# Rings of congruence preserving functions

• C. J. Maxson
• Frederik Saxinger
Open Access
Article

## Abstract

Let $$C_0 (G)$$ denote the near-ring of congruence preserving functions of the group G. We investigate the question “When is $$C_0 (G)$$ a ring?”. We obtain information externally via the lattice structure of the normal subgroups of G and internally via structural properties of the group G.

## Keywords

Congruence preserving functions 1-affine complete Normal subgroup lattice

## Mathematics Subject Classification

08A30 16Y30 20E15

## 1 Introduction: background and notation

Let $$G=\langle G,+,0\rangle$$ be a group, written additively but not necessarily abelian, with neutral element 0. The structure of the near-ring $$C_0 (G)=\langle C_0 (G),+,\circ \rangle$$ of zero fixing congruence preserving functions on G has been the topic of several previous investigations [1, 4]. In this paper we initiate the study of characterizing those groups G such that $$C_0 (G)$$ is a ring.

This investigation also has roots in universal algebra. Recall that a unary polynomial function $$p:G\rightarrow G$$ is a function that can be written in the form $$p(x):=a_0 +k_0 x+a_1 +k_1 x +\cdots +a_{n-1}+k_{n-1}x+a_n ,x\in G, n$$ a nonnegative integer, $$a_0 ,a_1 ,\dots ,a_n \in G, k_0 ,\dots ,k_{n-1}\in \mathbb {Z}$$. We let $$P_0 (G)=\langle P_0 (G),+,\circ \rangle$$ denote the near-ring of zero preserving polynomial functions on G (See [1, 11].)

The near-ring $$C_0 (G)$$ is a subnear-ring of the near-ring $$M_0 (G):=\{f:G\rightarrow G\mid f(0)=0\}$$ of zero fixing self maps on G where, as usual, the operations in $$M_0 (G)$$ are pointwise addition of functions and composition of functions. Let $$\text {Inn }(G)$$ denote the semigroup (under composition) of inner automorphisms of the group G and let I(G) denote the subnear-ring of $$M_0 (G)$$ generated additively by $$\text {Inn }(G)$$. From the definitions we see $$P_0 (G)=I(G)$$.

Now let $$f\in C_0 (G)$$. Then by definition, for each congruence, $$\rho$$ of G, for each $$x,y\in G,\,x\rho y$$ implies $$f(x)\rho f(y)$$. As is well-known, there is a lattice isomorphism between the congruence lattice, $$\text {Con}(G)$$, of congruences on G and the lattice, $$\eta (G)$$, of normal subgroups of G. Thus $$f\in M_0 (G)$$ is congruence preserving if f is compatible with every normal subgroup of G. That is for $$x,y\in G$$ we have $$f\in C_0 (G)$$ if and only if $$x+H=y+H$$ implies $$f(x)+H=f(y)+H$$, for each $$H\in \eta (G)$$.

Recall for any subgroup H of G the normal closure of H in G, denoted by $$\overline{H}$$, is defined by $$\overline{H}=\bigcap \{N\unlhd G\mid H \subseteq N\}$$. For $$x\in G$$ we write $$\overline{x}$$ for $$\overline{\langle x\rangle }$$, the principal closure of x. Thus we get the following characterization of congruence preserving functions: Let $$f\in M_0 (G)$$. Then $$f\in C_0 (G)$$ if and only if $$f(x)-f(y)\in \overline{x-y}$$ for each $$x,y\in G$$.

Using the above definitions one finds that every zero fixing unary polynomial on G is congruence preserving so $$I(G)=P_0 (G)\subseteq C_0 (G)\subseteq M_0 (G)$$. If every congruence preserving function is also a polynomial then the group G is said to be 1-affine complete. Finite abelian 1-affine complete groups have been characterized by Nöbauer [14]. All 1-affine complete groups of order up to 100 can be found in [17]. In relation to our problem under consideration, $$C_0 (G)$$ will be a ring if $$P_0 (G)$$ is a ring and G is 1-affine complete.

It is known when $$I(G)=P_0 (G)$$ is a ring [8]. A group G is said to be a 2-Engel group if $$[x,[x,y]]=0$$ for all $$x,y\in G$$. Equivalently (see [8, 12, 13]) G is a 2-Engel group if every element of G commutes with all of its conjugates. A 2-Engel group G is nilpotent of class at most 3 and, for finite groups, a 2-Engel group of class 3 must be a group of order 3$$^n$$. The smallest 2-Engel group of class 3 is the Burnside group, B(3, 3), of order 3$$^7$$.

### Theorem 1.1

(Chandy) The near-ring I(G) is a ring if and only if G is a 2-Engel group. Moreover, I(G) is a commutative ring if and only if G is of class 2.

In the remainder of this paper we restrict to finite groups G. In the next section we first show when considering $$C_0 (G)$$ we may restrict to p-groups, p a prime. Using results of Nöbauer [14] we completely characterize those finite abelian groups G such that $$C_0 (G)$$ is a ring.

In Sect. 3 we turn to nonabelian groups and find several necessary conditions for $$C_0 (G)$$ to be a ring. In this section we focus on certain properties of the lattice $$\eta (G)$$ of normal subgroups. In the final section we consider internal properties of the group G and we conclude with a complete answer to our question for groups of order $$p^n ,\,1\le n\le 5,\,p>2$$.

## 2 The abelian case

We begin this section by showing that, for finite groups, we can restrict to p-groups, where as usual p denotes a prime integer. We use results of Nöbauer, [14], to obtain this restriction.

We recall from [14] that a direct sum $$G=G_1 \oplus \cdots \oplus G_n$$ is said to be skew-free if every congruence $$\rho$$ of G is of the form $$\rho =\rho _1 +\cdots +\rho _n$$ where the $$\rho _i$$ are congruences on the $$G_i ,\,i=1,2,\dots ,n$$. In particular when G is a finite nilpotent group with direct sum of its unique Sylow subgroups $$G=S_{p_1}(G)\oplus \cdots \oplus S_{p_n}(G)$$ then G is skew-free. Recall from the introduction that $$\text {Con}(G)$$ is lattice isomorphic to $$\eta (G)$$, the lattice of normal subgroups of G.

### Theorem 2.1

[14, Satz 1] Let $$G=A\oplus B$$ be skew-free. Then the map $$\psi : C_0(G) \rightarrow C_0(A)+C_0(B)$$ given by $$\psi (\rho )=(\rho _A,\rho _B)$$, $$\rho \in C_0(G)$$ is a near-ring isomorphism.

From straight forward calculations one finds the following theorem and corollary.

### Theorem 2.2

Let G be a finite nilpotent group, $$G=S_{p_1}(G)\oplus \cdots \oplus S_{p_n}(G)$$. Then $$C_0 (G)\cong C_0 (S_{p_1}(G))\oplus \cdots \oplus C_0 (S_{p_n}(G))$$.

### Corollary 2.3

Let G be a finite group with Sylow decomposition as in Theorem 2.2. Then $$C_0 (G)$$ is a ring if and only if $$C_0 (S_{p_i}(G))$$ is a ring for each $$i\in \{1,2,\dots ,n\}$$.

Hence we only need to consider groups, G, of prime power order when investigating the structure of the near-ring, $$C_0 (G)$$, of zero fixing congruence preserving functions on G.

We now turn to abelian groups. We need a further result of Nöbauer [14].

### Theorem 2.4

[14, Satz 3,4] Let p be a prime, $$\mathbb {Z}_{p^\alpha }$$ be the cyclic group of order $$p^\alpha$$ and let $$A=\mathbb {Z}_{p^{\alpha _1}} \oplus \cdots \oplus \mathbb {Z}_{p^{\alpha _n}}$$ with $$\alpha _1 \ge \alpha _2 \ge \cdots \ge \alpha _n$$. Then A is 1-affine complete if and only if one of the following conditions holds:
1. (a)

$$n>1,\,\alpha _1 =\alpha _2 ,\,p$$ arbitrary,

2. (b)

$$n>1,\,\alpha _1 =\alpha _2 +1,\,p=2$$ or

3. (c)

$$n=1,\,\alpha _1 =1,\,p=2$$.

### Theorem 2.5

Let A be an abelian group of prime power order. The following are equivalent:
1. (1)

A is 1-affine complete;

2. (2)

$$C_0 (A)$$ is a ring;

3. (3)

$$C_0 (A)$$ is a commutative ring.

### Proof

Since A is abelian, A is a 2-Engel group of nilpotency class at most 2 so $$I(A)=P_0 (A)$$ is a commutative ring. If A is 1-affine complete then $$C_0 (A)$$ is a ring. Thus (1) implies (3).

It is clear that (3) implies (2) so it remains to show (2) implies (1). We assume A is not 1-affine complete and show $$C_0 (A)$$ is not a ring. From Theorem 2.4 we know the form A has to be so that A is not 1-affine complete.
1. Case (i)
$$|A|=p^m$$ for some prime $$p>2$$. In this case we must have $$A\cong \mathbb {Z}_{p^{\alpha _1}} \oplus \cdots \oplus \mathbb {Z}_{p^{\alpha _n}}$$ with $$n=1$$ or $$\alpha _1 >\alpha _2 \ge \cdots \ge \alpha _n$$. We let $$\mathbb {Z}_{p^{\alpha _1}} =\langle g\rangle$$ and let $$D=\{x\in A\mid p^{\alpha _1 -1}x=0\}=\{(a_1 ,\dots ,a_n )\mid a_1 \in \langle pg\rangle \}$$. Define $$c:A\rightarrow A$$ by
\begin{aligned} c(x)= {\left\{ \begin{array}{ll} (0,\dots ,0), &{} x\in D\\ (p^{\alpha _1 -1}g,0,\dots ,0), &{} x\notin D. \end{array}\right. } \end{aligned}
We show $$c\in C_0 (A)$$. To this end, let $$x,y\in A$$. If $$x,y\in D$$ or $$x,y\notin D$$ then $$c(x)-c(y)=(0,\dots ,0)\in \overline{x-y}$$ so we take $$x\notin D,\,y\in D$$. Thus $$c(x)-c(y)=(p^{\alpha -1}g,0,\dots ,0)$$. Now $$p^{\alpha _1 -1}(x-y)=(p^{\alpha _1 -1}kg,0,\dots ,0),\,p\not \mid k$$ so $$x\notin D$$. But then $$(p^{\alpha _1 -1}g,0,\dots ,0)\in \overline{x-y}$$ so $$c\in C_0 (A)$$. Further, $$[c(\text {id}+\text {id})]((g,0,\dots ,0))=c(\text {id}((g,0,\dots ,0))+\text {id}((g,0,\dots ,0)))=c((2g,0,\dots ,0))=(p^{\alpha _1 -1}g,0,\dots ,0)$$ while $$[c\circ \text {id}+c\circ \text {id}]((g,0,\dots ,0))=2c((g,0,\dots ,0))=2(p^{\alpha _1-1}g,0,\dots ,0)\ne (p^{\alpha _1-1}g,0,\dots ,0)$$. Hence $$C_0(G)$$ is not a ring.

2. Case (ii)
$$|A|=2^m$$. In this case $$A=\mathbb {Z}_{2^{\alpha _1}}\oplus \cdots \oplus \mathbb {Z}_{2^{\alpha _n}}$$ with either $$n>1$$ and $$\alpha _1 -1>\alpha _2 \ge \cdots \ge \alpha _n$$ or $$n=1$$ and $$\alpha _1 >1$$. Again we handle both cases together and as above we let $$\mathbb {Z}_{2^{\alpha _1}}=\langle g\rangle$$. Let $$D=\{x\in A\mid 2^{\alpha _1 -2}x=0\}=\{(a_1 ,\dots ,a_n )\mid a_1 \in \langle 4g\rangle \}$$ and further let $$\bar{D}=D\cup (D+(2g,0,\dots ,0))$$. Define $$c:A\rightarrow A$$ by
\begin{aligned} c(x)= {\left\{ \begin{array}{ll} (0,0,\dots ,0), &{} x\in \bar{D}\\ (2^{\alpha _1 -2}g,0,\dots ,0), &{} x\notin \bar{D}. \end{array}\right. } \end{aligned}
As above we show $$c\in C_0 (A)$$. Let $$x,y\in A,\,x\notin \bar{D},\,y\in \bar{D}$$. Then $$c(x)-c(y)=(2^{\alpha _1 -2}g,0,\dots ,0)$$. Now since $$x\notin \bar{D},\,x=(kg,0,\dots ,0)$$ where $$k\in \{1,3\}$$. Thus $$2^{\alpha _1 -2}(x-y)=(2^{\alpha _1 -2}kg,0,\dots ,0)$$ so $$(2^{\alpha _1 -2}g,0,\dots ,0)\in \overline{x-y}$$. Using $$f=\text {id},\,h=3\cdot \text {id}$$ we have $$[c(f+g)]((g,0,\dots ,0)) = [c(\text {id}+3\cdot \text {id})]((g,0,\dots ,0))=c(\text {id}((g,0,\dots ,0))+3\cdot \text {id}((g,0,\dots ,0)))=c((4g,0,\dots ,0))=(0,\dots ,0)\ne c((g,0,\dots ,0))+c((3g,0,\dots ,0))$$. So $$C_0 (A)$$ is not a ring. $$\square$$

We now have a characterization of those finite abelian groups A for which $$C_0 (A)$$ is a ring.

### Corollary 2.6

Let A be a finite abelian group. The following are equivalent:
1. (1)

A is 1-affine complete;

2. (2)

$$C_0 (A)$$ is a ring;

3. (3)

$$C_0 (A)$$ is a commutative ring.

### Proof

As in Theorem 2.5 (1) implies (3) and (3) implies (2). Using the Sylow decomposition of A we see that (2) implies (1) follows from Theorem 2.5 and Nöbauer [14, Lemma 5]. $$\square$$

In the next section we give several necessary conditions for $$C_0 (G)$$ to be a ring.

## 3 Lattice conditions

We start with some conditions on the congruence lattice, $$\text {Con}(G)$$. Since $$\text {Con}(G)$$ is lattice isomorphic to the normal subgroup lattice, $$\eta (G)$$, we often state our properties in terms of normal subgroups.

We recall that G must be a 2-Engel group, and thus nilpotent of class at most 3, for $$C_0 (G)$$ to be a ring.

Our first lattice concept is that of splitting pair. This property has been used previously [3, 5, 15]. Let $$D,E\in \eta (G),\,D\subset G,\,\{0\}\subset E$$. The pair (DE) is called a splitting pair if for each $$N\in \eta (G),\,N \subseteq D$$ or $$N \supseteq E$$. If G has a splitting pair then G splits.

Now let (DE) be a splitting pair for G and let $$0\ne b\in E$$. Define $$f:G\rightarrow G$$ by
\begin{aligned} f(x)= {\left\{ \begin{array}{ll} 0, &{} x\in D\\ b, &{} \text {otherwise.} \end{array}\right. } \end{aligned}
We show $$f\in C_0 (G)$$. Let $$x,y\in G$$ and let $$H\in \eta (G)$$ with $$x+H=y+H$$. If $$E\subseteq H$$ then since $$f(x)-f(y) \in \{-b,0,b\} \subseteq E$$ we get $$f(x)+H=f(y)+H$$. If $$H\subseteq D$$ and $$x\in D$$ then $$x+H\subseteq D$$ and so $$y+H\subseteq D$$ which means $$y\in D$$. By symmetry if $$x\notin D$$ then $$y\notin D$$, hence in both cases $$f(x)+H=f(y)+H$$. This establishes that $$f\in C_0 (G)$$. Now if $$C_0 (G)$$ is a ring then for $$v\notin D,\,[f\circ (\text {id}+\text {id})](v)=[(f\circ \text {id}+f\circ \text {id})](v)$$ or $$f(2v)=2f(v)$$. If $$2v \notin D$$ then $$b=2b$$, which contradicts $$b\ne 0$$. Therefore $$2v \in D$$ and further $$0=2b$$. Since $$0\ne b$$ was arbitrary in E we get $$E\cong (\mathbb {Z}_2 )^n$$ for some $$n>0$$. This establishes:

### Theorem 3.1

Let G be a finite group such that $$2\not \mid | G|$$. If G splits then $$C_0 (G)$$ is not a ring.

### Proof

From the above discussion, when G splits then G has a subgroup $$E\cong (\mathbb {Z}_2 )^n ,\,n>0$$ which is a contradiction since $$2\not \mid | G|$$. $$\square$$

In particular if G is a finite p-group, $$p>2$$, and G splits, then $$C_0 (G)$$ is not a ring. The situation is different in the non-split case as the next examples illustrate. These examples and some of the calculations have been done with GAP using the package Sonata [2].

### Example 3.2

1. (1)

Group with GAP index $$3^7 /6010$$. $$G=\langle e_1 ,e_2 ,e_3 ,e_4 ,c_1 ,c_2 ,c_3 \rangle ,\,3e_i =3c_j =0,\,[e_i ,c_j ]=[c_k ,c_j ]=0,\,[e_1 ,e_2 ]=c_1 ,\,[e_1 ,e_3 ]=c_2 ,\,[e_2 ,e_3 ]=c_3$$ otherwise $$[e_l ,e_m ]=0,\,i=1,2,3,4,\,j,k=1,2,3$$. Thus G is a group of exponent 3, nilpotent of class 2 with $$G^\prime =\langle c_1 ,c_2 ,c_3 \rangle \subseteq Z(G)$$ [9]. From GAP, G does not split but is 1-affine complete so $$C_0 (G)$$ is a ring since nilpotent of class 2 means G is 2-Engel.

2. (2)
GAP index $$3^7 /6576$$. $$G=\langle e_1 ,e_2 ,e_3 ,e_4 ,c_1 ,c_2 ,c_3 \rangle ,\,3e_i =3c_i =0,\,[e_i ,c_j ]=[c_k ,c_j ]=0,\,i=1,2,3,4,\,j,k=1,2,3$$ with $$[e_1 ,e_2 ]=c_1 ,\,[e_1 ,e_3 ]=[e_2 ,e_4 ]=c_2 ,\,[e_3 ,e_4 ]=c_3$$, otherwise $$[e_m ,e_l ]=0$$. Again G is of exponent 3, nilpotent of class 2, with $$G^\prime =\langle c_1 ,c_2 ,c_3 \rangle \subseteq Z(G)$$ [9]. Using GAP, G does not split and is not 1-affine complete. We show $$C_0 (G)$$ is not a ring. For $$x\in G,\,x=\alpha e_1 +\beta e_2 +\gamma e_3 +\delta e_4 +d$$ where $$d\in G^\prime$$. Define $$f:G\rightarrow G$$ by
\begin{aligned} f(x)= {\left\{ \begin{array}{ll} 2(c_1 +c_2 +c_3 ), &{} \text {if }\beta +\gamma =3,\\ 2c_3 , &{} \text {if }\beta =0,\gamma \ne 0,\\ 2c_1 , &{} \text {if }\beta \ne 0,\gamma =0,\\ 2c_1 +c_2 +2c_3 , &{} \text {if }\beta =\gamma \ne 0,\\ 0, &{} \text {otherwise.} \end{array}\right. } \end{aligned}
Calculations show that $$f\in C_0 (G)$$. Now $$f\circ (\text {id}+\text {id})(e_3 )=f(2e_3 )=2c_3$$ while $$(f\circ \text {id}+f\circ \text {id})(e_3 )=2c_3 +2c_3 =c_3 \ne 2c_3$$. Thus $$C_0 (G)$$ is not a ring.

Above we denoted the normal closure of $$x\in G$$ by $$\overline{x}$$. For $$x\in G$$, let $$P_0 (G)x=\{p(x)\mid p\in P_0 (G)\}$$ and $$C_0 (G)x=\{c(x)\mid c\in C_0 (G)\}$$.

### Lemma 3.3

Let G be a group and let $$x\in G$$.
1. (1)

$$P_0 (G)x=\overline{x}$$.

2. (2)

$$P_0 (G)x=C_0 (G)x$$. If further G is 2-Engel then

3. (3)

$$\overline{x}$$ is abelian;

4. (4)

$$\langle C_0 (G),+\rangle$$ is an abelian group.

5. (5)

If G is nilpotent of class at most 2 then $$\overline{x}=\langle x\rangle +[x,G]$$.

### Proof

1. (1)

Clearly $$P_0 (G)x\subseteq \overline{x}$$. On the other hand, $$P_0 (G)x$$ is a normal subgroup of G containing x, so $$\overline{x}\subseteq P_0 (G)x$$.

2. (2)

One has $$P_0 (G)x\subseteq C_0 (G)x$$. For $$c\in C_0 (G),\,c(x)-c(0)\in \overline{x-0}$$ so $$c(x)\in \overline{x}$$. Thus $$C_0 (G)x\subseteq \overline{x}=P_0 (G)x$$.

3. (3)

When G is 2-Engel, $$P_0 (G)$$ is a ring so with 1) we get that $$P_0 (G)x$$ is an abelian group.

4. (4)

Follows from 2) since G is 2-Engel.

5. (5)

In [10] Ecker shows $$p\in P_0 (G)$$ has the form $$p(x)=kx+[x,g]$$ for some integer k and $$g\in G$$ when G is nilpotent of class at most 2. Thus $$P_0 (G)x\subseteq \langle x\rangle +[x,G]$$. But $$\langle x\rangle +[x,G]\subseteq \overline{x}=P_0 (G)x$$. $$\square$$

We next give a characterization of those groups G for which $$C_0 (G)$$ is a ring. The usefulness of this result is somewhat limited since it requires knowledge of all $$c\in C_0 (G)$$.

### Theorem 3.4

Let G be a finite 2-Engel group. Then $$C_0 (G)$$ is a ring if and only if $$\left. c\right| _{\overline{x}}\in \text {End}(\overline{x})$$ for each $$c\in C_0 (G)$$ and $$x\in G$$.

### Proof

Let cfg be arbitrary in $$C_0 (G)$$ and let $$x\in G$$. From Lemma 3.3 $$\langle C_0 (G),+\rangle$$ is an abelian group. Suppose $$\left. c\right| _{\overline{x}}\in \text {End}(\overline{x})$$ for each $$x\in G$$. Then $$c\circ (f+g)(x)=c(f(x)+g(x))=c(f(x))+c(g(x))=(c\circ f+c\circ g)(x)$$, since $$f(x), g(x) \in \bar{x}$$ by (1) and (2) of Lemma 3.3. Thus $$C_0 (G)$$ is a ring.

For the converse let $$a,b\in \overline{x}$$. Thus by 1) and 2) of Lemma 3.3 there exist $$h,l\in C_0 (G),\,a=h(x),\,b=l(x)$$. Now let $$c\in C_0 (G)$$. It is clear that $$c(\overline{x})\subseteq \overline{x}$$. Moreover since $$C_0 (G)$$ is a ring we have $$c(a+b)=c(h(x)+l(x))=c(h(x))+c(l(x))=c(a)+c(b)$$ which shows $$c\in \text {End}(\overline{x})$$. $$\square$$

Therefore if one can construct a congruence preserving function that is not linear on some normal closure $$\overline{x},\,x\in G$$, where G is 2-Engel, then $$C_0 (G)$$ is not a ring. In the next example, using Theorem 3.4, we show that the result of Theorem 3.1 on groups which split is not true for $$p=2$$.

### Example 3.5

Let G be a semidirect product of $$\mathbb {Z}_4$$ and $$\mathbb {Z}_4$$: $$G=\langle x,y\mid 4x=4y=0,\,y+x=3x+y\rangle$$. We have $$Z(G)=\langle 2x,2y\rangle$$ and one verifies that $$D=Z(G)$$ and $$E=\langle 2x\rangle$$ is a splitting pair for $$\eta (G)$$. Define $$c:G\rightarrow G$$ by
\begin{aligned} c(w)= {\left\{ \begin{array}{ll} 0, &{} w\in Z(G),\\ 2x, &{} w\notin Z(G). \end{array}\right. } \end{aligned}
Let $$u,v\in G$$. If $$u,v\in Z(G)$$ or $$u,v\notin Z(G)$$ then $$c(u)-c(v)=0\in \overline{u-v}$$. If $$u\notin Z(G)$$ and $$v\in Z(G)$$ then $$u-v\notin Z(G)$$ and so $$\overline{u-v}\supseteq \langle 2x\rangle$$ since $$(Z(G),\langle 2x\rangle )$$ is a splitting pair. Thus $$c(u)-c(x)=2x\in \overline{u-v}$$ so $$c\in C_0 (G)$$ but $$c\notin P_0 (G)$$ since $$c(x)=2x$$ while $$c(y)=2x\ne 2y$$.

Using GAP one finds $$| P_0 (G)|=16$$ and $$| C_0 (G)|=32$$ so we have $$C_0 (G)=P_0 (G)+\langle c\rangle =\{p+lc\mid l\in \{0,1\},\,p\in P_0 (G)\}$$. For $$w\notin Z(G)$$, calculations show that c is linear on $$\overline{w}$$. Thus for all $$w\in G,\,\left. c\right| _{\overline{w}}\in \text {End}(\overline{w})$$. Thus for each $$p\in P_0 (G),\,p+c$$ is linear on each $$\overline{w}$$ so from Theorem 3.4 $$C_0 (G)$$ is a ring.

We turn to another lattice condition, a particular case of a splitting pair. If (DE) is a splitting pair for $$\eta (G)$$ and $$D=E$$ we say D is a cutting element and G cuts.

### Lemma 3.6

Let G be a finite p-group of nilpotency class at most 2 such that I is a cutting element for $$\eta (G)$$. Then $$I\subseteq Z(G)$$.

### Proof

Let T be the maximal cutting element for $$\eta (G)$$ which exists since G is finite and cutting elements form a chain in $$\eta (G)$$. We have $$T\supseteq I$$. If G is abelian then $$I\subseteq Z(G)=G$$ so we take G of class 2, hence $$G^\prime \subseteq Z(G)$$. If T is also a maximal element in $$\eta (G)$$ then G has a unique maximal normal subgroup. Thus from [16], G is cyclic, contrary to G being of class 2. Thus we suppose T is not a maximal element in $$\eta (G)$$. If $$T\subseteq G^\prime$$ then T, hence I, is contained in Z(G). To complete the proof we show $$G^\prime \subset T$$ cannot occur. Suppose $$G^\prime \subset T$$ and let $$N\in \eta (G)$$ be maximal with $$G^\prime \subseteq N \subset T$$. Since $$G/G^\prime$$ is abelian, G / N is also abelian. Therefore G / N has a unique minimal normal subgroup T / N. But this means that G / N is subdirectly irreducible and (from [7] p. 64) G / N is cyclic. However this contradicts the fact that T is the unique maximal cutting element but not a maximal element in $$\eta (G)$$. Thus we have $$I\subseteq Z(G)$$. $$\square$$

### Theorem 3.7

Let G be a finite nonabelian p-group such that G cuts. Then $$C_0 (G)$$ is not a ring.

### Proof

Let I be a cutting element. If G is of nilpotency class greater than 3 then G is not 2-Engel, hence $$C_0 (G)$$ is not a ring. Further if $$p>2$$ then from Theorem 3.1, $$C_0 (G)$$ is not a ring. Therefore $$p=2$$ and G is nilpotent of class 2, $$| G|=2^n ,\,n\ge 3$$. From Lemma 3.6 we get $$I\subseteq Z(G)$$.

Let $$T_1$$ be a transversal of G / I with $$0\in T_1$$. Let $$t\in T_1 -\{0\}$$ and define $$T_2 =(T_1 \backslash (t+t+I))\cup \{t+t\}$$. We note $$T_2$$ is a transversal of G / I with $$\{t,t+t\}\subseteq T_2$$. Suppose first that $$t+t\notin I$$ so $$0\in T_2$$. Let $$0\ne e$$ be in I and define $$h:G\rightarrow G$$ by
\begin{aligned} h(x)= {\left\{ \begin{array}{ll} e, &{} x\in t+t+I,\\ 0, &{} \text {otherwise.} \end{array}\right. } \end{aligned}
We note that $$h(0)=0$$. To show $$h\in C_0 (G)$$ it suffices to show for $$r\in T_2 -\{t+t\},\,d_1 ,d_2 \in I$$ that $$h(t+t+d_1 )-h(r+d_2 )\in \overline{t+t+d_1 -d_2 -r}$$, that is, we must show $$e\in \overline{t+t+d_1 -d_2 -r}$$. We first observe that $$t+t+d_1 -d_2 -r=t+t-r+d_1 -d_2 \notin I$$ since $$r\in T_2 -\{t+t\}$$. Therefore $$\overline{t+t+r+d_1 -d_2}\nsubseteq I$$ and, since I cuts $$\eta (G),\,I\subseteq \overline{t+t+r+d_1 -d_2}$$ giving the desired result that $$h\in C_0 (G)$$. But $$h(t+t)=e\ne h(t)+h(t)$$. Since h is not linear on $$\overline{t},\,C_0 (G)$$ is not a ring.

Suppose next we have $$0\ne t+t\in I$$. Using $$T_1$$ we define $$f:G\rightarrow G$$ by $$f(x)=j$$ where $$x=r+j,\,r\in T_1 ,\,j\in I$$. Since $$0\in T_1 ,\,f(0)=0$$. For $$x=r_1 +j_1 ,\,y=r_2 +j_2 ,\,f(x)-f(y)=j_1 -j_2$$. If $$r_1 =r_2 ,\,j_1 -j_2 \in \overline{r_1 -r_2 +j_1 -j_2}$$. If $$r_1 \ne r_2$$ then $$r_1 -r_2 +j_1 -j_2 \notin I$$ so $$I\subseteq \overline{r_1 -r_2 +j_1 -j_2}$$, hence $$f(x)-f(y)=j_1 -j_2 \in \overline{x-y}$$. Thus $$f\in C_0 (G)$$. Since $$t+t\in I,\,t+t=0+t+t$$ which means $$f(t+t)=t+t$$. But $$t=t+0$$ so $$f(t)=0=f(t)+f(t)\ne f(t+t)$$. This shows that $$C_0 (G)$$ is not a ring.

For the final case we have $$t+t\in I$$ and $$t+t=0$$. Define $$l:G\rightarrow G$$ by
\begin{aligned} l(x)= {\left\{ \begin{array}{ll} 0, &{} x\in I,\\ j, &{} x\notin I\text { and }x=r+j,\,r\in T_2 =T_1 . \end{array}\right. } \end{aligned}
For $$x\notin I,\,y\in I$$ say $$x=r+j,\,r\in T_2 ,\,j\in I$$ we have $$l(x)-l(y)=j$$. Moreover $$I\subseteq \overline{r+j-y}$$ since $$r+j-y\notin I$$. Thus $$l\in C_0 (G)$$. Since $$t+t\in I$$ for any $$0\ne i\in I,\,l(t+t+i)=0$$. Further since $$t\notin I,\,I\subseteq \overline{t}$$, hence $$t+i\in \overline{t}$$. Now $$l(t)+l(t+i)=0+i\ne 0$$ so l is not linear on $$\overline{t}$$. Thus in all cases we have found when G is cut, $$C_0 (G)$$ is not a ring. $$\square$$

## 4 Structural conditions

In this section we focus on group theoretical properties of a group G to determine when $$C_0 (G)$$ is a ring. We restrict to nilpotency class 2 and p-groups $$p\ge 3$$.

### Theorem 4.1

Let G be a nonabelian p-group, $$p>2$$ such that $$G^\prime$$ is cyclic. Then $$C_0 (G)$$ is not a ring.

### Proof

Let $$x\in G-Z(G)$$. Thus $$\{0\}\ne [x,G]\subseteq [G,G]$$. By hypothesis, $$G^\prime =\langle r\rangle$$ for some $$r\in G$$ so we have $$\overline{r}=\langle r\rangle =G^\prime$$.

Let $$\langle r^\prime \rangle$$ be the unique subgroup of order p in $$G^\prime$$. Since [xG] is a nonzero cyclic subgroup of $$G^\prime$$, $$[x, G^\prime ]$$ contains a cyclic subgroup of order p of $$G^\prime$$, hence we have $$\langle r^\prime \rangle$$ is a subgroup of $$[x,G^\prime ]$$. Moreover, for each $$g\in G,\,p(-g+r^\prime +g)=0$$ so $$-g+r^\prime +g$$ is in $$\langle r^\prime \rangle$$ which in turn leads to the fact that $$\langle r^\prime \rangle$$ is a normal subgroup of [xG]. Therefore, for $$x\in G-Z(G)$$, $$\overline{r^\prime }=\langle r^\prime \rangle \subseteq [x,G]\subseteq \overline{x}$$.

Define $$h:G\rightarrow G$$ by
\begin{aligned} h(x)= {\left\{ \begin{array}{ll} r^\prime , &{} x\notin Z(G),\\ 0, &{} x\in Z(G). \end{array}\right. } \end{aligned}
We show $$h\in C_0 (G)$$. To this end let $$u\notin Z(G),\,v\in Z(G)$$. Then $$h(u)-h(v)=r^\prime$$ which is in $$\overline{u-v}$$ since $$u-v\notin Z(G)$$. This gives $$h\in C_0 (G)$$. For $$w\notin Z(G),\,2w=w+w\notin Z(G)$$ since $$p>2$$ so if $$2w \in Z(G)$$, we would have $$w \in Z(G)$$, a contradiction. From this observation, $$h(w+w)=r^\prime \ne 2r^\prime =h(w)+h(w)$$. Therefore $$C_0 (G)$$ is not a ring. $$\square$$

We actually have a little more.

### Corollary 4.2

If G is a nonabelian p-group, $$p>2$$, such that $$G^\prime$$ is cyclic, then $$\eta (G)$$ splits.

### Proof

From the proof of Theorem 4.1 we get if $$N\trianglelefteq G$$ and $$N\nsubseteq Z(G)$$ then for $$x\in N-Z(G),\,N\supseteq \overline{x}\supseteq \overline{r^\prime }$$. Thus $$\langle Z(G),\overline{r^\prime }\rangle$$ is a splitting pair for $$\eta (G)$$. $$\square$$

### Corollary 4.3

Let G be a p-group, $$p>2$$, of nilpotency class 2. If G is 2-generated (generated by 2 elements) then $$C_0 (G)$$ is not a ring.

### Proof

If $$G=\langle x,y\rangle$$ then one finds $$G^\prime =\langle [x,y]\rangle$$ so $$G^\prime$$ is cyclic [6]. The result now follows from the above theorem. $$\square$$

We remark that Example 3.5 shows that Corollary 4.3 does not hold for $$p=2$$.

Recall that a group G is abelian by cyclic, or G is said to be an extension of an abelian group by a cyclic group if there exists an abelian normal subgroup A of G such that G / A is cyclic. For finite G one may always take A to be a maximal abelian normal subgroup.

### Theorem 4.4

Let G be a nonabelian p-group, $$p>2$$, of nilpotency class 2 which is abelian by cyclic. Then $$C_0 (G)$$ is not a ring.

### Proof

We let A be a maximal abelian normal subgroup such that $${G}/{A}\cong \mathbb {Z}_{p^k},\,k$$ a positive integer. Let $${G}/{A}=\langle b+A\rangle$$ so $$G=\langle A,b\rangle$$. Since G is nonabelian there exists $$a_1 \in A$$ such that $$[a_1 ,b]\ne 0$$. Every $$x \in G$$ can be decomposed into a sum of the form $$x=a+\beta b+c$$, with $$a\in A,\, \beta \in \mathbb {Z},\,c\in G^\prime$$. (Recall the basic assumption that G is of class 2 so $$[G,G]\subseteq Z(G)$$.)

Using this decomposition we define $$f:G\rightarrow G$$ by
\begin{aligned} f(x)= {\left\{ \begin{array}{ll} [b,a_1], &{} p\not \mid \beta ,\\ 0, &{} p\mid \beta . \end{array}\right. } \end{aligned}
Let $$x \in G$$. We note $$[x,a_1 ]=\beta [b,a_1 ]$$ so when $$p\not \mid \beta ,\,[b,a_1 ]\in \overline{x}$$. Also, $$f\in C_0 (G)$$. For if $$u,v\in G,\,u=a+\beta b+c,\,v=a^\prime +\beta ^\prime b+c^\prime ,\,a,a^\prime \in A,\,\beta ,\beta ^\prime \in \mathbb {Z},\,c,c^\prime \in [G,G]$$ with $$p\not \mid \beta$$ and $$p\mid \beta ^\prime$$ then $$p\not \mid (\beta -\beta ^\prime )$$ so $$[b,a_1]\in \overline{u-v}$$. Thus $$f(u)-f(v)=[b,a_1]\in \overline{u-v}$$. For $$u=a+\beta b+c$$ with $$p\not \mid \beta ,\,f(u)+f(u)=[b,a_1 ]+[b,a_1 ]$$ while $$f(u+u)=f(2u)=[b,a_1 ]$$. Using Theorem 3.4, $$C_o (G)$$ is not a ring. $$\square$$

### Corollary 4.5

Let G be a nonabelian p-group, $$p>2$$, of class 2 such that there exists $$g\in G$$ with $${G}/{C_G (g)}$$ cyclic. Then $$C_0 (G)$$ is not a ring.

### Proof

Let $${G}/{C_G (g)}=\langle b+C_G (g)\rangle$$ so $$G=\langle C_G (g),b\rangle$$. For $$x\in G,\,x=w+\beta b+c,\,w\in C_G (g),\,\beta \in \mathbb {Z},\,c\in G^\prime$$ and since $$b\notin C_G (g),\,[b,g]\ne 0$$. Now $$[x,g]=\beta [b,g]$$. The remainder of the proof is as above and is omitted. $$\square$$

As we did following Theorem 4.1, we again show that under the hypothesis of Theorem 4.4, $$\eta (G)$$ splits.

### Theorem 4.6

Let G be a p-group, nilpotent of class 2, which is abelian by cyclic. Then $$\eta (G)$$ splits.

### Proof

As above we let A be a maximal abelian normal subgroup with $${G}/{A}=\langle b+A\rangle$$. Then $$G=\langle A,b\rangle$$ and $$b\notin A$$ so there exists $$a_1 \in A,\,[a_1 ,b]\ne 0$$. Let $$A=\langle a_1 ,a_2 ,\dots ,a_n \rangle$$ so $$G=\langle a_1 ,\dots ,a_n ,b\rangle$$. Let $$A_0 =\langle a_1 ,\dots ,a_n ,pb\rangle$$. If $$pb=0$$ then $$A_0 =A$$ and the same type of argument works. We first show that $$A_0$$ is a normal subgroup of G. Let $$g=a+\beta b+c\in G,\,a\in A,\,\beta \in \mathbb {Z},\,c\in G^\prime$$. It suffices to show $$-g+pb+g\in A_0$$. To this end, $$-g+pb+g=-c-\beta b-a+pb+a+\beta b+c=-\beta b-a+pb+a+\beta b=-\beta b+pb+[pb,a]+\beta b=pb+[pb,a]\in A_0$$. Therefore $$A_0 \trianglelefteq G$$.

Further $$A_0 \subset G$$. For if $$A_0 =G$$ then $$b\in A_0=\langle a_1, a_2, \dots , a_n, pb\rangle$$. From this, $$b=a+\alpha (pb)+c$$ where $$a \in A$$ and c is a sum of commutators. Since A is a maximal abelian normal subgroup we have $$Z(G) \subseteq A$$ and since G is nilpotent of class 2, $$[G,G] \subseteq Z(G) \subseteq A$$. Thus $$c\in A$$. Hence $$(1-\alpha p)b\in A$$ which implies $$b \in A$$ since $$1-\alpha p$$ is invertible modulo p, a contradiction. Thus $$A_0 \ne G$$.

Now let $$N\in \eta (G)$$ such that $$N\nsubseteq A_0$$. For $$n\in N-A_0 ,\,n=a+\delta b+c ,\,a\in A,\,p\not \mid \delta ,\,c\in G^\prime$$, hence $$[n,a_1 ]=\delta [b,a_1 ]$$ and since $$p\not \mid \delta ,\,0\ne [b,a_1 ]\in \overline{n}\subseteq N$$. From this, $$\langle [b,a_1 ]\rangle \subseteq N$$ for each $$N\in \eta (G)$$ such $$N\nsubseteq A_0$$. This shows that $$(A_0 ,\langle [b,a_1 ]\rangle )$$ is a splitting pair for $$\eta (G)$$. $$\square$$

In Theorems 4.6 and 4.1 one has the situation where G has a partition $$G=X\cup (G-X)$$ with the property that $$\bigcap \{ \overline{u} \mid u \in X\} \ne \{0\}$$ and, for each $$u\in X$$, for each $$v\in G-X,\,u-v\in X$$. It is an open question if this condition implies the splitting of $$\eta (G)$$.

We apply the above results to p-groups, $$p>2$$, of small order. Let G be a group of order $$p^n ,\,p>2,\,1\le n\le 5$$. When $$n=1$$ or $$n=2$$, G is abelian so $$C_0 (G)$$ is a ring if and only if $$G\cong \mathbb {Z}_p +\mathbb {Z}_p$$. Thus we take $$n\ge 3$$ and since the abelian case is known from Theorems 2.4 and 2.5 we restrict to nonabelian groups.

### Theorem 4.7

Let G be a nonabelian p-group, $$p>2$$ of order $$p^n ,\,3\le n\le 5$$ such that G is nilpotent of class 2. Then $$C_0 (G)$$ is not a ring.

### Proof

1. (i)

$$n=3.$$ Since g is nonabelian we have $$| Z(G)|=p$$. Thus $$\{0\}\ne G^\prime \subseteq Z(G)$$, hence $$G^\prime$$ is cyclic and the result follows from Theorem 4.1.

2. (ii)

$$n=4.$$ Let $$\Phi (G)$$ denote the Frattini subgroup of G. We know $$G^\prime \subseteq \Phi (G)$$ and if $$|{G}/{\Phi (G)}|=p^k$$ then G is generated by k elements [16]. If $$|\Phi (G)|=p$$ then $$G^\prime$$ is cyclic while if $$|\Phi (G)|=p^2$$ then G is generated by 2 elements. Using Theorem 4.1 and Corollary 4.3 we see that $$C_0 (G)$$ is not a ring. If $$|\Phi (G)|=p^3$$ then G has a unique maximal normal subgroup which cuts G. The result now follows from Theorem 3.7.

3. (iii)

$$n=5$$. If $$|\Phi (G)|=p$$ or $$p^3$$ or $$p^4$$ then as in the above case we have $$C_0 (G)$$ is not a ring. It remains to consider $$|\Phi (G)|=p^2$$. We must have $$G^\prime =\Phi (G)$$ for if $$G^\prime \ne \Phi (G)$$ then $$G^\prime$$ is cyclic and we are finished. Thus we have $$G^\prime =\Phi (G)\subseteq Z(G)$$ since G is of class 2. For $$x\in G-Z(G)$$ define $$\varphi _x :G\rightarrow G$$ by $$\varphi _x (w)=[x,w]$$. Since G is of class 2, $$\varphi _x$$ is an endomorphism of G and $$\ker \varphi _x =C_G (x)\supseteq \langle x\rangle +Z(G)$$ while $$Im\varphi _x =[x,G]\subseteq \overline{x}$$. We have $$|\langle x\rangle +Z(G)|=p|Z(G)|$$ so if $$G^\prime \subset Z(G)$$ then $$|Z(G)|=p^3$$ and $$|\ker \varphi _x |=|C_G (x)|=|\langle x\rangle +Z(G)|=p^4$$. But this means G is abelian by cyclic so the result follows from Theorem 4.4. Thus we take $$|G^\prime |=|\Phi (G)|=|Z(G)|=p^2$$. Thus $$|\ker \varphi _x |=|C_G (x)|\ge |\langle x\rangle +Z(G)|=p^3$$. If $$|C_G (x)|=p^4$$ then the result follows from Corllary 4.5, so we take $$|\ker \varphi _x |=|C_G (x)|=p^3$$. But then $$|Im\varphi _x |=|{G}/{\ker \varphi _x}|=p^2$$. Thus $$|[x,G]|=p^2$$, so $$[x,G]=G^\prime$$ which means $$G^\prime \subseteq \overline{x}$$ for each $$x\notin Z(G)$$. Thus, if $$N\trianglelefteq G$$ and $$N\nsubseteq Z(G)$$ then $$N\supseteq G^\prime$$. This shows that Z(G) cuts $$\eta (G)$$ and so $$C_0 (G)$$ is not a ring. $$\square$$

$$\square$$

In conclusion we have found that when G is a finite abelian p-group then $$C_0 (G)$$ is a ring if and only if G is 1-affine complete. For nonabelian p-groups, $$p=2$$, we have seen that $$C_0 (G)$$ can be a ring properly containing $$P_0 (G)$$. For $$p>2$$ we have several classes for which $$C_0 (G)$$ is not a ring. In fact, for $$p>2$$ the authors have no example of a nonabelian p-group for which $$C_0 (G)$$ is a ring unless G is 1-affine complete. We thus close with the following.

### Conjecture

For finite nonabelian p-groups $$G,\,p>2,\,C_0 (G)$$ is a ring if and only if G is 1-affine complete.

## Notes

### Acknowledgements

Open access funding provided by Austrian Science Fund (FWF). Portions of this research were done while the first author was visiting the Institute for Algebra at Johannes Kepler Universität-Linz. He wishes to express his appreciation for the financial assistance and the gracious hospitality provided. Both authors wish to thank the Austrian Science Fund (FWF), Project FWF24077, for financial support.

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