Proof. Obviously, the set of the regular elements is non-empty and closed under the multiplication. Let's think the multiplication table of this set. It is a finite square where every row only contains distinct elements (any equation $ax = ay$ reduces to $x = y$ ). Hence, for every regular element $a$ , the square $a^2$ determines another $a'$ such that $a^2a' = a$ . This implies $a'(a^2a')(aa') = a'a(aa')$ , i.e. $(a'a)(aa')^2 = (a'a)(aa')$ , and since $a'a$ is regular, we obtain that $(aa')^2 = aa'$ . So $aa'$ is idempotent, and because it also is regular, it must be the unity of the ring: $aa' = 1$ . Thus we see that $R$ has a unity which is a regular element and that $a$ has a multiplicative inverse $a'$ , also regular. Consequently the regular elements form a group.