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Prf: Suppose <math>1_a</math> and <math>1_b</math> are both identity elements of <math>G</math>. Then because <math>1_a</math> is an identity, by the identity axiom <math>1_a\cdot 1_b = 1_b</math>. But because <math>1_b</math> is an identity, by the above theorem <math>1_a\cdot 1_b = 1_a</math>. Thus <math>1_a = 1_b</math>. | Prf: Suppose <math>1_a</math> and <math>1_b</math> are both identity elements of <math>G</math>. Then because <math>1_a</math> is an identity, by the identity axiom <math>1_a\cdot 1_b = 1_b</math>. But because <math>1_b</math> is an identity, by the above theorem <math>1_a\cdot 1_b = 1_a</math>. Thus <math>1_a = 1_b</math>. | ||
| + | |||
| + | === Each element has a unique inverse === | ||
| + | Thm: Let <math>\langle G, \cdot \rangle</math> be a group and <math>a</math> be an arbitrary element of <math>G</math>. Then the inverse element of <math>a</math> is unique. | ||
| + | |||
| + | Prf: Suppose <math>a^{-1}_1</math> and <math>a^{-1}_2</math> are both inverses of <math>a</math>. Then by the inverse axiom <math>a^{-1}_1\cdot a = 1</math> and <math>a^{-1}_2\cdot a = 1</math>. Thus <math>a^{-1}_1\cdot a = a^{-1}_2\cdot a</math>. Let <math>a^{-1}</math> be an arbitrary inverse of <math>a</math> and postmultiply both sides by it. Then <math>(a^{-1}_1\cdot a)\cdot a^{-1} = (a^{-1}_2\cdot a)\cdot a^{-1}</math> and by associativity <math>a^{-1}_1\cdot (a\cdot a^{-1}) = a^{-1}_2\cdot (a\cdot a^{-1})</math>. But because each element commutes with its inverse and by the inverse axiom <math>a\cdot a^{-1} = 1</math> so <math>a^{-1}_1\cdot 1 = a^{-1}_2\cdot 1</math>. Because the identity commutes with all elements and by the identity axiom <math>a^{-1}_1 = a^{-1}_2</math>. | ||
Revision as of 11:52, 14 May 2008
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: 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 $.
