Groups
Introduction to Groups
Definition 1.1. A group is a set G equipped with a binary operation \cdot : G \times G \to G satisfying the following properties:
Associativity: For all a, b, c \in G, (a \cdot b) \cdot c = a \cdot (b \cdot c).
Identity: There exists an element e \in G such that for all a \in G, e \cdot a = a \cdot e = a.
Inverses: For each a \in G, there exists b \in G such that a \cdot b = b \cdot a = e.
If the operation is commutative (i.e., a \cdot b = b \cdot a for all a, b \in G), the group is called abelian.
Example 1.2. The set of integers \mathbb{Z} under addition forms an abelian group, with identity element 0 and inverse -a for each a \in \mathbb{Z}.
Theorem 1.3. Let G be a group. The identity element e \in G is unique.
Proof. Suppose e and e' are both identity elements. Then, for any a \in G, we have a \cdot e = a and e' \cdot a = a. Consider e \cdot e'. Since e is an identity, e \cdot e' = e'. Since e' is an identity, e \cdot e' = e. Thus, e = e'. ◻
Rings
Definition 1.4. A ring is a set R equipped with two binary operations, addition (+) and multiplication (\cdot), satisfying:
(R, +) is an abelian group with identity 0.
Multiplication is associative: (a \cdot b) \cdot c = a \cdot (b \cdot c) for all a, b, c \in R.
Distributivity: a \cdot (b + c) = a \cdot b + a \cdot c and (b + c) \cdot a = b \cdot a + c \cdot a for all a, b, c \in R.
A ring is commutative if multiplication is commutative. A ring has a multiplicative identity if there exists 1 \in R such that 1 \cdot a = a \cdot 1 = a for all a \in R.
Example 1.5. The set of integers \mathbb{Z} with standard addition and multiplication is a commutative ring with multiplicative identity 1.
Fields
Definition 1.6. A field is a commutative ring with a multiplicative identity 1 \neq 0 in which every non-zero element a \in F has a multiplicative inverse, i.e., there exists b \in F such that a \cdot b = 1.
Example 1.7. The sets \mathbb{Q}, \mathbb{R}, and \mathbb{C} are fields under standard addition and multiplication. The set \mathbb{Z} is not a field, as elements like 2 have no multiplicative inverse in \mathbb{Z}.
Proposition 1.8. Every field is an integral domain, i.e., a commutative ring with 1 \neq 0 and no zero divisors (if a \cdot b = 0, then a = 0 or b = 0).
Proof. Let F be a field, and suppose a \cdot b = 0 with a \neq 0. Since F is a field, a has a multiplicative inverse a^{-1}. Multiply both sides of a \cdot b = 0 by a^{-1}: a^{-1} \cdot (a \cdot b) = a^{-1} \cdot 0. This gives (a^{-1} \cdot a) \cdot b = 0, so 1 \cdot b = 0, hence b = 0. Thus, F is an integral domain. ◻