Local nearrings on finite non-abelian $2$-generated $p$-groups

TL;DR

This paper classifies finite non-abelian, non-metacyclic 2-generated p-groups of nilpotency class 2 with cyclic commutator subgroup that serve as additive groups of local nearrings, revealing the structure of their non-invertible elements.

Contribution

It provides a detailed description of such p-groups as additive groups of local nearrings, including the index of non-invertible elements subgroup.

Findings

Subgroup of non-invertible elements has index p in the additive group.

Characterization of these p-groups as additive groups of local nearrings.

Structural properties of the groups with cyclic commutator subgroup.

Abstract

Finite non-abelian non-metacyclic $2$-generated $p$-groups (${p>2}$) of nilpotency class $2$ with cyclic commutator subgroup which are the additive groups of local nearrings are described. It is shown that the subgroup of all non-invertible elements is of index $p$ in its additive group.

Equations

$$G=⟨a,\text{leavevmode}b,\text{leavevmode}c|a^{p^m}=b^{p^n}=c^{p^d}=1,\text{leavevmode}{a}^b=ac,\text{leavevmode}{c}^a=c^b=c⟩$$

$$(a^k{b}^l{)}^r=a^{kr}{b}^{lr}{c}^{-kl\left(\genfrac{}{}{0ex}{}{r}{2}\right)}$$

$$(a^k{b}^l{)}^r=a^{kr}{b}^{lr}{c}^{-kl\left(\genfrac{}{}{0ex}{}{r}{2}\right)}.$$

$$(a^k{b}^l{)}^{(r+1)}=a^{kr}{b}^{lr}{a}^k{b}^l{c}^{-kl\left(\genfrac{}{}{0ex}{}{r}{2}\right)}=a^{k(r+1)}{b}^{l(r+1)}{c}^{-klr}{c}^{-kl\left(\genfrac{}{}{0ex}{}{r}{2}\right)}$$

$$=a^{k(r+1)}{b}^{l(r+1)}{c}^{-kl(r+(\genfrac{}{}{0ex}{}{r}{2}))}=a^{k(r+1)}{b}^{l(r+1)}{c}^{kl\left(\genfrac{}{}{0ex}{}{r+1}{2}\right)}.$$

$$x^{p^m}=(a^{\alpha }{b}^{\beta }{c}^{\gamma }{)}^{p^m}=(a^{\alpha }{b}^{\beta }{)}^{p^m}{c}^{\gamma p^m}=a^{\alpha p^m}{b}^{\beta p^m}{c}^{\gamma p^m-\alpha \beta \left(\genfrac{}{}{0ex}{}{p^m}{2}\right)}$$

$$={a^{p^m}}^{\alpha }{b}^{p^m\beta }{c}^{p^m(\gamma -\alpha \beta \frac{(p^m-1)}{2})}=1$$

$$x^{p^{m-1}}=a^{p^{m-1}\alpha }{b}^{p^{m-1}\beta }{c}^{p^{m-1}(\gamma -\alpha \beta \frac{(p^{m-1}-1)}{2})}\ne 1,$$

$$⟨x,b⟩=⟨a^{\alpha }{b}^{\beta }{c}^{\gamma },b⟩=⟨a^{\alpha },b⟩=⟨a,b⟩=G$$

$$b^{-1}xb=b^{-1}(a^{\alpha }{b}^{\beta }{c}^{\gamma })b=(ac{)}^{\alpha }{b}^{\beta }{c}^{\gamma }=(a^{\alpha }{b}^{\beta }{c}^{\gamma })c^{\alpha }=x{c}^{\alpha }.$$

$$\Phi (G)=(⟨a^p⟩\times ⟨c⟩)⋊⟨b^p⟩,$$

$$|a^A|=|G|-|\Phi (G)|=p^{2m+d-2}(p^2-1),$$

$$xb=a\alpha (x)+b\beta (x)+c\gamma (x).$$

$$xy=a(x_1{y}_1+y_2\alpha (x))+b(x_2{y}_1+y_2\beta (x))+c(-x_1{x}_2(\genfrac{}{}{0ex}{}{y_1}{2})$$

$$-\left(\genfrac{}{}{0ex}{}{y_2}{2}\right)\alpha (x)\beta (x)-x_2{y}_1{y}_2\alpha (x)+x_3{y}_1+y_2\gamma (x)+x_1{y}_3\beta (x)-x_2{y}_3\alpha (x)),$$

$$+x_3\alpha (y)+\gamma (x)\beta (y)+x_1\beta (x)\gamma (y)-x_2\alpha (x)\gamma (y)(\mathrm{mod}{p}^d).$$

$$0=0\cdot b=a\alpha (0)+b\beta (0)+c\gamma (0),$$

$$0\cdot x=0\cdot (a{x}_1+b{x}_2+c{x}_3)=(0\cdot a)x_1+(0\cdot b)x_2+(0\cdot c)x_3=0.$$

$$xc=-xa-xb+xa+xb=-c{x}_3-b{x}_2-a{x}_1-c\gamma (x)-b\beta (x)-a\alpha (x)$$

$$+a{x}_1+b{x}_2+c{x}_3+a\alpha (x)+b\beta (x)+c\gamma (x)$$

$$=-b{x}_2-a{x}_1-b\beta (x)-a\alpha (x)+a{x}_1+b{x}_2+a\alpha (x)+b\beta (x)$$

$$=-b{x}_2+c{x}_1\beta (x)-b\beta (x)-a{x}_1-a(\alpha (x)-x_1)+b{x}_2+a\alpha (x)+b\beta (x)$$

$$=c{x}_1\beta (x)-b(x_2+\beta (x))-a\alpha (x)+b{x}_2+a\alpha (x)+b\beta (x)$$

$$=c{x}_1\beta (x)-b(x_2+\beta (x))-a\alpha (x)-c{x}_2\alpha (x)+a\alpha (x)+b{x}_2+b\beta (x)$$

$$=c(x_1\beta (x)-x_2\alpha (x))-b(x_2+\beta (x))+b{x}_2+b\beta (x)=c(x_1\beta (x)-x_2\alpha (x)).$$

$$xy=(a{x}_1+b{x}_2+c{x}_3)y_1+(a\alpha (x)+b\beta (x)+c\gamma (x))y_2$$

$$+(c{x}_1\beta (x)-x_2\alpha (x))y_3.$$

$$(a{x}_1+b{x}_2)y_1=a{x}_1{y}_1+b{x}_2{y}_1-c{x}_1{x}_2\left(\genfrac{}{}{0ex}{}{y_1}{2}\right),$$

$$(a\alpha (x)+b\beta (x))y_2=a{y}_2\alpha (x)+b{y}_2\beta (x)-c\left(\genfrac{}{}{0ex}{}{y_2}{2}\right)\alpha (x)\beta (x)$$

$$b{x}_2{y}_1+a{y}_2\alpha (x)=a{y}_2\alpha (x)+b{x}_2{y}_1-c{x}_2{y}_1{y}_2\alpha (x).$$

Full text

Keywords: finite $p$-group, local nearring

2010 Mathematics Subject Classification: 16Y30

An abbreviated title: Local nearrings

This work was partially supported by the grant 346300 for IMPAN from the Simons Foundation and the matching 2015-2019 Polish MNiSW fund.

1 Introduction

Nearrings are a generalization of associative rings in the sense that with the respect to addition they need not be commutative and only one distributive law is assumed. In this paper the concept “nearring” means a left distributive nearring with a multiplicative identity. The reader is referred to the books by Meldrum [6] or Pilz [8] for terminology, definitions and basic facts concerning nearrings.

Following [3], the nearring with identity will be called local, if the set of all non-invertible elements forms a subgroup of its additive group. The main results concerning local nearrings are summarized in [11].

In [4] it is shown that every non-cyclic abelian $p$-group of order $p^{n}>4$ is the additive group of a zero-symmetric local nearring which is not a ring. As it was noted in [5], neither a generalized quaternion group nor a non-abelian group of order $8$ can be the additive group of a local nearring.

Therefore the structure of the non-abelian finite $p$-groups which are the additive groups of local nearrings is an open problem [2].

It was proved that every non-metacyclic Miller–Moreno $p$-group of order $p^{n}>8$ is the additive group of a local nearring and the multiplicative group of such a nearring is the group of order $p^{n-1}(p-1)$ [9]. In this paper finite non-abelian non-metacyclic $2$-generated $p$-groups ($p>2$) of nilpotency class $2$ with cyclic commutator subgroup are studied.

2 Preliminaries

Let $G$ be a finite non-abelian non-metacyclic $2$-generated $p$-group ($p>2$) of nilpotency class $2$ with cyclic commutator subgroup.

Denote by $G^{\prime}$ and $Z(G)$ the commutator subgroup and the centre of $G$, respectively.

Let $a$ and $b$ be generators for $G$ such that $G/G^{\prime}=\langle aG^{\prime}\rangle\times\langle bG^{\prime}\rangle$, $aG^{\prime}$ has order $p^{m}$ and $bG^{\prime}$ has order $p^{n}$. Then $c=[a,b]$ generates $G^{\prime}$, $c$ has order $p^{d}$ with $1\leq d\leq n\leq m$, and $c\in Z(G)=\langle a^{p^{m}},b^{p^{n}},c\rangle$.

Suppose that $\langle a\rangle\cap G^{\prime}=\langle b\rangle\cap G^{\prime}=1$. Then

[TABLE]

and each element of $G$ can be uniquely written in the form $a^{x_{1}}b^{x_{2}}c^{x_{3}}$, $x_{1}\in C_{p^{m}}$, $x_{2}\in C_{p^{n}}$, $x_{3}\in C_{p^{d}}$. Therefore the group $G$ with $p>2$ will be denoted by $G(p^{m},p^{n},p^{d})$.

Lemma 1**.**

For any natural numbers $k$ and $l$ the equality $[a^{k},b^{l}]=c^{kl}$ holds.

Proof.

Since $b^{-1}ab=ac$, it follows that $b^{-l}ab^{l}=ac^{l}$. Therefore, $b^{-l}a^{k}b^{l}=(ac^{l})^{k}=a^{k}c^{kl}$, thus $a^{-k}b^{-l}a^{k}b^{l}=c^{kl}$. ∎

Corollary 1**.**

Let the group $G(p^{m},p^{n},p^{d})$ be additively written. Then for any natural numbers $k$ and $l$ the equalities $-ak-bl+ak+bl=c(kl)$ and $bl+ak=-c(kl)+ak+bl$ hold.

Lemma 2**.**

For any natural numbers $k$, $l$ and $r$ the equality

[TABLE]

holds.

Proof.

For $r=1$, there is nothing to prove. By induction on $r$, we derive

[TABLE]

Replacing $r$ by $r+1$ in equality $(*)$, we have

[TABLE]

[TABLE]

Thus, equality $(*)$ holds for an arbitrary $r$. ∎

Corollary 2**.**

Let the group $G(p^{m},p^{n},p^{d})$ be additively written. Then for any natural numbers $k$, $l$ and $r$ the equality $(ak+bl)r=akr+blr-ckl\binom{r}{2}$ holds.

Obviously, the exponent of $G(p^{m},p^{n},p^{d})$ is equal to $p^{m}$ for ${1\leq d\leq n\leq m}$.

Lemma 3**.**

If $x$ is an element of order $p^{m}$ of $G(p^{m},p^{n},p^{d})$, then there exist generators $a$, $b$, $c$ of this group such that $a=x$ and ${a^{p^{m}}=b^{p^{n}}=c^{p^{d}}=1}$, $a^{b}=ac$, $c^{a}=c^{b}=c$.

Proof.

Indeed, for each $x\in G(p^{m},p^{n},p^{d})$ there exist positive integers $\alpha$, $\beta$ and $\gamma$ such that $x=a^{\alpha}b^{\beta}c^{\gamma}$. Thus, we have

[TABLE]

[TABLE]

by Lemma 2. Since $|a|=p^{m}$ and $1\leq d\leq n\leq m$, where $m>1$ and $p>2$, it follows that the exponent of $G(p^{m},p^{n},p^{d})$ equals $p^{m}$.

If

[TABLE]

then either $(\alpha,p)=1$, or $(\beta,p)=1$ for $m=n$, or $(\gamma,p)=1$ for $m=n=d$. So, without loss of generality, we can assume that $(\alpha,p)=1$. Then

[TABLE]

and

[TABLE]

Furthermore, substituting $c^{\alpha}$ instead of $c$ for generators $x$ and $b$ of $G(p^{m},p^{n},p^{d})$, we have similar expressions as for generators $a$ and $b$, thus replacing the element $a$ by $x$. ∎

The following assertion concerning the automorphisms group of $G(p^{m},p^{n},p^{d})$ is a direct consequence of statement (B1) [7].

Lemma 4**.**

Let $G=G(p^{m},p^{n},p^{d})$ and let $\mbox{Aut}(G)$ be the automorphism group of $G$. Then the following statements hold:**

if $m=n$, then $|\mbox{Aut}(G)|=p^{2d+4m-5}(p^{2}-1)(p-1)$;

• 2)

if $m>n$, then $|\mbox{Aut}(G)|=p^{2d+3n+m-2}(p-1)^{2}$.

An information about a group of automorphisms of $G(p^{m},p^{m},p^{d})$ is given by the following lemma.

Lemma 5**.**

Let $G=G(p^{m},p^{m},p^{d})$ and let there exist a subgroup $A$ of $\mbox{Aut}G$ of order $p^{2m+d-2}(p^{2}-1)$, where $m,\leavevmode\nobreak\ d>1$ with odd $p$. If an element $g\in G$ of order $p^{m}$ and $A$ contains Sylow normal $p$-subgroup, then $G\neq g^{A}\cup\Phi(G)$.

Proof.

Assume that $G=g^{A}\cup\Phi(G)$. Then $G={(\langle a\rangle\times\langle c\rangle)\rtimes\langle b\rangle}$ with generators $a$, $b$ of order $p^{m}$ and a central commutator $c=[a,b]$ of order $p^{d}$ by the definition. Hence

[TABLE]

and thus all elements of order $p^{m}$ are contained in $g^{A}$. Furthermore, $a=g^{u}$ for some $u\in A$, hence $g^{A}=a^{A}$, i. e. $G=a^{A}\cup\Phi(G)$. Since $|G|=p^{2m+d}$ and $\Phi(G)=p^{2m+d-2}$, it follows that

[TABLE]

and so the centralizer $C_{A}(a)$ of $a$ in $A$ equals $1$. In particular, $(a\langle c^{p}\rangle)^{A}=(a\langle c^{p}\rangle)^{B}=a\langle c^{p}\rangle$ for the normal subgroup $B=C_{A}(a\langle c^{p}\rangle)$ of order $p^{d-1}$ in $A$.

Considering the factor-group $\bar{G}=G/\langle c^{p}\rangle$ and $\bar{A}=A/B$. Taking into consideration, that $|\bar{a}^{\bar{A}}|=p^{2m-1}(p^{2}-1)$, we have $\bar{G}=\bar{a}^{\bar{A}}\cup\Phi(\bar{G})$. Since $|\Phi(\bar{G})|=|Z(\bar{G})|$ and $xy=yx$ for all $x\in\Phi(\bar{G})$, $y\in\bar{G}$, we have $\Phi(\bar{G})=Z(\bar{G})$. Therefore, $\bar{G}$ is a Miller–Moreno group. Since $\bar{G}=\bar{a}^{\bar{A}}\cup Z(\bar{G})$, the latter equality is impossible by [9, Lemma 7]. This contradiction completes the proof. ∎

3 Nearrings with identity on $G(p^{m},p^{n},p^{d})$

First recall some basic concepts of the theory of nearrings.

Definition 1**.**

A set $R$ with two binary operations “$+"$ and “$\cdot"$ is called a (left) nearring if the following statements hold:

1. (1)

$(R,+)=R^{+}$* is a (not necessarily abelian) group with neutral element [math];* 2. (2)

$(R,\cdot)$* is a semigroup;* 3. (3)

$x(y+z)=xy+xz$* for all $x$, $y$, $z\in R$.*

If $R$ is a nearring, then the group $R^{+}$ is called the additive group of $R$. If in addition $0\cdot x=0$, then the nearring $R$ is called zero-symmetric and if the semigroup $(R,\cdot)$ is a monoid, i.e. it has an identity element $i$, then $R$ is a nearring with identity $i$. In the latter case the group $R^{*}$ of all invertible elements of the monoid $(R,\cdot)$ is called the multiplicative group of $R$.

The following assertion is well-known.

Lemma 6**.**

Let $R$ be a finite nearring with identity $i$. Then the exponent of $R^{+}$ is equal to the additive order of $i$ which coincides with additive order of every element of $R^{*}$.

As a direct consequence of Lemmas 3 and 6 we have the following corollary.

Corollary 3**.**

Let $R$ be a nearring with identity $i$ whose group $R^{+}$ is isomorphic to a group $G(p^{m},p^{n},p^{d})$. Then ${R^{+}=\langle a\rangle+\langle b\rangle+\langle c\rangle}$ with elements $a$, $b$ and $c$, satisfying relations $ap^{m}=bp^{n}=cp^{d}=0$, $-b+a+b=a+c$ and $-a+c+a=-b+c+b=c$ with $1\leq d\leq n\leq m$, where $a=i$.

The following statement [10, Lemma 1] establishes the connection between the automorphism group of the additive group of the nearring with identity and its multiplicative group.

Lemma 7**.**

Let $R$ be a nearring with identity $i$. Then there exists a subgroup $A$ of the automorphism group $\mbox{Aut}R^{+}$ which is isomorphic to $R^{*}$ and satisfying the condition $i^{A}=\{i^{a}\mid a\in A\}=R^{*}$.

The subgroup $A$ defined in Lemma 7 is called the automorphism group of the group $R^{+}$ associated with the group $R^{*}$.

The following statement [11, Theorem 54] concerns the structure of $L$ which is the subgroup of all non-invertible elements of finite local nearring $R$. Let $\Phi(G)$ denote the Frattini subgroup of $G$.

Theorem 1**.**

Let $R$ be a local nearring of order $p^{n}$ and let $G(R)=R^{+}\rtimes R^{*}$ be a group associated with $R$. Then $H=R^{+}\rtimes(i+L)$ is a Sylow normal $p$-subgroup of $G(R)$ and $L=R^{+}\cap\Phi(H)$. In particular, if $L$ is non-abelian, then its center is non-cyclic.

Considering $\Phi(R^{+})\leq\Phi(H)$, we have the following corollary.

Corollary 4**.**

$\Phi(R^{+})\leq L=\Phi(H)\cap R^{+}$.

Let $R$ be a nearring with identity $i$ whose group $R^{+}$ is isomorphic to a group $G(p^{m},p^{n},p^{d})$. Thus it follows from Corollary 3 that ${R^{+}=\langle a\rangle+\langle b\rangle+\langle c\rangle}$ with elements $a$, $b$ and $c$, satisfying relations $ap^{m}=bp^{n}=cp^{d}=0$, $-b+a+b=a+c$ and $-a+c+a=-b+c+b=c$ with $1\leq d\leq n\leq m$, where $a=i$ and each element $x\in R$ is uniquely written in the form ${x=ax_{1}+bx_{2}+cx_{3}}$ with coefficients $0\leq x_{1}<\!p^{m}$, $0\leq x_{2}<\!p^{n}$ and $0\leq x_{3}<\!p^{d}$.

Furthermore, we can assume $xa=ax=x$ for each $x\in R$ and there exist uniquely defined mappings ${\alpha\colon R\to\mathbb{Z}_{p^{m}}}$, ${\beta\colon R\to\mathbb{Z}_{p^{n}}}$ and ${\gamma\colon R\to\mathbb{Z}_{p^{d}}}$ such that

[TABLE]

Lemma 8**.**

If $x=ax_{1}+bx_{2}+cx_{3}$ and $y=ay_{1}+by_{2}+cy_{3}$ are arbitrary elements of $R$, then

[TABLE]

[TABLE]

where mappings $\alpha\colon R\to\mathbb{Z}_{p^{m}}$, $\beta\colon R\to\mathbb{Z}_{p^{n}}$ and $\gamma\colon R\to\mathbb{Z}_{p^{d}}$ satisfy the conditions:**

• (0)

$\alpha(0)\equiv 0\;(\!\!\mod p^{m})$, $\beta(0)\equiv 0\;(\!\!\mod p^{n})$ and ${\gamma(0)\equiv 0\;(\!\!\mod p^{d})}$ if and only if the nearring $R$ is zero-symmetric;**

• (1)

$\alpha(xy)\equiv x_{1}\alpha(y)+\alpha(x)\beta(y)\;(\!\!\mod p^{m}\;)$;**

• (2)

$\beta(xy)\equiv x_{2}\alpha(y)+\beta(x)\beta(y)\;(\!\!\mod p^{n}\;)$;**

• (3)

$\gamma(xy)\equiv-x_{1}x_{2}\binom{\alpha(y)}{2}-\alpha(x)\beta(x)\binom{\beta(y)}{2}-x_{2}\alpha(x)\alpha(y)\beta(y)$**

[TABLE]

Proof.

If $R$ is a zero-symmetric nearring, then

[TABLE]

thus $\alpha(0)\equiv 0\;(\!\!\mod p^{m})$, ${\beta(0)\equiv 0\;(\!\!\mod p^{n})}$ and ${\gamma(0)\equiv 0\;(\!\!\mod p^{d})}$. On the other hand, if the last congruences hold, then $0\cdot b=a\cdot 0+b\cdot 0+c\cdot 0=0$. Since $a$ is the multiplicative identity in $R$, we have $0\cdot a=a\cdot 0=0$. Moreover, from the equality $c=-a-b+a+b$ and the left distributive law it follows that $0\cdot c=-0\cdot a-0\cdot b+0\cdot a+0\cdot b=0$, hence

[TABLE]

This proves statement (0).

Next, using (**) and Corollary 1, we obtain

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Therefore

[TABLE]

[TABLE]

Corollary 2 implies that

[TABLE]

[TABLE]

and

[TABLE]

By the left distributive law, we have

[TABLE]

[TABLE]

Finally, the associativity of multiplication for all $x$, $y\in R$ implies that

[TABLE]

Thus

[TABLE]

and $yb=a\alpha(y)+b\beta(y)+c\gamma(y)$ by formula (**). Substituting the last expression in the right part of equality 1), we get

[TABLE]

[TABLE]

[TABLE]

Comparing the coefficients $a$, $b$ and $c$ in 2) and 3) by equality 1), we derive statements (1)–(3) of the lemma. ∎

4 Local nearrings on $G(p^{m},p^{n},p^{d})$

Let $R$ be a local nearring with identity $i$ whose group $R^{+}$ is isomorphic to the group $G(p^{m},p^{n},p^{d})$. Then ${R^{+}=\langle a\rangle+\langle b\rangle+\langle c\rangle}$ with elements $a$, $b$ and $c$, satisfying relations $ap^{m}=bp^{n}=cp^{d}=0$, $-b+a+b=a+c$ and $-a+c+a=-b+c+b=c$ with $1\leq d\leq n\leq m$, where $a=i$ and each element $x\in R$ is uniquely written in the form ${x=ax_{1}+bx_{2}+cx_{3}}$ with coefficients $0\leq x_{1}<\!p^{m}$, $0\leq x_{2}<\!p^{n}$ and $0\leq x_{3}<\!p^{d}$.

We show that the set $L$ of all non-invertible elements of $R$ is a subgroup of index $p$ in $R^{+}$.

Theorem 2**.**

The following statements hold:**

${L=\langle a\cdot p\rangle+\langle b\rangle+\langle c\rangle}$* and, in particular, the subgroup $L$ is of index $p$ in $R^{+}$ and $|R^{*}|=p^{m+n+d-1}(p-1)$;* 2. 2.

$x=ax_{1}+bx_{2}+cx_{3}$* is an invertible element if and only if ${x_{1}\not\equiv 0\;(\!\!\mod p\;)}$.*

Proof.

Assume that $|R^{+}:L|=p^{t}$, $t>1$. Since $R=R^{*}\cup L$, it follows that

[TABLE]

According to Lemma 7, the group $R^{*}$ is isomorphic to the subgroup $A$ of the automorphism group of $R^{+}$ and so $|R^{*}|$ divides $|\mbox{Aut}R^{+}|$. According to statement 1) of Lemma 4 it is possible only if $t=2$ and $m=n$.

Assume that $|R^{+}:L|=p^{2}$ and $m=n$. If $d=1$, then it is impossible because of [9, Theorem 2]. Now let $d>1$. Since $|R^{+}:\Phi(R^{+})|=p^{2}$ and Corollary 4, we have $L=\Phi(R^{+})$. Hence by Lemma 7, we get $R^{+}=a^{A}\cup\Phi(R^{+})$, which is impossible by Lemma 5. This contradiction shows that our assumption is false and so $|R^{+}:L|=p$.

It is clear that $R/L$ is a nearfield and so the factor-group $R^{+}/L^{+}$ is an elementary abelian $p$-group. Thus for $a\notin L$ we have $ap\in L$ and so ${L=\langle a\cdot p\rangle+\langle b\rangle+\langle c\rangle}$. Therefore $R^{*}=R\setminus L$ and hence

[TABLE]

∎

Applying statement (1) of Theorem 2 to Lemma 8, we get the following formula for multiplying elements $x=ax_{1}+bx_{2}+cx_{3}$ and $y=ay_{1}+by_{2}+cy_{3}$ in the local nearring $R$.

Corollary 5**.**

If $x$, $y\in R$ with $1\leq d\leq n\leq m$ and $xb=a\alpha(x)+b\beta(x)+c\gamma(x)$, then

[TABLE]

[TABLE]

where mappings $\alpha\colon R\to\mathbb{Z}_{p^{m}}$, $\beta\colon R\to\mathbb{Z}_{p^{n}}$ and $\gamma\colon R\to\mathbb{Z}_{p^{d}}$ and the following statements hold:**

• (0)

$\alpha(0)\equiv 0\;(\!\!\mod p^{m})$, $\beta(0)\equiv 0\;(\!\!\mod p^{n})$ and ${\gamma(0)\equiv 0\;(\!\!\mod p^{d})}$ if and only if the nearring $R$ is zero-symmetric;**

• (1)

$\alpha(x)\equiv 0\;(\!\!\mod p)$;**

• (2)

if $\beta(x)\equiv 0\;(\!\!\mod p)$, then $x_{1}\equiv 0\;(\!\!\mod p)$;

• (3)

$\alpha(xy)\equiv x_{1}\alpha(y)+\alpha(x)\beta(y)\;(\!\!\mod p^{m}\;)$;**

• (4)

$\beta(xy)\equiv x_{2}\alpha(y)+\beta(x)\beta(y)\;(\!\!\mod p^{n}\;)$;**

• (5)

$\gamma(xy)\equiv-x_{1}x_{2}\binom{\alpha(y)}{2}-\alpha(x)\beta(x)\binom{\beta(y)}{2}-x_{2}\alpha(x)\alpha(y)\beta(y)$**

[TABLE]

Proof.

Indeed, statements (0), (3)–(5) repeat statements (0)–(4) of Lemma 8. Since $L=\langle a\cdot p\rangle+\langle b\rangle+\langle c\rangle$ by Theorem 2 and $L$ is an $(R,R)$-subgroup in $R$ by statement 2) [1, Lemma 3.2], it follows that $xb\in L$ and hence $\alpha(x)\equiv 0\;(\!\!\mod p)$, proving statement (1). Taking $y=c$, we have $xc=c(x_{1}\beta(x)-x_{2}\alpha(x))$. Thus, if $\beta(x)\equiv 0\;(\!\!\mod p)$, then $xc=0\;(\!\!\mod p\;)$, and so $x\in L$. Thus $x_{1}\equiv 0\;(\!\!\mod p)$ by Theorem 2, proving statement (2). ∎

The following theorem shows that conditions given in Theorem 2 are sufficient for existing of finite local nearrings on $G(p^{m},p^{n},p^{d})$. Moreover, each group $G(p^{m},p^{n},p^{d})$ is the additive group of a nearring with identity.

Theorem 3**.**

For each prime $p$ and positive integers $m$, $n$ and $d$ with $1\leq d\leq n\leq m$ there exists a local nearring $R$ whose additive group $R^{+}$ is isomorphic to the group $G(p^{m},p^{n},p^{d})$.

Proof.

Let $R$ be an additively written group $G(p^{m},p^{n},p^{d})$ with generators $a$, $b$ and $c$ satisfying the relations $|a|=p^{m}$, $|b|=p^{n}$, $|c|=p^{d}$, $b^{-1}ab=ac$ and $a^{-1}ca=b^{-1}cb=c$. Then ${G=\langle a\rangle+\langle b\rangle+\langle c\rangle}$ and each element $x\in R$ is uniquely written in the form $x=ax_{1}+bx_{2}+cx_{3}$ with coefficients $0\leq x_{1}<\!p^{m}$, $0\leq x_{2}<\!p^{n}$ and $0\leq x_{3}<\!p^{d}$. In order to define a multiplication “$\cdot$” on $R$ in such a manner that $(R,+,\cdot)$ is a local nearring.

Assume that $1\leq d\leq n\leq m$ and let the mappings from Corollary 5 be defined by the congruences ${\alpha(x)\equiv 0\;(\!\!\mod p^{m})}$, $\beta(x)\equiv x_{1}\;(\!\!\mod p^{n})$ and $\gamma(x)\equiv 0\;(\!\!\mod p^{d})$ for each $x\in G$. Then

[TABLE]

It suffices to show that the mappings $\alpha:G\to\mathbb{Z}_{p^{m}}$, $\beta:G\to\mathbb{Z}_{p^{n}}$ and $\gamma:G\to\mathbb{Z}_{p^{d}}$ with respect to the multiplication “$\cdot$” satisfy statements (0)–(5) of Corollary 5.

Indeed, $\alpha(0)\equiv 0\;(\!\!\mod p^{m})$, $\beta(0)\equiv 0\;(\!\!\mod p^{n})$ and ${\gamma(0)\equiv 0\;(\!\!\mod p^{d})}$ by the definition. Since ${0\cdot y=a\cdot 0+b\cdot 0+c\cdot 0=0}$ for each $y\in G$, this implies that a multiplication “$\cdot$” is zero-symmetric and so, proving statement (0) of Corollary 5. Indeed, we have $\alpha(x)\equiv 0\;(\!\!\mod p)$ and ${x_{1}\equiv 0\;(\!\!\mod p)}$, if ${\beta(x)\equiv 0\;(\!\!\mod p)}$, so that statements (1) and (2) of Corollary 5 hold. Clearly, we derive statements (3)–(5) of Corollary 5. ∎

As corollary we have the following assertion.

Corollary 6**.**

Each group $G(p^{m},p^{n},p^{d})$ is the additive group of a nearring with identity.