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.
$\blacksquare$