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# 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 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 $a^{{\prime}}$ such that $a^{2}a^{{\prime}}=a$. This implies $a^{{\prime}}(a^{2}a^{{\prime}})(aa^{{\prime}})=a^{{\prime}}a(aa^{{\prime}})$, i.e. $(a^{{\prime}}a)(aa^{{\prime}})^{2}=(a^{{\prime}}a)(aa^{{\prime}})$, and since $a^{{\prime}}a$ is regular, we obtain that $(aa^{{\prime}})^{2}=aa^{{\prime}}$. So $aa^{{\prime}}$ is idempotent, and because it also is regular, it must be the unity of the ring: $aa^{{\prime}}=1$. 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.

## Mathematics Subject Classification

13G05*no label found*16U60

*no label found*

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