Unity and Negative form Subgroup of Units

Theorem
Let $\left({R, +, \circ}\right)$ be a ring with unity.

Then:
 * $\left({\left\{{1_R, -1_R}\right\}, \circ}\right) \le U_R$

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 the ring axioms, $\left({R, +}\right)$ is a 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:

This exhausts all the ways we can form a ring product between $1_R$ and $-1_R$.

That is:
 * $\forall x, y \in \left\{{1_R, -1_R}\right\}: x \circ y \in \left\{{1_R, -1_R}\right\}$

Hence the result from the Two-Step Subgroup Test.