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:

\((M0)\)   $:$   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.

$\blacksquare$