Group Theory OldKiwi - Rhea

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:

1. Associativity: $ a\cdot(b\cdot c) = (a\cdot b)\cdot c $ $ \forall a,b,c \in G $
2. Identity: $ \exists e\in G $ such that $ e\cdot a = a $ $ \forall a \in G $
3. 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: By the inverse axiom $ \exists c^{-1}\in G $. Thus $ c^{-1}(ca) = c^{-1}(cb) $ and by associativity $ (c^{-1}c)a = (c^{-1}c)b $ so $ 1\cdot a = 1\cdot b $ and $ a = b $.

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: By the inverse axiom $ \exists c^{-1}\in G $. Thus $ (ac)c^{-1} = (bc)c^{-1} $ and by associativity $ a(cc^{-1}) = b(cc^{-1}) $. By commutativity of the inverse $ a\cdot 1 = b\cdot 1 $ so by commutativity of the identity $ a = b $.

Unique solution to linear equation (left)

Thm: Let $ \langle G, \cdot \rangle $ be a group and $ a, b $ be arbitrary elements of $ G $ and $ x $ be an indeterminant over $ G $. Then there exists a unique solution to the equation $ ax=b $.

Prf: Existance -- $ \exists a^{-1}\in G $ by the inverse axiom and $ a^{-1}b\in G $ by closure. Furthermore, $ a(a^{-1}b) = (aa^{-1})b $ by associativity and $ 1\cdot b $ by the commutative inverse, which is $ b $ by the identity axiom. Thus $ x = a^{-1}b $ is a solution. Uniqueness -- assume there exist two solutions $ x = x_0 $ and $ x = x_1 $ to the equation. Because $ x_0 $ is a solution, $ ax_0 = b $, and because $ x_1 $ is a solution, $ ax_1 = b $. By substitution, $ ax_0 = ax_1 $, so by the left cancellation law $ x_0 = x_1 $. Thus the solution is unique.

Unique solution to linear equation (right)

Thm: Let $ \langle G, \cdot \rangle $ be a group and $ a, b $ be arbitrary elements of $ G $ and $ x $ be an indeterminant over $ G $. Then there exists a unique solution to the equation $ xa=b $.

Prf: Existance -- $ \exists a^{-1}\in G $ by the inverse axiom and $ ba^{-1}\in G $ by closure. Furthermore, $ (ba^{-1})a = b(a^{-1}a) $ by associativity and $ b\cdot 1 $ by the inverse axiom, which is $ b $ by the commutative identity. Thus $ x = ba^{-1} $ is a solution. Uniqueness -- assume there exist two solutions $ x = x_0 $ and $ x = x_1 $ to the equation. Because $ x_0 $ is a solution, $ x_0a = b $, and because $ x_1 $ is a solution, $ x_1a = b $. By substitution, $ x_0a = x_1a $, so by the right cancellation law $ x_0 = x_1 $. Thus the solution is unique.

Inverse of a product

Thm: Let $ \langle G, \cdot \rangle $ be a group and $ a, b $ be arbitrary elements of $ G $. Then $ (ab)^{-1} = b^{-1}a^{-1} $.

Prf: By the inverse axiom $ (ab)(ab)^{-1} = 1 $, so $ x = (ab)^{-1} $ is a solution to the linear equation $ (ab)x = 1 $. But premultiplying both sides of this equation by $ a^{-1} $ gives $ a^{-1}(ab)x = a^{-1}\cdot 1 $ so $ (a^{-1}a)bx = a^{-1}\cdot 1 $ by associativity and $ 1\cdot bx = a^{-1}\cdot 1 $ by the inverse. Then by the identity $ bx = a^{-1} $. Similarly, premultiplying both sides of this equation by $ b^{-1} $ gives $ b^{-1}(bx) = b^{-1}a^{-1} $, so by associativity $ (b^{-1}b)x = b^{-1}a^{-1} $ and by the inverse axiom $ 1\cdot x = b^{-1}a^{-1} $ and by the identity $ x = b^{-1}a^{-1} $. Thus $ x = b^{-1}a^{-1} $ and $ x = (ab)^{-1} $ are both solutions to the equation $ (ab)x = 1 $. However, by the left equation theorem, this equation has a unique solution. Thus $ (ab)^{-1} = b^{-1}a^{-1} $

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} $.