Let \(G\) be any set and let \(\cdot\) be a binary operation on \(G\) that satisfies the following:
- Associativity: For all \(a, b, c \in G\), \(a \cdot (b \cdot c) = (a \cdot b) \cdot c\)
- Identity: There exists \(e \in G\) such that \(a \cdot e = e \cdot a = a\) for all \(a \in G.\) We shall use \(1\) to represent the identity.
- Inverses: For each \(a \in G\) there exists an element \(b\) such that \(a \cdot b = b \cdot a = 1\). We shall use \(a^{-1}\) to denote the inverse of \(a.\)
We call the pair \((G, \cdot)\) a group.
Definition of a binary operation
Commutativty: If \((G,\cdot)\) is a group and \(g_1 \cdot g_2 = g_2 \cdot g_1\) for all \(g_1, g_2 \in G\) then \((G,\cdot)\) is called an abelian group.
Example \(\downarrow\) \((\mathbb{Z},+)\) is a group.
Addition is associative. The identity is \(0.\) Given any element \(n \in \mathbb{Z},\) \(-n\) is the inverse.
Example \(\downarrow\) \((\mathbb{Q}\backslash\{0\},\times)\) is a group.
Multiplication is associative. The identity is \(1.\) Given any rational number \(\frac{a}{b}\) where \(a, b \in \mathbb{Z},\) \(a \neq 0\) and \(b \neq 0,\) the inverse is \(\frac{b}{a}.\)
Example \(\downarrow\) \((\mathbb{N},+)\) is not a group.
The set of natural numbers does not contain the element \(0\) so there is no identity.
Example \(\downarrow\) \((\mathbb{Z},-)\) is not a group.
Subtraction is not associative. Take for example the expression \(5 - 2 - 1.\)
\[(5 - 2) - 1 = 3 - 1 = 2\]
\[5 - (2 - 1) = 5 - 1 = 4\]