Group

\[ g_1 \cdot (g_2 \cdot g_3) = (g_1 \cdot g_2) \cdot g_3 \qquad \forall g_1,g_2,g_3 \in G \]

There exists an element \(e \in G\) (called the identity) such that \[ g \cdot e = e \cdot g = g \qquad \forall g \in G. \]

For every \(g \in G\) there exists an element \(h \in G\) (called an inverse of \(g\)) such that \[ g \cdot h = h \cdot g = e. \]

\[ g \cdot h = h \cdot g \qquad \forall g,h \in G. \] A group that satisfies this axiom is called an abelian group (or commutative group).

A group has exactly one identity element.

Every element of a group has exactly one inverse.

Let \(g,h \in G\). Then \((g \cdot h)^{-1} = h^{-1} \cdot g^{-1}\).

Let \(g \in G\). Then \((g^{-1})^{-1} = g\).

Let \(G\) be a group and \(g,h,k \in G\).

• If \(g \cdot h = g \cdot k\), then \(h = k\) (left cancellation).
• If \(h \cdot g = k \cdot g\), then \(h = k\) (right cancellation).

This rule lets us simplify equations by cancelling common factors on the left or right.

For each integer \(n \in \mathbb{N}\), we can define \(g^{n}\) for the product of g with itself n times. ²² By convention, \(g^{0} = e\); and if \(n < 0\), then \(g^{n}\) is product of \(g^{-1}\) with itself \(|n|\).

We call the smallest integer \(n \geq 1\) the order of \(g\) when \(g^{n} = e\).

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The collection of all possible permutations of a set \(X\) forms a with operation of Composition of functions is called Symmetric group, and we denote it with \(S_x\).

When the set \(X\) consists of the first \(n \in \mathbb{N}\) integers, i.e., \(X = \{1, 2, \dots, n\}\), the symmetric group is denoted as \(S_n\).

The set of \(\mathbb{Z} = \{\dots,-2,-1,0,1,2,\dots\}\) is a group with addition operator. ⁵⁵ It’s not group within product operator. (It’s not closed by inverse law. For example \(3 \in \mathbb{Z} \quad 3^{-1}= \frac{1}{3} \notin \mathbb{Z}\)). \(\mathbb{Q},\mathbb{R}\) and \(\mathbb {C}\) are a group with this operator? This is also a cyclic group by \(\langle 1 \rangle\).

The group operation is addition modulo \(n\): add the numbers normally, then take the remainder when divided by \(n\). ⁶⁶ Same as Group of integers, It’s not a group within product operator, unless \(n \in \mathbb{P}\). Why? We will prove it in further sections. \(\mathbb{Q}, \mathbb{R}\) and \(\mathbb {C}\) are a group with this operator?

Let \(\zeta = e^{\frac{2\pi i}{n}} \in \mathbb{C}\), then multiplication turns the set \(\{1,\zeta,\zeta^2,\dots,\zeta^{n-1}\}\) into a cyclic group of order \(n\)

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The General Linear Group, denoted \(\text{GL}_n(F)\), is the group of all \(n \times n\) invertible matrices with entries from a field \(F\) (like \(\mathbb{R}, \mathbb{Q}, \text{ or } \mathbb{C}\)).

• Group Operation: Matrix multiplication.
• Condition for Inclusion: A matrix \(A\) is in \(\text{GL}_n(F)\) if and only if \(\det(A) \neq 0\). This ensures the matrix has an inverse, satisfying the Inverse Axiom.