Finite unitary ring with minimal non-nilpotent group of units

TL;DR

This paper characterizes finite unitary rings with a minimal non-nilpotent group of units, showing their size is a power of 2 and classifying their structure under certain conditions.

Contribution

It provides a classification of finite unitary rings with minimal non-nilpotent unit groups, including their size and structural isomorphisms.

Findings

Cardinality of the ring is a power of 2.

If the unit group modulo the Jacobson radical is not a p-group, the ring is isomorphic to specific matrix rings.

Proper subgroups of the unit group are all nilpotent.

Abstract

Let $R$ be a finite unitary ring such that $R=R_0[R^*]$ where $R_0$ is the prime ring and $R^*$ is not a nilpotent group. We show that if all proper subgroups of $R^*$ are nilpotent groups, then the cardinal of $R$ is a power of prime number 2. In addition, if $(R/Jac(R))^*$ is not a $p-$group, then either $R\cong M_2(GF(2))$ or $R\cong M_2(GF(2))\oplus A$ where $M_2(GF(2))$ is the ring of $2\times 2$ matrices over the finite field $GF(2)$ and $A$ is a direct sum of finite field $GF(2)$.

Equations

$$R=R_1⨁R_2⨁...⨁R_k,$$

$$(x{m}_1)m_2=m_2(x{m}_1)=(m_{2}x)m_1=m_1{m}_{2}x.$$

$$R/Jac(R)\cong R/M_1\oplus ...\oplus R/M_k.$$

$$R=R_1\oplus R_2\oplus \cdots \oplus R_k,$$

Full text

\par\parfootnotetext: E-mail Address: mohsen@ufam.edu.br; mamini1356@yahoo.com

Finite unitary ring with minimal non-nilpotent group of units

Mohsen Amiri, Mostafa Amini

*Departamento de Matem tica-ICE-UFAM, 69080-900, Manaus-AM, Brazil

Department of Mathematics, Payame Noor University, Tehran, Iran*

Abstract. Let $R$ be a finite unitary ring such that $R=R_{0}[R^{*}]$, where $R_{0}$ is the prime ring and $R^{*}$ is not a nilpotent group. We show that if all proper subgroups of $R^{*}$ are nilpotent groups, then the cardinality of $R$ is a power of 2. In addition, if $(R/Jac(R))^{*}$ is not a $p$-group, then either $R\cong M_{2}(GF(2))$ or $R\cong M_{2}(GF(2))\oplus A$, where $M_{2}(GF(2))$ is the ring of $2\times 2$ matrices over the finite field $GF(2)$ and $A$ is a direct sum of copies of the finite field $GF(2)$.

2010 MSC :16P10, 16U60, 20D15.

Keywords : Finite ring, Group of units, Nilpotent groups.

1 Introduction

The relations between rings and their groups of units are an interesting research subject. In [4], Groza has shown that if $R$ is a finite ring and at most one simple component of the semi-simple quotient ring $R$ by it’s Jacobson radical $Jac(R)$ is the field of order 2, then the group of units $R^{*}$ is a nilpotent group if and only if $R$ is a direct sum of two-sided ideals that are homomorphic images of group algebras of the form $SP$, where $S$ is a particular commutative finite ring, $P$ is a finite $p$-group, and $p$ is a prime number. More recently, Dolzan improved some results of Groza and described the structure of an arbitrary finite ring with a nilpotent group of units, see [1].

Let $X$ be a class of groups. We say that a group $G$ is a minimal non-$X$-group, if $G\not\in X$, and all proper subgroups of $G$ belong to $X.$ Minimal non-$X$-groups have been studied for various classes of groups $X$. For example, minimal non-abelian groups were studied by Miller and Moreno in [5], while Schmidt in [7] studied minimal non-nilpotent groups and he characterized such finite groups. The natural question is what we can say about a finite ring whose group of units is a minimal non-$X$-group. In this paper, we study finite rings with minimal non-nilpotent groups of units and we prove that the order of such rings is $2^{n}$ for some positive integer $n$. More precisely, we prove the following theorem:

Theorem 1

.* Let $R$ be a unitary ring of finite cardinality $2^{a}m$, $m$ be an odd number and $R^{*}$ a minimal non-nilpotent group. Then, $|R|=2^{a}$. Also, if $(R/Jac(R))^{*}$ is not a $p$-group, then either $R\cong M_{2}(GF(2))$ or $R\cong M_{2}(GF(2))\oplus A$, where $A$ is a direct sum of copies of the finite field of order two.*

In this paper, from now on $R$ denotes a ring with an identity $1\neq 0$ and for an arbitrary finite set $X$, $|X|$ denotes the number of elements in $X$. We denote the group of units of $R$ by $R^{*}$, the order of element $x$ in $R^{*}$ by $o(x)$, and the group generated by $x$ by $\langle x\rangle$. The ring of $n\times n$ matrices over a ring $R$ is denoted by $M_{n}(R)$, and the set of elements that commute with every element of $R$ is denoted by $Z(R)$. The centralizer of the subset $X$ of $R$ is the set of all elements of $R$ which commute with every element of $X$ is denoted by $C_{R}(X)$. Also, for any pair $a,b\in R$, $[a,b]=ab-ba$ is the Lie product of $a$ and $b$ and $R_{0}[S]$ denotes the subring of $R$ which is generated by $S\subseteq R$ over $R_{0}$, where $R_{0}$ is the prime subring of $R$. The characteristic of $R$ is denoted by $CharR$ and $GF(p^{m})$ is the finite field of order $p^{m}$, where $p$ is a prime number.

2 Results

We begin with the following useful lemma:

Lemma 2

.* Let $R$ be a unitary finite local ring with a nontrivial minimal ideal $I$ and we assume that $Jac(R)$ is a commutative ideal. Then $Jac(R)\subseteq ann_{R}(I)$.*

**Proof. **

We have $Jac(R)I\subseteq I$, and since $I$ is minimal, we deduce that $Jac(R)I=0$ or $Jac(R)I=I$. Since $I\neq 0$, by Nakayama’s Lemma, we conclude that $Jac(R)I\neq I$. Therefore $Jac(R)I=0$.* $\Box$*

Remark 1

. Let $R=A\oplus B$ be a finite ring where $A$ and $B$ are two ideals in $R$. Then $R^{*}=A^{*}\oplus B^{*}$ and $1=1_{A}+1_{B}$ where $1_{A}$ and $1_{B}$ are the identity elements of $A$ and $B$, respectively. It is clear that $A^{*}+1_{B}\leq R^{*}$ and that $A^{*}+1_{B}\cong A^{*}$.

Lemma 3

.* Let $R$ be a finite ring. If $|R|$ is an odd number, then $R=R_{0}[R^{*}]$.*

**Proof. **

By Lemma 1.1 from [4], the proof is clear.* $\Box$*

Minimal non-nilpotent groups are characterized by Schmidt as follows:

Theorem 4

.(see (9.1.1) of [6])* Assume that every maximal subgroup of a finite group $G$ is nilpotent but $G$ itself is not nilpotent. Then:*

(i) $G$ is soluble.

(ii) $|G|=p^{m}q^{n}$ where $p$ and $q$ are unequal prime numbers.

(iii) There is a unique Sylow $p$-subgroup $P$ and a Sylow $q$-subgroup $Q$ is cyclic. Hence $G=QP$ and $P\trianglelefteq G$.

Let $R$ be a finite ring with $|R|$ an odd number. Then, in the following theorem, we show that $R^{*}$ is not a minimal non-abelian group.

Theorem 5

.* Let $R$ be a finite ring of order $m$ with $m$ an odd number. If every proper subgroup of $R^{*}$ is an abelian group, then $R$ is a commutative ring.*

**Proof. **

Consider the finite ring $R$ which is minimal subject to hypothesis of the theorem. Since every maximal subgroup of $R^{*}$ is abelian, $R^{*}$ is a minimal non-abelian group. By Lemma of [2](p.512), every unitary ring of order $p$ or $p^{2}$ for prime number $p$, is a commutative ring. So we may assume that $|R|\not\in\{p,p^{2}\}$. Let $S$ be a proper subring of $R$. From Lemma 3, it follows that $R=R_{0}[R^{*}]$, and hence $S^{*}\neq R^{*}$. By assumption, $S^{*}$ is an abelian group and by Lemma 3, $S=S_{0}[S^{*}]$ is a commutative ring. So every proper subring of $R$ is a commutative ring. Let $|R|=p_{1}^{\alpha_{1}}...p_{k}^{\alpha_{k}}$ be the canonical decomposition of $|R|$ to the prime numbers $p_{i}$. Then we know that

[TABLE]

where each ideal $R_{i}$ is of order $p_{i}^{\alpha_{i}}$. Let $H_{i}$ be a subgroup of $R^{*}$ such that $H_{i}\cong R_{i}^{*}$ for all $i$. If $k>1$, then $H_{i}$ is an abelian subgroup of $R^{*}$. By minimality of $R$, $R_{i}$ is a commutative ring for all $i$ and then $R$ is a commutative ring, which is a contradiction. So, suppose that $|R|=p^{\beta}$, where $p>2$ is a prime number. We have the following two cases with respect to the Jacobson radical, either $Jac(R)=0$ or $Jac(R)\neq 0$:

**Case 1. ** Let $Jac(R)=0$. By Wedderburn Structure Theorem, $R\cong\bigoplus_{i=1}^{t}M_{n_{i}}(D_{i})$ where $D_{i}$ is a finite field. If $t>1$, then by minimality of $R$ and Remark 1, every $M_{n_{i}}(D_{i})$ is commutative, and so $R$ is commutative, which is a contradiction. It follows that $t=1$, and so $R\cong M_{n}(D)$, where $D$ is a finite field and $n$ is a positive integer. Since $R$ is not a commutative ring, we have $n>1$. So, $(M_{n}(D))^{*}\cong R^{*}$ implies that $R^{*}$ is not a nilpotent group, hence $R^{*}$ is a minimal non-nilpotent group. But any Sylow $2$-subgroup of $R^{*}$ is neither cyclic nor normal, which is a contradiction by Theorem 4(iii).

Case 2. Let $Jac(R)\neq 0$. First suppose that $R^{*}$ is a nilpotent group. Since $-1\in R^{*}$ and $o(-1)=2$, we deduce that $2\mid|R^{*}|$. Since $Jac(R)\neq 0$, by Lemma 1.2 of [4], $1+Jac(R)$ is a $p$-group. Let $P\in Syl_{p}(R^{*})$ and $K$ be a subgroup of $R^{*}$ such that $R^{*}=PK$ and $P\cap K=1$. By assumption, $P$ and $K$ are abelian groups, and hence $R^{*}$ is an abelian group. Thus by Lemma 3, $R=R_{0}[R^{*}]$ is commutative, which is a contradiction. It follows that $R^{*}$ is not a nilpotent group, and it follows that $R^{*}$ is a minimal non-nilpotent group. By Theorem 4, $|R^{*}|=r^{m}q^{n}$ where $r,q$ are prime numbers. Also, since $2\mid|R^{*}|$ and by Lemma (1.2) of [4], $1+Jac(R)$ is a $p-$group, we may assume that $r=p$ and $q=2$. On the other hand, $({R}/{Jac(R)})^{*}={R^{*}+Jac(R)}/{Jac(R)}$, and so every proper subgroup of $({R}/{Jac(R)})^{*}$ is an abelian group. By minimality of $R$, ${R}/{Jac(R)}$ is a commutative ring and hence, we deduce that $[R,R]\subseteq Jac(R)$. Let $P\in Syl_{p}(R^{*})$ and $Q\in Syl_{2}(R^{*})$. By Theorem 4(iii), either $P\lhd R^{*}$ or $Q\lhd R^{*}$. We claim that $P\lhd R^{*}$. Otherwise, by Theorem 4(iii), $P=\langle z\rangle$ is a cyclic subgroup of $R^{*}$, where $z\in P$. Since $1+Jac(R)\leq P$, there is a positive integer $i$ such that $H=1+Jac(R)=\langle z^{i}\rangle$. Since $H\lhd R^{*}$ and $R^{*}$ is a non-nilpotent group, $H\neq P$. Since $HQ$ is an abelian subgroup of $R^{*}$ and $R=\langle z,Q\rangle$, we have $H\leq Z(R^{*})$, and so $H\leq Z(R)$. Consequently, $Jac(R)\subseteq Z(R)$. Since $[R,R]\subseteq Jac(R)$, we have $uv-vu\in Jac(R)$ for all $u,v\in R^{*}$. Thus, it follows that $uvu^{-1}v^{-1}-1\in Jac(R)\subseteq Z(R)$, and so $uvu^{-1}v^{-1}\in Z(R)$ for all $u,v\in R^{*}$. Hence $R^{*}$ is a nilpotent group, which is a contradiction. Therefore $P\lhd R^{*}$, as claimed. By Theorem 4(iii), $Q=\langle x\rangle$ for some $x\in Q$. Now, we claim that $R$ is a local ring. Let $\{M_{1},...,M_{k}\}$ be the set of all maximal ideals of $R$, where $k>1$. Since ${R}/{Jac(R)}={R}/{(M_{1}\cap...\cap M_{k})}\cong{R}/{M_{1}}\times....\times{R}/{M_{k}}$, we conclude that $({R}/{Jac(R)})^{*}\cong({R}/{M_{1}})^{*}\times....\times({R}/{M_{k}})^{*}$. Let $\overline{Q}\in Syl_{2}(({R}/{M_{1}}\times....\times{R}/{M_{k}})^{*})$. Then, since $2\mid|({R}/{M_{1}})^{*}|$ for all $1\leq i\leq k$, we see that $\overline{Q}$ is not a cyclic group. Also, we have $({R}/{Jac(R)})^{*}={R^{*}+Jac(R)}/{Jac(R)}={PQ+Jac(R)}/{Jac(R)}$, and so $Q$ is not a cyclic group, which is a contradiction. Hence $k=1$, as claimed. Let $M=Jac(R)$. Since $1+M$ is an abelian group, $M$ is a commutative ideal. Since $R/M$ is a finite field, $M$ is not a central ideal. So, there exists $w\in M$ such that $wx\neq xw$. By minimality of $R$, we have $R=R_{0}[w,x]$. Let $I$ be a minimal ideal of $R$. We follow the proof by separating two subcases, either $Z(R)\cap I\neq 0$ or $Z(R)\cap I=0$:

Subcase 1. Let $0\neq a\in Z(R)\cap I$. By Lemma 2, $M\subseteq ann_{R}(I)=\{r\in R:\ rs=0\ for\ all\ s\in I\}$. Since $R/M$ is a finite field, $M=ann_{R}(I)$. It follows from $a\in Z(R)$ that $I=Ra$ is two sided ideal. Since $R/M$ is a finite field, $(R/M)^{*}=\langle x+M\rangle$ for some $x\in R$ with $gcd(o(x+M),p)=1$. Let $y\in R\setminus M$. Therefore $y+M=x^{i}+M$ for some integer $0\leq i\leq n-1$. Then $y=x^{i}+s$ for some $s\in M$ and this, implies that $ya=x^{i}a+sa=x^{i}a$. So, we have $I=\{0,xa,...,x^{n}a\}\subseteq M$. Since $xx^{i}a=x^{i}ax$, $w(x^{i}a)=(x^{i}a)w$ and $R=R_{0}[x,w]$, we conclude that $x^{i}a\in Z(R)$, and so $I\subseteq Z(R)$. The minimality of $R$ implies that ${R}/{I}$ is a commutative ring, and consequently $[R,R]\subseteq I\subseteq Z(R)$. Thus, we have $uvu^{-1}v^{-1}-1\in I\subseteq Z(R)$ for all $u,v\in R^{*}$. It follows that $uvu^{-1}v^{-1}\in Z(R^{*})$, and hence $R^{*}$ is a nilpotent group, which is a contradiction.

Subcase 2. Let $Z(R)\cap I=0$ and $0\neq b\in I$. Since $R=R_{0}[w,x]$ and $M$ is commutative, we have $bw=wb$, and so $[b,x]\neq 0$. Therefore $R=R_{0}[b,x]$. We may assume that $b=w\in I$ and $m_{1},m_{2}\in M$. Since $M$ is a commutative ring and $xm_{1},\,m_{2}x\in M$, we have

[TABLE]

Since $R=R_{0}[x,w]$ and $wm_{1}m_{2}=m_{1}m_{2}w$, we conclude that $M^{2}\subseteq Z(R)$. If $M^{2}\neq 0$, then by minimality of $R$, ${R}/{M^{2}}$ is a commutative ring, and so $0\neq[R,R]\subseteq M^{2}\cap I$. Since $I$ is a minimal ideal and $M^{2}$ is an ideal, $I\subseteq M^{2}\subseteq Z(R)$, which is a contradiction. Hence $M^{2}=0$, and so by considering $R$ as a local ring, for all $s\in M\setminus\{0\}$, we have $M\subseteq ann_{R}(s)$. We claim that $I=M$, otherwise consider $l\in M\setminus I$. Since $R=R_{0}[x,w]$, we have $l=\sum n_{i}x^{i}+c$, where $c\in I$ and $n_{i}\in R_{0}$. Therefore $l-c=\sum n_{i}x^{i}\in M$. Then $\sum n_{i}x^{i}\in Z(R)$. If $l-c\neq 0$, then by minimality of $R$, we have that $R/R(l-c)$ is a commutative ring. By a similar argument to the one subcase 1, we have $R(l-c)\subseteq Z(R)$. From $0\neq[R,R]\subseteq R(l-c)\cap I$, where $I$ is a minimal ideal, follows that $I\subseteq R(l-c)\subseteq Z(R)$, which is a contradiction. Then $l-c=0$, which is a contradiction. Therefore $M=I$. Since $R/M$ is a finite field, we have $R=R^{*}\cup M$. Then $|R^{*}|=|R|-|M|=o(x)|P|$. Let $|R/M|=p^{m}$. Then $|R|=p^{m}|M|$, and so $|R|-|M|=(p^{m}-1)|M|$, consequently, $1+M=P$. Since $P\langle x^{2}\rangle$ is an abelian subgroup of $R^{*}$ and $R=R_{0}[w,x]$, we conclude that $x^{2}\in Z(R)$. If $p\mid o(x+h)$ for some $h\in P$, then $e=(x+h)^{\frac{o(x+h)}{p}}\in P=1+M$, and so $e-1\in M$. Since $R=R_{0}[x+h,w]$ and $M$ is commutative, $e-1\in Z(R)$. Therefore $M\cap Z(R)\neq 0$, which is a contradiction. It follows that $gcd(o(x+h),p)=1$ for all $h\in M$. By a similar argument to the one above, $(x+h)^{2}\in Z(R)$ for all $h\in M$. Then $x^{2}+xh+hx+h^{2}\in Z(R)$. Since $h^{2}=0$ and $x^{2}\in Z(R)$, we have $xh+hx\in Z(R)\cap M=0$. Therefore $xh=-hx$ for all $h\in M$. Let $0\neq h\in M$. Since $({R}/{M})^{*}=\langle x+M\rangle$, we deduce that $x+1+M=x^{t}+M$ for some integer $t$, and then $q=x^{t}-x-1\in M$. Since $M$ is a commutative ideal, $qw=wq$. It follows from $qx=xq$ and from $R=R_{0}[x,w]$ that $q=x^{t}-x-1\in M\cap Z(R)=0$. So, $x+1=x^{t}$. Since $x+1=x^{t}\not\in Z(R)$, we deduce that $t$ is an odd number. Then, we have $x^{t}h=(-1)^{t}hx^{t}=-hx^{t}$. Therefore, $(x+1)h=-h(x+1)$. But $(x+1)h=xh+h=-hx+h=-h(x+1)$. Hence $2h=0$, and so $h=0$ for all $h\in M$, which is a contradiction.* $\Box$*

Theorem 6

.* Let $R$ be a finite ring of order $p^{m}$ where $p$ is an odd prime number. If every proper subgroup of $R^{*}$ is nilpotent, then $R$ is a commutative ring.*

**Proof. **

Consider the finite ring $R$ which is minimal with respect to these conditions, but it is not commutative. Then $R^{*}$ is a minimal non-nilpotent group. By a similar argument to the one in case 1 of the previous theorem, we may assume that $Jac(R)\neq 0$. By Theorem 4, $|R^{*}|=r^{m}q^{n}$ where $r$ and $q$ are prime numbers. By Lemma 1.2 from [4], $1+Jac(R)$ is a $p$-group and then $r=p$. Since $-1\in R^{*}$ and $o(-1)=2$, we have $q=2$. Let $P\in Syl_{p}(R^{*})$ and $Q\in Syl_{2}(R^{*})$. Let $I$ be a minimal ideal of $R$ that is contained in $Jac(R)$. Then $I^{2}=0$, and hence $I$ is commutative. By minimality of $R$, we have that $R/I$ is commutative, so $[R,R]\subseteq I$. We have two cases with respect to $I\cap Z(R)$, either $I\cap Z(R)\neq 0$ or $I\cap Z(R)=0$

**Case 1. ** Suppose that $I\cap Z(R)\neq 0$. If $0\neq c\in I\cap Z(R)$, then by a similar argument to the one in the subcase 1 of case 2, in the above theorem, $I\subseteq Z(R)$, and then $1+I\leq Z(R^{*})$. Since $uvu^{-1}v^{-1}-1\in I$, we have $(R^{*})^{\prime}\leq Z(R^{*})$, and so $R^{*}$ is a nilpotent group, which is a contradiction.

**Case 2. ** Let $I\cap Z(R)=0$. First suppose that $1+I\neq P$. Since $1+I\lhd P$, there exists $c\in I\setminus\{0\}$ such that $1+c\in Z(P)$. Also, $(1+I)Q$ is a proper nilpotent subgroup of $R^{*}$. So $1+I\leq C_{R^{*}}(Q)$, and hence $c\in Z(R)$, which is a contradiction. Therefore $1+I=P$ is an abelian subgroup of $R^{*}$. By Theorem 4, either $P$ is cyclic or $Q$ is cyclic. Since $1+I=P\lhd R^{*}$, by Theorem 4, $Q$ is cyclic. Since $P$ is an abelian group and $Q$ is cyclic, every proper subgroup of $R^{*}$ is an abelian group, which is a contradiction by Theorem 5.

• $\Box$*

If $2\mid|R|$, then the above theorem is no longer valid. For example, let $R$ be the set of all $2\times 2$ matrices over the finite field $GF(2)$. Then $R^{*}\cong S_{3}$, where $S_{3}$ is the symmetric group of order $6$ and clearly, $S_{3}$ is a minimal non-abelian group. For simplicity, let $\Delta$ be the set of all rings $R$ in which either $R\cong M_{2}(GF(2))$ or $R\cong M_{2}(GF(2))\oplus A$, where $A$ is a direct sum of copies of the finite field of order two.

Remark 2

. Let $Sl(2,GF(2^{m}))$ be the kernel of the homomorphism $det(M_{n}(GF(2^{m}))\longrightarrow GF(2^{m})^{*}.$ We recall that when $m>1$, then $Sl(2,GF(2^{m}))^{*}=((Sl(2,GF(2^{m})))^{*})^{\prime}$, and hence for $n>1$ and $m>1$, we have that $(M_{n}(D))^{*}$ is not a minimal non-nilpotent group.

Theorem 7

.* Let $R$ be a unitary non-commutative ring of finite cardinality $2^{\beta}$ such that every proper subgroup of $R^{*}$ is a nilpotent group.*

(a) If $Jac(R)=0$, then $R\in\Delta$.

(b) If $Jac(R)\neq 0$, then $(R/Jac(R))^{*}$ is a cyclic $p$-group for some odd prime number $p$.

**Proof. **

(a) We proceed by induction on $\beta$. Since $R$ is a simple artinian ring, by the structure theorem of Artin-Wedderburn, we have $R\cong\bigoplus_{i=1}^{t}M_{n_{i}}(D_{i}),$ where every $D_{i}$ is a finite field. If $t=1$, then $R\cong M_{n_{1}}(D)$, and clearly, $n_{1}=2$ and $D\cong GF(2)$, so $R\in\Delta$. Let $t>1$ and $n_{i}>1$ for some $1\leq i\leq t$. By Remark 2, if $M_{n_{i}}(D_{i})$ is a minimal non-abelian group, then, $n_{i}=2$ and $D_{i}=GF(2)$. If for some $j\neq i$, we have $n_{j}>1$, then $R^{*}$ is not a minimal non-abelian group, which is a contradiction. Therefore $M_{n_{j}}(D_{j})\cong D_{j}$ for all $j\neq i$. Let $H\leq R^{*}$ such that $H\cong(M_{n_{i}}(GF(2)))^{*}$. If $|D_{j}^{*}|>1$ for some $j\neq i$, then $R^{*}\neq H$, and hence $H$ is a non-nilpotent proper subgroup of $R^{*}$, which is a contradiction. Consequently, $D_{j}\cong GF(2)$, and hence $R\in\Delta$.

(b) Suppose for a contradiction that $(R/Jac(R))^{*}$ is not a $p$-group. We may assume that $|R|$ is minimal such that $(R/Jac(R))^{*}$ is not a $p$-group. By Theorem 4, $R^{*}=PQ$, where $P\lhd G$ and $Q$ is a cyclic Sylow subgroup. Let $I\subseteq Jac(R)$ be a minimal ideal of $R$. It is easy to see that $char(I)=2$ and $I^{2}=0$. Therefore, $1+I$ is an elementary abelian $2$-group. Also, $(R/I)/(Jac(R/I))\cong R/Jac(R)$ implies that $((R/I)/(Jac(R/I)))^{*}$ is not a $p$-group, so by minimality of $R$, $(R/I)^{*}$ is a nilpotent group. Let $p>2$ be a prime number such that $p\mid|R^{*}|$. Clearly, $2p\mid|(R/Jac(R))^{*}$. Let $\{M_{1},...,M_{k}\}$ be the set of all maximal ideals of $R$. Then, we have

[TABLE]

Since $R/M_{i}$ is a simple ring, $R/M_{i}\cong M_{n_{i}}(GF(2^{m_{i}}))$ for some positive integers $n_{i}$ and $m_{i}$. If $n_{i}>1$ for some i, then $R/M_{i}\cong M_{n_{i}}(GF(2^{m_{i}}))$. Let $x$ and $y$ be two arbitrary elements of $R^{*}$ such that $xy\neq yx$ and $gcd(o(x),o(y))=1$. Since $(R/I)^{*}$ is a nilpotent group, we have $xy-yx\in I\subseteq Jac(R)$. Hence $n_{i}=1$ for all $i$, and so $R/M_{i}$ is a finite field. But $gcd(|(R/M_{i})^{*}|,2)=gcd(2^{m_{i}}-1,2)=1$ for all $i$, so $2\nmid|(R/Jac(R))^{*}$, which is a contradiction.

$\Box$**

Here we give an example for the statement of Theorem 7 part (b). Let $GF(2)[x,y]$ be the free ring generated with two elements $x$ and $y$ over finite field $GF(2)$. Let $H$ be the ideal generated by $\{x^{2},y^{3}+y+1,xy-y^{2}x\}$. Let $R=GF(2)[x,y]/H$, and let $I$ be the ideal generated with $x+H$ in $R$. Since $xy-yx\not\in H$, $R$ is a non-commutative ring. Let $L$ be the ideal generated by $t^{3}+t+1$ in $Z_{2}[t]$. Since $R/I\cong Z_{2}[t]/L$ and $L$ is a maximal ideal, $R/I$ is a finite field of order $8$. Let $(R/I)^{*}=\langle u+I\rangle$. It is easy to check that $I=\{0,x+H,ux+H,...,u^{7}x+H\}$. Clearly, $1+I$ is an elementary abelian $2$-group and $R^{*}=(1+I)\langle u\rangle$ is a minimal non-nilpotent group.

Now we are ready to prove Theorem 1.

**Proof. **

By Theorem 6, $\alpha_{1}\geq 1$. Let $|R|=2^{\alpha_{1}}p_{2}^{\alpha_{2}}...p_{k}^{\alpha_{k}}$ be the canonical decomposition of $|R|$ to the prime numbers $p_{i}$. Then

[TABLE]

where each ideal $R_{i}$ is of order $p_{i}^{\alpha_{i}}$. By Theorem 6, $R_{2}\oplus\cdots\oplus R_{k}$ is a commutative ring, and hence $(R_{1})^{*}$ is a minimal non-nilpotent group. Consequently, $k=1$. The rest of proof is clear by Theorem 7.

• $\Box$*