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
It has been established that $$1_R \in U_R$$.

We need to show that $$-1_R \in U_R$$.

From the ring axioms, $$\left({R, +}\right)$$ is a group, therefore $$1_R \in R \Longrightarrow -1_R \in R$$.

From Negative Product, $$-1_R \circ -1_R = 1_R \circ 1_R = 1_R$$.

Thus $$-1_R$$ has a 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 $$-1_R$$, thus we see that $$\forall x, y \in \left\{{1_R, -1_R}\right\}: x \circ y \in \left\{{1_R, -1_R}\right\}$$ and thus $$\left({\left\{{1_R, -1_R}\right\}, \circ}\right) \le \left({U_R, \circ}\right)$$.