Unity and Negative form Subgroup of Units
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Theorem
Let $\struct {R, +, \circ}$ be a ring with unity.
Then:
- $\struct {\set {1_R, -1_R}, \circ} \le U_R$
That is, the set consisting of the unity and its negative forms a subgroup of the group of units.
Proof
From Unity is Unit:
- $1_R \in U_R$
It remains to be shown that $-1_R \in U_R$.
From Ring Axiom $\text A$: Addition forms an Abelian Group, $\struct {R, +}$ is an abelian group.
Therefore:
- $1_R \in R \implies -1_R \in R$.
From Product of Ring Negatives:
- $-1_R \circ -1_R = 1_R \circ 1_R = 1_R$
Thus $-1_R$ has a ring product inverse (itself) and therefore $-1_R \in U_R$.
We have:
\(\ds 1_R \circ 1_R\) | \(=\) | \(\ds 1_R\) | ||||||||||||
\(\ds 1_R \circ -1_R\) | \(=\) | \(\ds -1_R\) | ||||||||||||
\(\ds -1_R \circ 1_R\) | \(=\) | \(\ds -1_R\) | ||||||||||||
\(\ds -1_R \circ -1_R\) | \(=\) | \(\ds 1_R\) |
This exhausts all the ways we can form a ring product between $1_R$ and $-1_R$.
That is:
- $\forall x, y \in \set {1_R, -1_R}: x \circ y \in \set {1_R, -1_R}$
Hence the result from the Two-Step Subgroup Test.
$\blacksquare$