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, Terminology, and Notes
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).
A right-sided definition is also possible where the element is postmultiplied rather than premultiplied by the identity and inverse in the group axioms. The first two theorems below can be proved in an analogous way for a right-sided definition so the definitions are equivalent.
Theorems
Element commutes with inverse
Thm: Let $ \langle G, \cdot \rangle $ be a group. Then $ \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: Let $ \langle G, \cdot \rangle $ be a group. Then $ \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 $.
Identity is unique
Thm: Let $ \langle G, \cdot \rangle $ be a group. Then its identity element is unique.
Prf: Suppose $ 1_a $ and $ 1_b $ are both identity elements of $ G $. Then because $ 1_a $ is an identity, by the identity axiom $ 1_a\cdot 1_b = 1_b $. But because $ 1_b $ is an identity, by the above theorem $ 1_a\cdot 1_b = 1_a $. Thus $ 1_a = 1_b $.
Each element has a unique inverse
Thm: Let $ \langle G, \cdot \rangle $ be a group and $ a $ be an arbitrary element of $ G $. Then the inverse element of $ a $ is unique.
Prf: Suppose $ a^{-1}_1 $ and $ a^{-1}_2 $ are both inverses of $ a $. Then by the inverse axiom $ a^{-1}_1\cdot a = 1 $ and $ a^{-1}_2\cdot a = 1 $. Thus $ a^{-1}_1\cdot a = a^{-1}_2\cdot a $. Let $ a^{-1} $ be an arbitrary inverse of $ a $ and postmultiply both sides by it. Then $ (a^{-1}_1\cdot a)\cdot a^{-1} = (a^{-1}_2\cdot a)\cdot a^{-1} $ and by associativity $ a^{-1}_1\cdot (a\cdot a^{-1}) = a^{-1}_2\cdot (a\cdot a^{-1}) $. But because each element commutes with its inverse and by the inverse axiom $ a\cdot a^{-1} = 1 $ so $ a^{-1}_1\cdot 1 = a^{-1}_2\cdot 1 $. Because the identity commutes with all elements and by the identity axiom $ a^{-1}_1 = a^{-1}_2 $.
Left Cancellation Law Holds
Thm: Let $ \langle G, \cdot \rangle $ be a group and $ a, b, c $ be arbitrary elements of $ G $ such that $ ca = cb $. Then $ a = b $.
Prf:
Right Cancellation Law Holds
Thm: Let $ \langle G, \cdot \rangle $ be a group and $ a, b, c $ be arbitrary elements of $ G $ such that $ ac = bc $. Then $ a = b $.
Prf:
Each element of a finite group has a finite order
Thm: Let $ \langle G, \cdot \rangle $ be a finite group with $ \mid G\mid = m $ and $ a $ be an arbitrary element of $ G $. Then $ \exists n\in\mathbb{Z}^+ $ such that $ a^n = 1 $.
Prf: Consider the sequence $ a, a^2, a^3, \ldots, a^{m+1} $. By closure, every element of this sequence is in $ G $, but there are $ m+1 $ elements in this sequence and $ m $ elements in $ G $. Thus there must be at least 2 elements in this sequence which are identical, so $ \exists i,j\in\{1, 2, \ldots, m+1\} $ such that $ a^i = a^j $ and $ i \ne j $. Assume without loss of generality that $ i < j $. Then by associativity $ a^j = a^{j - i} a^i $. Substituting this on the right hand side, and using the identity axiom on the left hand side: $ 1 \cdot a^i = a^{j-i} a^i $. By the right cancellation law $ 1 = a^{j-i} $.
