Multiplicative Group of Field is Abelian Group

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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:

\((\text M 0)\)   $:$   Closure under product      \(\ds \forall x, y \in F:\) \(\ds x \circ y \in F \)      
\((\text M 1)\)   $:$   Associativity of product      \(\ds \forall x, y, z \in F:\) \(\ds \paren {x \circ y} \circ z = x \circ \paren {y \circ z} \)      
\((\text M 2)\)   $:$   Commutativity of product      \(\ds \forall x, y \in F:\) \(\ds x \circ y = y \circ x \)      
\((\text M 3)\)   $:$   Identity element for product      \(\ds \exists 1_F \in F, 1_F \ne 0_F: \forall x \in F:\) \(\ds x \circ 1_F = x = 1_F \circ x \)      $1_F$ is called the unity
\((\text M 4)\)   $:$   Inverse elements for product      \(\ds \forall x \in F^*: \exists x^{-1} \in F^*:\) \(\ds x \circ x^{-1} = 1_F = x^{-1} \circ x \)      

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.

$\blacksquare$