regular elements of finite ring

Theorem.

If the finite ring $R$ has regular elements, then it has a unity.  All regular elements of $R$ form a group under the ring multiplication and with identity element the unity of $R$.  Thus the regular elements are exactly the units of the ring; the rest of the elements are the zero and the zero divisors.

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 distinct elements (any equation  $ax=ay$  reduces to  $x=y$).  Hence, for every regular element $a$, the square $a^2$ determines another $a^{\prime }$ such that  $a^2{a}^{\prime }=a$.  This implies  $a^{\prime }(a^2{a}^{\prime })(a{a}^{\prime })=a^{\prime }a(a{a}^{\prime })$,  i.e.  $(a^{\prime }a){(a{a}^{\prime })}^2=(a^{\prime }a)(a{a}^{\prime })$,  and since $a^{\prime }a$ is regular (http://planetmath.org/ZeroDivisor),  we obtain that  ${(a{a}^{\prime })}^2=a{a}^{\prime }$.  So $a{a}^{\prime }$ is idempotent, and because it also is , it must be the unity of the ring (http://planetmath.org/Unity):  $a{a}^{\prime }=1$.  Thus we see that $R$ has a unity which is a regular element and that $a$ has a multiplicative inverse $a^{\prime }$, also regular.  Consequently the regular elements form a group.