Contents
Definition (left-sided)
A group $ \langle G, \cdot \rangle $ is a set G and a Binary Operation_OldKiwi $ \cdot $ on G (closed over G by definition) such that the group axioms hold:
- Associativity: $ a\cdot(b\cdot c) = (a\cdot b)\cdot c $ $ \forall a,b,c \in G $
- Identity: $ \exists e\in G $ such that $ e\cdot a = a $ $ \forall a \in G $
- Inverse: $ \forall a\in G $ $ \exists a^{-1}\in G $ such that $ a^{-1}\cdot a = e $
Notation
Groups written additively use + to denote their Binary Operation_OldKiwi, 0 to denote their identity, $ -a $ to denote the inverse of element $ a $, and $ na $ to denote $ a + a + \ldots + a $ ($ n $ terms).
Groups written multiplicatively use $ \cdot $ or juxtaposition to denote their Binary Operation_OldKiwi, 1 to denote their identity, $ a^{-1} $ to denote the inverse of element $ a $, and $ a^n $ to denote $ a \cdot a \cdot \ldots \cdot a $ ($ n $ terms).
Theorems
Element commutes with inverse
Thm: $ \forall a\in G $ $ a\cdot a^{-1} = a^{-1}\cdot a = 1 $
Prf: Let $ a $ be an arbitrary element of $ G $. Since $ a^{-1}\in G $, it has an inverse $ (a^{-1})^{-1} $ in $ G $ such that $ (a^{-1})^{-1}\cdot a^{-1} = 1 $ by the inverse axiom. But $ 1\cdot a^{-1} = a^{-1} $ by the identity axiom, so substituting into the previous equation: $ (a^{-1})^{-1}\cdot (1\cdot a^{-1}) = 1 $. But by the inverse axiom, $ 1 = a^{-1}\cdot a $, so substituting again: $ (a^{-1})^{-1}((a^{-1}\cdot a)\cdot a^{-1}) = 1 $ and by associativity $ ((a^{-1})^{-1}\cdot a^{-1})\cdot(a\cdot a^{-1}) = 1 $. But $ ((a^{-1})^{-1}\cdot a^{-1}) = 1 $ and $ 1\cdot(a\cdot a^{-1}) = a\cdot a^{-1} $, so $ a\cdot a^{-1} = 1 $. Since $ a^{-1}\cdot a = 1 $ is given by the inverse axiom, $ a\cdot a^{-1} = a^{-1}\cdot a = 1 $.
Identity commutes with all elements
Thm: $ \forall a\in G $ $ a\cdot 1 = 1\cdot a = a $
Prf: $ 1\cdot a = a $ by the identity axiom, but $ 1 = a\cdot a^{-1} $ by the previous theorem. Substituting: $ (a\cdot a^{-1})\cdot a = a $ and by associativity $ a\cdot(a^{-1}\cdot a) = a $. By the inverse axiom $ a^{-1}\cdot a = 1 $, so substituting again $ a\cdot 1 = a $. Thus $ a\cdot 1 = 1\cdot a = a $.
