# Multiplicative Group of Field is Abelian Group

## Theorem

Let $\struct {F, +, \times}$ be a field.

Let $F^* := F \setminus \set 0$ be the set $F$ less its zero.

The algebraic structure $\struct {F^*, \times}$ is an abelian group.

## Proof 1

From the field axioms:

 $(M0)$ $:$ Closure under product $\displaystyle \forall x, y \in F:$ $\displaystyle x \circ y \in F$ $(M1)$ $:$ Associativity of product $\displaystyle \forall x, y, z \in F:$ $\displaystyle \left({x \circ y}\right) \circ z = x \circ \left({y \circ z}\right)$ $(M2)$ $:$ Commutativity of product $\displaystyle \forall x, y \in F:$ $\displaystyle x \circ y = y \circ x$ $(M3)$ $:$ Identity element for product $\displaystyle \exists 1_F \in F, 1_F \ne 0_F: \forall x \in F:$ $\displaystyle x \circ 1_F = x = 1_F \circ x$ $1_F$ is called the unity $(M4)$ $:$ Inverse elements for product $\displaystyle \forall x \in F^*: \exists x^{-1} \in F^*:$ $\displaystyle x \circ x^{-1} = 1_F = x^{-1} \circ x$

Hence the result.

$\blacksquare$

## Proof 2

Recall that a field is a non-trivial commutative division ring.

The result follows from Non-Zero Elements of Division Ring form Group.

$\blacksquare$