Groups vs Fields

A group is reversible moves (associativity, identity, inverses); a field is a world where all four operations work. Includes the subgroup test and $(\mathbb{Z},+)$ vs $(\mathbb{Q},+,\times)$ .

(a+b)+c = a+(b+c),\ \ \exists\,0,\ \ \forall a\,\exists(-a)

a(b+c) = ab+ac,\qquad 2\mathbb{Z} \le \mathbb{Z},\ \ \mathbb{Q} \le \mathbb{R}

Examples & Exercises

Check that $(\mathbb{Z}, +)$ is a group from the three axioms.

0 \in \mathbb{Z},\qquad a + (-a) = 0

The

(\mathbb{Z}_5, \times)

table — the additive group

(\mathbb{Z},+)

satisfies the same three axioms.

Show proof

Closure: the sum of two integers is an integer. Associativity is inherited from integer addition.
The identity is $0 \in \mathbb{Z}$ with $a + 0 = a$ . The inverse of $a$ is $-a \in \mathbb{Z}$ , satisfying $a + (-a) = 0$ .

Prove $(\mathbb{Z}, \times)$ is not a group.

2x = 1 \Rightarrow x = \frac{1}{2} \notin \mathbb{Z}

The

(\mathbb{Z}_6, \times)

table — 2 has no inverse.

Show proof

The multiplicative identity would have to be $1$ , since $1 \cdot a = a$ .
An inverse of $2$ would need $x \in \mathbb{Z}$ with $2x = 1$ , but $x = \frac{1}{2} \notin \mathbb{Z}$ . So the inverse axiom fails, and $(\mathbb{Z}, \times)$ is not a group.

Prove $a \cdot 0 = 0$ in any field, from the axioms.

a \cdot 0 = a \cdot (0 + 0) = a \cdot 0 + a \cdot 0

Cancellation in the additive group is the key.

Show proof

By distributivity, $a \cdot 0 = a \cdot (0 + 0) = a \cdot 0 + a \cdot 0$ .
Adding $-(a \cdot 0)$ to both sides (additive inverses exist) gives $0 = a \cdot 0$ .

1. Number Systems