Groups vs Fields | Real Analysis

Check the rules of reversible moves in a table. Compare with a world where all four operations work.

A group rests on three promises. Grouping must not change results. A do-nothing element must exist. Every move needs an undoing move. A field adds division on top of that.

For example, pedaling a bicycle forward can be undone by pedaling back. Staying still counts as a move too. Chaining moves gives another move. A set of moves obeying these rules is a group.

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 $a + (-a) = 0$ .
Only the right identity and right inverse are checked here; since addition on $\mathbb{Z}$ is commutative, the left versions $0 + a = a$ and $(-a) + a = 0$ follow automatically.

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 $2x = 1$ .
This is impossible by an argument inside $\mathbb{Z}$ : for every integer $x$ , $2x$ is even, whereas $1$ is odd, so the equation $2x = 1$ has no solution in $\mathbb{Z}$ (consistently, solving in $\mathbb{Q}$ gives $x = \frac{1}{2} \notin \mathbb{Z}$ ).
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, $a \cdot 0 = a \cdot (0 + 0) = a \cdot 0 + a \cdot 0$ .
Put $b = a \cdot 0$ , so the identity above reads $b = b + b$ .
Add $-b$ to both sides (additive inverses exist): the left side becomes $-b + b = 0$ .
The right side needs associativity to rebracket: $-b + (b + b) = (-b + b) + b = 0 + b = b$ .
Hence $0 = b$ , i.e. $0 = a \cdot 0$ .