Groups
A group is an ordered pair (G,\star) where G is a set and \star is an associative binary relation on that set, such that
- There exists an element e in G known as the identity such that for all a \in G,
a \star e = e \star a = a
- For each a \in G, there exists an element a^{-1} \in G called the inverse of a, such that
a \star a^{-1} = a^{-1} \star a = e
Groups have many properties, but perhaps most importantly
Let (G,\star) be a group. If \star is commutative on G, then we call the group abelian.
Is the ordered pair (\mathbb{Z},+) a group?
We know the addition operator is associative, so we just need to prove that there exists an identity element in G, and an inverse element for each a \in \mathbb{Z}.
- The element 0 is the identity of (\mathbb{Z},+), as for every a \in \mathbb{Z}
a + 0 = 0 + a = a
- For each a \in G, a^{-1} = -a, because
a + (-a) = (-a) + a = 0
Therefore (\mathbb{Z},+) is a group.
Groups have the following properties
Given a group G under an binary operation \star, the following hold
(1) The identity element e of G is unique.
(2) For each a \in G, there exist a unique a^{-1}.
(3) (a^{-1})^{-1} = a for all a \in G.
(4) (a \star b)^{-1} = b^{-1} \star a^{-1} for all a,b \in G.
(1) Suppose there exists two identity elements e_1 and e_2. By the definition of an identity element, e_1 \star e_2 = e_1 and e_1 \star e_2 = e_2. Thus e_1 = e_2, so the identity is unique.
(2) Suppose b and c are both inverses of a and let e be the identity of G. By the definition of an inverse element a \star b = e and c \star a = e. Note the following relation
c = c \star e = c \star (a \star b) = (c \star a) \star b = e \star b = b
(3) Rather straightforwardly a \star a^{-1} = a^{-1} \star a = e. So by the definition of an inverse element, (a^{-1})^{-1} = a.
(4) Let c = (a \star b)^{-1}. Therefore (a \star b) \star c = e. Using algebra
\begin{align*} a^{-1} \star (a \star b) \star c &= a^{-1} \star e\\ (a^{-1} \star a) \star (b \star c) &= a^{-1}\\ (b \star c) &= a^{-1}\\ c &= b^{-1} \star a^{-1} \end{align*}Additionally
Let G be a group under an binary operation \star, and a,b \in G. The equations a \star x = b and y \star a = b have unique solutions for all x,y \in G. In particular
(1) If a \star u = a \star v then u = v
(2) If u \star b = v \star b then u = v
(1) The existence of a solution to a \star x = b is given by x = a^{-1} \star b. This is unique as proved in the previous proposition.
(2) The proof is nearly identical for (y \star a) = b, where y = b \star a^{-1}.