \ there \ exists \ y \in G \ such \ that \ xy = yx = e \ \ (inverse \ element).$$

Furthermore, if a property

$$(5) \ xy = yx \forall \ x,y \in G \ \ (commutativity)$$

is also valid, we say that $(G, \cdot)$ is a commutative or Abelian group.

**Example 5:** $(\mathbf{Z}, +), (\mathbf{Q}, +), (\mathbf{R}, +), (\mathbf{C}, +)$ are Abelian groups.

**Example 6: **Is $(\mathbf{Q^{*}}, \cdot)$ an Abelian group? $(\mathbf{Q^{*}} = \mathbf{Q} \setminus \{0\})$

**Solution: **We need to check the mentioned $5$ properties:

Closure: $x \cdot y \in \mathbf{Q^{*}}, \ \forall x, y \in \mathbf{Q^{*}}$

Associativity: $x(yz) = (xy)z \ \forall x, y, z \in \mathbf{Q^{*}}$

Neutral element: $1 \cdot x = x \cdot 1 = x \ \forall x \in \mathbf{Q^{*}}$ and $1 \in \mathbf{Q^{*}}$

Inverse element: $ xx^{-1} = x^{-1}x = 1 \forall x \in \mathbf{Q^{*}}$

Commutativity: $xy = yx \forall \ x,y \in \mathbf{Q^{*}}$

Therefore, all the properties are valid so we can conclude that $(\mathbf{Q^{*}}, \cdot)$ is an Abelian group.

**Note 3: **Try to prove that $(\mathbf{R^{*}}, \cdot)$ and $(\mathbf{C^{*}}, \cdot)$ are also Abelian groups.

In the next chapter we will focus on a very important example of a group – a permutation group.

## Permutation group

Let $S_{n}$ be a set of all bijections $f: \{1, 2, \cdots, n\} \rightarrow \{1, 2, \cdots, n\}$. Let us show that $S_{n}$ is a group with operation of composition of functions:

(1) Composition of bijections is a bijection. In other words, $\circ$ is a binary operation on $S_{n}$.

(2) Composition of functions is associative.

(3) A neutral element for composition of functions is identity function. Furthermore, identity function is bijection.

(4) Every bijective function $f$ has an inverse $f^{-1}$ and that inverse is bijection also.

Composition of functions is generally not commutative. In other words, $(S_{n}, \circ)$ is a group but not an Abelian group. We call it a **permutation group**.

Bijection $f: \{1, 2, \cdots, n\} \rightarrow \{1, 2, \cdots, n\}$ is called a **permutation** and it is denoted by

$$\begin{pmatrix} 1 & 2 & 3 & \cdots & n \\ f(1) & f(2) & f(3) & \cdots & f(n) \end{pmatrix}.$$