Automorphisms of n-Ary Groups

Abstract

Isotopies and autotopies of n-ary groups are described. As a consequence, we obtain various characterizations of the group of automorphisms of n-ary groups. We also determine the number of automorphisms of a given n-ary group.

1 Introduction

Polyadic groups (called also n-ary groups) are a natural generalization of classical groups. The first important paper on n-ary groups was written (under the inspiration of Emmy Nöther) by Dörnte in 1928 (cf. [2]). Another important publication is the Post’s article [16] which shows strong relationships with classical groups but also gives many significant differences between n-ary groups ($n> 2$) and ordinary groups. Today we know many such differences. For example, the intersection of two n-ary subgroups may be the empty set; there are n-ary groups without neutral elements (cf. [2]), but there are n-groups containing many neutral elements (cf. [4]). Also there are groups in which all elements are neutral.

Many authors have attempted to classify n-ary groups according to their isomorphism. Criteria were found to determine if the groups are isomorphic (for example, [8, 9]). However, isotopies between n-ary groups have not been studied so far. Also, little is known about the automorphisms and autotopies of n-ary groups (cf. [14]). The number of autotopies of a given n-ary group (quasigroup) is very large. An n-ary quasigroup of order four has at least $2^{[\frac{n}{2}]}+2$ and not more than $6\cdot 4^n$ autotopies (cf. [10]). For n-groups of greater order, the result is unknown. We only know the exact number of autotopies and automorphisms for some n-ary quasigroups. For n-ary quasigroups linear over a commutative group it is given in [15].

In this article (inspired by [15]) we find a criterion for when two n-ary groups are isotopic. Then we characterize the autotopies of such groups and determine the number of all automorphisms of n-ary groups.

2 Preliminaries

Throughout the paper, we assume that $n>2$. We will also use the abbreviated form of the notation. Instead of ’an n-ary group’ we will write ’an n-group’. The sequence of elements $x_i,\ldots ,x_j$ we will be written as $x_i^j$, remembering that for $j<i$ it is an empty symbol.

The nonempty set G with one n-ary operation $f:G^n\rightarrow G$ is called an n-groupoid and is denoted by (G, f). An n-groupoid is an n-quasigroup if for all $a_1^n,b\in G$ and each $i\in \overline{1,n}=\{1,2,\ldots ,n\}$ there is exactly one $x_i\in G$ such that $f(a_1^{i-1},x_i,a_{i+1}^n)=b$. If the operation f is associative, i.e.

$$\begin{aligned} f(f(a_1^n),a_{n+1}^{2n-1})=f(a_1^j,f(a_{j+1}^{j+n}),a_{j+n+1}^{2n-1}) \end{aligned}$$

holds for all $a_1^{2n-1}\in G$ and all $1\leqslant j<n$, then an n-groupoid (G, f) is called an n-semigroup. An n-quasigroup with an associative operation is called an n-group. For $n=2$ we obtain quasigroup, semigroup, and group, respectively.

Note by the way that in the definition of the n-group (G, f), assuming the associativity of the operation f, it is enough to postulate the existence of a solution of the equation $f(a_1^{i-1},x_i,a_{i+1}^n)=b$ (cf. [16]).

In [7] the following characterization of n-groups is proved.

Theorem 2.1

An n-semigroup (G, f) is an n-group if and only if for every $a\in G$ there exists an element ${\bar{a}}\in G$ such that

$$\begin{aligned} f({\bar{a}},a,a,\ldots ,a,y)=f(y,a,a,\ldots ,a,{\bar{a}})=y \end{aligned}$$

for all $y\in G$.

Other characterizations of this type can be found in [5].

We say that an n-groupoid (G, f) has a neutral element if there exist $e\in G$ such that $f(e,\ldots ,e,x,e,\ldots ,e)=x$ no matter where x occurs. An n-quasigroup containing neutral elements (one of more) is called an n-loop. An element $a\in G$ with the property $f(a,\ldots ,a)=a$ is called idempotent. The set of all idempotents of an n-groupoid (G, f) is denoted by $E_{(G,f)}$. Generally, n-groups do not have such elements, but there are groups that have several neutral elements and groups that contain only neutral elements

Theorem 2.2

The set of all neutral elements of an n-semigroup (G, f) is empty or is a commutative n-group contained in (G, f).

Proof

Let e be a neutral element of an n-semigroup (G, f). Then $f(e,\ldots ,e)=e$. Moreover, e is the neutral element of a semigroup $(G,\cdot )$, where $x y=f(x,e,\ldots ,e,y)$. In this case $f(x_1^n)=x_1 x_2\cdots x_n$. If a is another neutral element of (G, f), then $x=f(x,a,\ldots ,a)=x a\cdots a=xa^{n-1}$ in $(G,\cdot )$. Thus $x=f(a,x,a,\ldots ,a)=a x a^{n-2}$ implies $x a=a x$ for all $x\in G$.

Let $a_1,\ldots ,a_n$ be (not necessarily different) neutral elements of (G, f). Then $f(f(a_1^n),\ldots ,f(a_1^n),x,f(a_1^n),\ldots ,f(a_1^n))$, calculated in $(G,\cdot )$, is equal to x. Hence $f(a_1^n)$ is a neutral element. Thus, the set E of all neutral elements of (G, f) is a commutative n-semigroup contained in (G, f). Since the equation $f(x,a_2^n)=a_1$ is solved by $x=f_{(\cdot )}(a_1,{\mathop {a_n}\limits ^{(n-2)}},\ldots ,{\mathop {a_2}\limits ^{(n-2)}})\in E$, (E, f) is a commutative n-group. $\square$

A binary groupoid $(G,\cdot )$ with the operation $x y=f(x,a_2^{n-1},y)$, where $a_2^{n-1}\in G$ are fixed, is called a retract of (G, f) and is denoted by $\mathrm{ret}_{a_2^n}(G,f)$. The symbol $(G,\cdot )=\mathrm{ret}_a(G,f)$ means that $xy=f(x,a,\ldots ,a,y)$.

It is clear that retracts of n-semigroups are semigroups, retracts of n-groups are groups. Using these retracts and their automorphisms, we can find a very useful characterization of n-ary groups known as Hosszú Theorem (cf. [13]). Since two years later Gluskin proved this theorem in a more general form (cf. [11]), this theorem is known also as Gluskin–Hosszú or Hosszú–Gluskin Theorem (cf. [6]).

Theorem 2.3

An n-groupoid (G, f) is an n-group if and only if

• (i) on G one can define a binary operation $\cdot$ such that $(G,\cdot )$ is a group,

• (ii) there exist an automorphism $\varphi$ of $(G,\cdot )$ and $b\in G$ such that $\varphi (b)=b$,

• (iii) $\varphi ^{n-1}(x) b=b x$ for all $x\in G$,

• (iv) $f(x_1^n)=x_1\varphi (x_2)\varphi ^2(x_3)\cdots \varphi ^{n-1}(x_n) b$ for all $x_1^n\in G$.

In this case, we say that an n-group (G, f) is $\langle \varphi ,b\rangle$-derived from the group $(G,\cdot )$ and denote this fact by $(G,f)=\mathrm{der}_{\varphi ,b}(G,\cdot )$. All retracts of $(G,f)=\mathrm{der}_{\varphi ,b}(G,\cdot )$ are isomorphic to $(G,\cdot )$.

An n-groupoid (G, f) is medial if

$$\begin{aligned} f(f(x_{11}^{1n}),f(x_{21}^{2n}),\ldots ,f(x_{n1}^{nn}))=f(f(x_{11}^{n1}),f(x_{12}^{2n}),\ldots ,f(x_{1n}^{nn})) \end{aligned}$$

holds for all $x_{ij}\in G$ and $i,j\in \overline{1,n}$.

The following two characterizations of medial n-groups (proved in [3]) will be used later.

Lemma 2.4

An n-group (G, f) is medial if and only if for some $a\in G$ the retract $\mathrm{ret}_a(G,f)$ is commutative.

Corollary 2.5

An n-group (G, f) is medial if and only if it is $\langle \varphi ,b\rangle$-derived from an abelian group.

3 Isotopies and Isomorphisms

Suppose (G, f) and (H, g) are two n-groupoids. If there are bijections $\alpha _1,\ldots ,\alpha _{n},\delta$ from G to H, such that

$$\begin{aligned} \delta (f(x_1,x_2,\ldots ,x_n))=g(\alpha _{1}(x_1),\alpha _2(x_2),\ldots ,\alpha _n(x_n)), \end{aligned}$$

(1)

then we say that these two n-groupoids are isotopic. In this case, we also say that $T = (\alpha _1,\alpha _2,\ldots ,\alpha _{n},\delta )$ is an isotopy or an isotopism, or a weak isomorphism between (G, f) and (H, g). Then we write $T:(G,f)\rightarrow (H,g)$ and say that (H, g) and (G, f) are isotopic or that (H, g) is an isotope of (G, f). An isotopism with $\alpha _1=\alpha _2=\cdots =\alpha _n=\delta$ is an isomorphism.

An isomorphism saves all properties of n-groupoids. Unfortunately, an isotopism does not do that. For example, an n-groupoid isotopic to a commutative n-groupoid may not be commutative; an n-groupoid isotopic to an idempotent n-groupoid may not be idempotent. An n-groupoid isotopic to an n-group may not be an n-group, but it will always be an n-quasigroup. To confirm the above statements, it is enough to consider two 5-groupoids: $({\mathbb {Z}}_5,f)=\mathrm{der}_0({\mathbb {Z}}_5,+)$ and $({\mathbb {Z}}_5,g)$ with $g(x_1^5)=(x_1-x_2+x_3+x_4+x_5)(\mathrm{mod}\,5)$.

The isotopy $T = (\alpha _1,\alpha _2,\ldots ,\alpha _n,\varepsilon )$, where $\varepsilon$ is the identity mapping, is called a principal isotopism.

Proposition 3.1

Every isotope of n-groupoid (G, f) is isomorphic to a certain principal isotope of (G, f).

Proof

Let (H, g) be an isotope of an n-groupoid (G, f). Then there are bijections $\alpha _i,\delta :G\rightarrow H$ satisfying (1). Consider the n-ary operation

$$\begin{aligned} h(x_1^n)=f\Big (\alpha _1^{-1}\delta (x_1),\alpha _2^{-1}\delta (x_2),\ldots ,\alpha _n^{-1}\delta (x_n)\Big ). \end{aligned}$$

Since all $\alpha _i^{-1}\delta$ are bijections from G onto G, the n-groupoid (G, h) is a principal isotope of (G, f).

Also,

$$\begin{aligned} \begin{array}{rl} \delta (h(x_1,\ldots ,x_n))&{}=\delta f\big (\alpha _1^{-1}\delta (x_1),\alpha _2^{-1}\delta (x_2),\ldots ,\alpha _n^{-1}\delta (x_n)\big )\\[4pt] &{}=g\big (\delta (x_1),\delta (x_2),\ldots ,\delta (x_n)\big ). \end{array} \end{aligned}$$

Thus, (H, g) is isomorphic to the principal isotope (G, h). $\square$

In a binary case ($n=2$), the isotopic groups are isomorphic (cf. [17]). Unfortunately, for n-groups $(n>2)$ this is not true. These two $(n+1)$-groups $({\mathbb {Z}}_n,f)=\mathrm{der}_0({\mathbb {Z}}_n,+)$ and $({\mathbb {Z}}_n,g)=\mathrm{der}_1({\mathbb {Z}}_n,+)$ are isotopic but not isomorphic. The first is idempotent, the second has not idempotents. However, for n-groups we have the following result.

Theorem 3.2

Two n-groups are isotopic if and only if their retracts are isomorphic.

Proof

Let n-groups (G, f) and (H, g) be isotopic. Then, by (1), for retracts $(G,\circ )=\mathrm{ret}_{a_2^{n-1}}(G,f)$ and $(H,\diamond )=\mathrm{ret}_{b_2^{n-1}}(H,g)$, with $b_i=\alpha _i(a_i)$, $1<i<n$, we have

$$\begin{aligned} \delta (x_1\circ x_n)=\delta (f(x_1,a_2^{n-1},x_n))=g(\alpha _1(x_1),b_2^{n-1},\alpha _n(x_n))=\alpha _1(x_1)\diamond \alpha _n(x_n). \end{aligned}$$

So, $(G,\circ )$ and $(H,\diamond )$ are isotopic groups. Hence they are isomorphic.

Conversely, let the retracts of n-groups (G, f) and (H, g) be isomorphic. By Hosszú Theorem, there are groups and their automorphisms such that $(G,f)=\mathrm{der}_{\varphi ,a}(G,*)$ and $(H,g)=\mathrm{der}_{\psi ,b}(H,\cdot )$. All retracts of (G, f) are isomorphic to $(G,*)$, all retracts of (H, g) to $(H,\cdot )$. So, there is an isomorphism $\delta :(G,*)\rightarrow (H,\cdot )$. Thus, the mappings $\alpha _1=\delta$, $\alpha _2=\psi ^{-1}\delta \varphi$, $\alpha _3=\psi ^{-2}\delta \varphi ^2,\ldots , \alpha _{n-1}=\psi ^{2-n}\delta \varphi ^{n-2}$ and $\alpha _n=R_b^{-1}\psi ^{1-n}\delta \varphi ^{n-1}P_a$, where $R_b(x)=x\cdot b$, $P_a(x)=x*a$, are bijections. Also,

$$\begin{aligned} \begin{array}{rlll} \delta (f(x_1^n))&{}=\delta (x_1*\varphi (x_2)*\varphi ^2(x_3)*\cdots *\varphi ^{n-2}(x_{n-1})*\varphi ^{n-1}(x_n*a))\\[4pt] &{}=\delta (x_1)\cdot \delta \varphi (x_2)\cdot \delta \varphi ^2(x_3)\cdot \cdots \cdot \delta \varphi ^{n-2}(x_{n-1})\cdot \delta \varphi ^{n-1}P_a(x_n)\\[4pt] &{}=\alpha _1(x_1)\cdot \psi \alpha _2(x_2) \cdot \cdots \cdot \psi ^{n-2}\alpha _{n-1}(x_{n-1})\cdot \psi ^{n-1}R_b\alpha _n(x_n)\\[4pt] &{}=g(\alpha _1(x_1),\alpha _2(x_2),\ldots ,\alpha _{n-1}(x_{n-1}),\alpha _n(x_n)). \end{array} \end{aligned}$$

This means that $T=(\alpha _1,\alpha _2,\ldots ,\alpha _n,\delta )$ is the isotopy between (G, f) and (H, g). This completes the proof. $\square$

As a consequence of Theorem 3.2 and Corollary 2.5 we obtain

Corollary 3.3

An n-group isotopic to a medial n-group also is medial.

Corollary 3.4

All n-groups of the same prime order are isotopic but not necessarily isomorphic.

All n-groups (for fixed n and order of G) can be divided into disjoint classes containing isotopic n-groups. Each class can be represented by an n-group $\mathrm{der}(G,\cdot )$. The number of such classes coincides with the number of non-isomorphic groups of a given order. For order $n\leqslant 30$ there is

$$\begin{aligned} 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4 \end{aligned}$$

such classes, but for $|G|=64$ we have 267 such classes (for each $n>2$).

Using Theorem 3.2 we can prove

Proposition 3.5

Any isotopism $T:(G,f)\rightarrow (H,g)$ of n-groups $(G,f)\!=\!\mathrm{der}_{\varphi ,a}(G,\!\star )$ and $(H,g)=\mathrm{der}_{\psi ,b}(H,\cdot )$ can be expressed in the form

$$\begin{aligned} T=(h,\,\psi ^{-1}h\varphi ,\,\psi ^{-2}h\varphi ^2,\ldots ,\psi ^{-(n-2)}h\varphi ^{n-2}, \,R_{b^{-1}}\psi ^{-(n-1)}h\varphi ^{n-1}R_a^{\star },\,h), \end{aligned}$$

where $h:(G,\star )\rightarrow (H,\cdot )$ is a group isomorphism, $R_c^{\star }$ and $R_c$—right translations in $(G,\star )$ and $(H,\cdot )$, respectively.

An isomorphism of n-groups can be characterized by the following proposition.

Proposition 3.6

Let $(G,\star )$ and $(H,\cdot )$ be groups with neutral elements $e^{\star }$ and e, respectively. Then $h:G\rightarrow H$ such that $h(e^{\star })=e$ is an isomorphism of n-groups $(G,f)=\mathrm{der}_{\varphi ,a}(G,\star )$ and $(H,g)=\mathrm{der}_{\psi ,b}(H,\cdot )$ if and only if it is an isomorphism of groups $(G,\star )$ and $(H,\cdot )$, $h(a)=b$ and $h\varphi =\psi h$.

Proof

Any isomorphism is an isotopism, so, by Proposition 3.5, we get $h=\alpha _2$. This gives $\psi h=h\varphi$. Now $h=\alpha _n$ implies $h(a)=b$. Therefore,

$$\begin{aligned} e= & {} h(e^{\star })=h(a^{-1}\star \varphi (e^{\star })\star \cdots \star \varphi ^{n-1}(e^{\star })\star a)=hf(a^{-1},e^{\star },\ldots ,e^{\star })\\= & {} g(h(a^{-1}),h(e^{\star }),\ldots ,h(e^{\star }))=g(h(a^{-1}),e,\ldots ,e)=h(a^{-1})\cdot b. \end{aligned}$$

Thus $h(a^{-1})=b^{-1}$ in $(H,\cdot )$.

Since, by Hosszú Theorem, $\varphi (a^{-1})=a^{-1}$ and $\varphi ^{n-1}(x)\star a=a\star x$, we obtain

$$\begin{aligned} \begin{array}{rll} h(x\star y)=&{}h(x\star \varphi (a^{-1})\star \varphi ^2(e^{\star })\star \cdots \star \varphi ^{n-2}(e^{\star })\star a\star y)\\ =&{}hf(x,a^{-1},e^{\star },\ldots ,e^{\star },y)=g(h(x),h(a^{-1}),h(e^{\star }),\ldots ,h(e^{\star }),h(y))\\ =&{}g(h(x),b^{-1},e,\ldots ,e,h(y))\\ =&{}h(x)\cdot \psi (b^{-1})\cdot \psi ^2(e)\cdot \cdots \cdot \psi ^{n-2}(e)\cdot \psi ^{n-1}h(y)\cdot b\\ =&{}h(x)\cdot b^{-1}\cdot \psi ^{n-1}h(y)\cdot b=h(x)\cdot b^{-1}\cdot b\cdot h(y)=h(x)\cdot h(y). \end{array}. \end{aligned}$$

This shows that h is a group isomorphism.

The converse statement is obvious. $\square$

Theorem 3.7

Every isotopism $T:\! G\!\rightarrow \! H$ of n-groups $(G,f)=\mathrm{der}(G,\star )$ and $(H,g)=\mathrm{der}(H,\cdot )$ can be uniquely presented in the form

$$\begin{aligned} T=(L_{a_1}I_{c_1}\beta ,L_{a_2}I_{c_2}\beta ,\ldots ,L_{a_n}I_{c_n}\beta ,R_{c_n}\beta ), \end{aligned}$$

(2)

where $a_1,\ldots ,a_n\in H$, $c_i=a_1a_2\cdots a_i$, $I_{c_i}(x)=c_i^{-1}xc_i$ and $\beta :G\rightarrow H$ is an isomorphism of groups $(G,\star )$ and $(H,\cdot )$.

Proof

Direct calculations show that such defined T is an isotopism of these n-groups.

To prove the converse statement, let $T=(\alpha _1,\alpha _2,\ldots ,\alpha _n,\delta )$ be an isotopism of n-groups $(G,f)=\mathrm{der}(G,\star )$ and $(H,g)=\mathrm{der}(G,\cdot )$, i.e. let

$$\begin{aligned} \delta f(x_1^n)=\delta (x_1\star x_2\star \cdots \star x_n)=\alpha _1(x_1)\alpha _2(x_2)\cdots \alpha _n(x_n), \end{aligned}$$

(3)

for bijections $\alpha _1,\ldots ,\alpha _n,\delta :G\rightarrow H$.

For the sake of clarity, let $e^*$ be the identity of $(G,\star )$, $a_i=\alpha _i(e^*)$ and $c_i=a_1a_2\cdots a_i$, $i\in \overline{1,n}$.

From (3) we obtain $\delta (e^*)=a_1a_2\cdots a_n$ and

$$\begin{aligned} \delta (x)=a_1a_2\cdots a_{i-1}\alpha _i(x)a_{i+1}\cdots a_n=R_{a_{i+1}\cdots a_n}L_{a_1\cdots a_{i-1}}\alpha _i(x) \end{aligned}$$

for every $x\in G$. Thus

$$\begin{aligned} \alpha _i(x)=L_{a_1\cdots a_{i-1}}^{-1}R_{a_{i+1}\cdots a_n}^{-1}\delta (x)= L_{(a_1\cdots a_{i-1})^{-1}}R_{(a_{i+1}\cdots a_n)^{-1}}\delta (x). \end{aligned}$$

Hence, (3) can be rewritten in the form

$$\begin{aligned} \begin{array}{rl} \delta f(x_1^n)=&{}R_{(a_2\cdots a_n)^{-1}}\delta (x_1)L_{a_1^{-1}}R_{(a_3\cdots a_n)^{-1}}\delta (x_3)\\ &{}\cdots L_{(a_1\cdots a_{n-2})^{-1}}R_{a_n^{-1}}\delta (x_{n-1})L_{(a_1\cdots a_n)^{-1}}\delta (x_n)\\ =&{}\delta (x_1)a_n^{-1}\cdots a_2^{-1} a_1^{-1}\delta (x_2)a_n^{-1}\cdots a_3^{-1} a_2^{-1}a_1^{-1}\delta (x_3)a_n^{-1}\cdots a_1^{-1}\\ &{}\cdots a_n^{-1}\cdots a_1^{-1}\delta (x_{n-1})a_n^{-1}\cdots a_1^{-1}\delta _n\\ =&{}\delta (x_1)d\cdot \delta (x_2)d\cdot \delta (x_3)d\cdots \delta (x_{n-1})d\cdot \delta (x_n), \end{array} \end{aligned}$$

where $d=a_n^{-1}\cdots a_2^{-1}a_1^{-1}$. From this, multiplying by d, we get

$$\begin{aligned} \delta f(x_1^n)d=\delta (x_1)d\cdot \delta (x_2)d\cdot \delta (x_3)d\cdots \delta (x_{n-1})d\cdot \delta (x_n)d, \end{aligned}$$

which shows that $\beta =R_d\delta :G\rightarrow H$ is an isomorphism of n-groups (G, f) and (H, g). Since $\beta (e^*)=\delta (e^*)d=a_1a_2\cdots a_nd=e$, where e is the identity of $(H,\cdot )$, $\beta$ is an isomorphism of groups ${\mathrm{ret}}_{e^*} (G,f)=(G,\star )$ and $\mathrm{ret}_e(H,g)=(H,\cdot )$ too.

From $\beta =R_d\delta$ we also have $\delta =R_d^{-1}\beta =R_{d^{-1}}\beta =R_{a_1a_2\cdots a_n}\beta$. Thus

$$\begin{aligned} \begin{array}{rl} \alpha _i=&{}L_{(a_1\cdots a_{i-1})^{-1}}R_{(a_{i+1}\cdots a_n)^{-1}}\delta =L_{(a_1\cdots a_{i-1})^{-1}}R_{(a_{i+1}\cdots a_n)^{-1}}R_{a_1a_2\cdots a_n}\beta \\[3pt] =&{}L_{(a_1\cdots a_{i-1})^{-1}}R_{a_i\cdots a_n}\beta =L_{a_i}L_{a_i^{-1}}L_{(a_1\cdots a_{i-1})^{-1}}R_{a_i\cdots a_n}\beta \\[3pt] =&{}L_{a_i}L_{(a_1\cdots a_i)^{-1}}R_{a_i\cdots a_n}\beta =L_{a_i}I_{a_1\cdots a_i}\beta , \end{array} \end{aligned}$$

where $I_a$ is the inner automorphism of $(H,\cdot )$.

Therefore,

$$\begin{aligned} T=(L_{a_1}I_{c_1}\beta ,L_{a_2}I_{c_2}\beta ,\ldots ,L_{a_n}I_{c_n}\beta ,R_{c_n}\beta ), \end{aligned}$$

where $c_i=a_1a_2\cdots a_i$, is an isotopy of n-groups (G, f) and (H, g).

This representation is unique. Indeed, if also

$$\begin{aligned} T=(L_{b_1}I_{d_1}\beta ,L_{b_2}I_{d_2}\beta ,\ldots ,L_{b_n}I_{d_n}\beta ,R_{d_n}\sigma ), \end{aligned}$$

for some $b_1,\ldots ,b_n\in H$, $d_i=b_1b_2\cdots b_i$ and an automorphism $\sigma$ of $(H,\cdot )$, then $R_{c_n}\beta =R_{d_n}\sigma$ and $L_{a_i}I_{c_i}\beta =L_{b_i}I_{d_i}\sigma$ for all $i\in \overline{1,n}$. Thus $L_{a_1}I_{c_1}\beta (e)=L_{b_1}I_{d_1}\sigma (e)$, which gives $a_1=b_1$ and $\beta =\sigma$. Consequently, $L_{a_2}I_{c_2}=L_{b_2}I_{d_2}$ implies $a_2=b_2$, and so on. $\square$

Corollary 3.8

Every isotopism $T:G\rightarrow H$ of n-groups $(G,f)=\mathrm{der}(G,\star )$ and $(H,g)=\mathrm{der}(H,\cdot )$, where $(H,\cdot )$ is an abelian group, can be uniquely presented in the form

$$\begin{aligned} T=(L_{a_1}\beta ,L_{a_2}\beta ,\ldots ,L_{a_n}\beta ,L_{c_n}\beta ), \end{aligned}$$

(4)

where $a_1,a_2,\ldots ,a_n\in H,$ $c_n=a_1a_2\cdots a_n$ and $\beta :G\rightarrow H$ is an isomorphism of groups $(G,\star )$ and $(H,\cdot )$.

Proof

In an abelian group $R_c=L_c$ and $I_c=\varepsilon$. $\square$

Lemma 3.9

Let $(G,f)=\mathrm{der}(G,\star )$ and $(H,g)=\mathrm{der}(H,\cdot )$ be two fixed n-groups. Mapping $h: G\!\rightarrow \! H$ is an isomorphism of these n-groups if and only if there is an isomorphism $\beta :(G,\star )\!\rightarrow \! (H,\cdot )$ such that $h=R_{h(e^*)}\beta$, where $e^*$ is the identity of $(G,\star )$, and $h(e^*)^n=h(e^*)$ belongs to the center of $(H,\cdot )$.

Proof

An isomorphism h can be considered as an isotopism (2) of which all components are equal. Thus $L_{a_1}I_{a_1}\beta (e^*)=L_{a_2}I_{a_1a_2}\beta (e^*)$ implies $a_1=a_2$. Analogously, $L_{a_2}I_{a_1a_2}\beta (e)=L_{a_3}I_{a_1a_2a_3}\beta (e^*)$ implies $a_2=a_3$, and so on. Hence, $h(x)=\beta (x)a$. Consequently, $h(e^*)=a$. Therefore, $h=R_{h(e^*)}\beta$ and $h(e^*)= hf(e^*,\ldots ,e^*)=h(e^*)^n.$ Since $h f(x,e^*,\ldots ,e^*)=h(x)=h f(e^*,x,e^*,\ldots ,e^*)$, $h(e^*)$ must be in the center of $(H,\cdot )$.

The converse statement is obvious. $\square$

Theorem 3.10

A mapping $h:G\rightarrow H$ is an isomorphism of n-groups $(G,f)=\mathrm{der}_{\varphi ,a}(G,*)$ and $(H,g)=\mathrm{der}_{\psi ,b}(H,\cdot )$ if and only if it can be expressed in the form $h=R_c\beta$, where $c=c^n$ belongs to the center of $(H,\cdot )$, $\beta :(G,*)\rightarrow (H,\cdot )$ is an isomorphism, $\beta \varphi =\psi \beta$, $\beta (a)=b$ and $\psi (c)=c$.

Proof

Any isomorphism $h:(G,f)\rightarrow (H,g)$ can be considered as an isotopism $P=(h,h,\ldots ,h):(G,f)\rightarrow (H,g)$ such that $P=S\circ A\circ T$, where $T =(\varepsilon ,\varphi ,\varphi ^2,\ldots ,\varphi ^{n-2},R_a^{*}\varphi ^{n-1},\varepsilon )$ is an isotopism between (G, f) and $\mathrm{der}(G,*)$, $A=(\theta ,\theta ,\ldots ,\theta )$, $\theta :\mathrm{der}(G,*)\rightarrow \mathrm{der}(H,\cdot )$ is an isomorphism of n-groups $\mathrm{der}(G,*)$ and $\mathrm{der}(H,\cdot )$, $S=(\varepsilon ,\psi ^{-1},\psi ^{-2},\ldots ,\psi ^{-(n-2)},R_{b^{-1}}\psi ^{-(n-1)},\varepsilon )$ is an isotopism between $\mathrm{der}(H,\cdot )$ and (H, g).

$$\begin{aligned} \begin{array}{ccccc} (G,f)&{}\;\;{\mathop {\longrightarrow }\limits ^{_T}}\;\; &{}\mathrm{der}(G,*) &{} &{} \\[4pt] {}^{_P}\!\downarrow &{} &{} \downarrow \! {}^{_A} &{} &{} \\[2pt] (H,g)&{}{\mathop {\longleftarrow }\limits ^{_S}} &{}\mathrm{der}(H,\cdot ) &{} &{} \\ \end{array} \end{aligned}$$

By Lemma 3.9, $\theta =R_c\beta$, $\beta :(G,*)\rightarrow (H,\cdot )$ is an isomorphism, $c=\theta (e^*)$, where $e^*$ is the neutral element of $(G,*)$. Then

$$\begin{aligned} P=(R_c\beta ,\,\psi ^{-1}R_c\beta \varphi ,\, \ldots ,\psi ^{2-n}R_c\beta \varphi ^{n-2},\,R_{b^{-1}}\psi ^{1-n}R_c\beta R_a^*\varphi ^{n-1},\,R_c\beta ). \end{aligned}$$

Since P is an isomorphism, all components of P should be equal. In particular, $R_c\beta =\psi ^{-1}R_c\beta \varphi =R_{b^{-1}}\psi ^{1-n}R_c\beta R_a^*\varphi ^{n-1}$. From the first equality, we obtain $R_c\beta =\psi ^{-k}R_c\beta \varphi ^k$ for all $1\leqslant k<n$. Thus, the first $n-1$ components are equal. From $R_c\beta =\psi ^{-1} R_c\beta \varphi$, for $x=e^*$, we also have $c=\psi ^{-1}(c)$. Hence, $\psi (c)=c$ and $\beta (x)c=\psi ^{-1}\beta \varphi (x)c$. So, $\psi \beta =\beta \varphi$.

The equality $R_c\beta =R_{b^{-1}}\psi ^{1-n}R_c\beta R_a^*\varphi ^{n-1}$, implies $cb=\psi ^{1-n}(\beta (a)c)$. Thus $\psi ^{n-1}(bc)=\beta (a)c$, which, by Hosszú Theorem, gives $bc=\beta (a)c$. So, $\beta (a)=b$.

Conversely, let $h=R_c\beta$, where c and $\beta$ are as in theorem. Then h is a bijection $G\rightarrow H$ and

$$\begin{aligned} \begin{array}{rlll} hf(x_1^n)=&{}R_c\beta (x_1*\varphi (x_2)*\varphi ^2(x_3)*\cdots *\varphi ^{n-1}(x_n)*a)\\[3pt] =&{}\beta (x_1)\cdot \beta \varphi (x_2)\cdot \beta \varphi ^2(x_3)\cdot \cdots \cdot \beta \varphi ^{n-1}(x_n)\cdot \beta (a)\cdot c\\[3pt] =&{}\beta (x_1)\cdot \psi \beta (x_2)\cdot \psi ^2\beta (x_3)\cdot \cdots \cdot \psi ^{n-1}\beta (x_n)\cdot b\cdot c^n\\[3pt] =&{}\beta (x_1)c\cdot \psi \beta (x_2)c\cdot \psi ^2\beta (x_3)c\cdot \cdots \cdot \psi ^{n-1}\beta (x_n)c\cdot b\\[3pt] =&{}R_c\beta (x_1)\cdot \psi R_c\beta (x_2)\cdot \psi ^2R_c\beta (x_3)\cdot \cdots \cdot \psi ^{n-1}R_c\beta (x_n)\cdot b\\[3pt] =&{}g(R_c\beta (x_1),R_c\beta (x_2),R_c\beta (x_3),\ldots ,R_c\beta (x_n))\\ =&{}g(h(x_1),h(x_2),h(x_3),\ldots ,h(x_n)). \end{array} \end{aligned}$$

This shows that h is an isomorphism between (G, f) and (H, g). $\square$

Corollary 3.11

A mapping $h:G\rightarrow H$ is an isomorphism of n-groups $(G,f)=\mathrm{der}_{\varphi ,a}(G,*)$ and $(H,g)=\mathrm{der}_{\psi ,b}(H,\cdot )$ if and only if it can be expressed in the form $h=R_c\beta$, where $\beta :(G,*)\rightarrow (H,\cdot )$ is an isomorphism, $\beta \varphi =\psi \beta$ and $c\in Z(H,\cdot )$ is such that $\beta (a)=\psi (c)\psi ^2(c)\cdots \psi ^{n-1}(c)a$.

4 Autotopies

If for an n-groupoid (G, f) there are bijections $\alpha _1,\ldots ,\alpha _{n},\delta :G\rightarrow G$ such that

$$\begin{aligned} \delta f(x_1,x_2,\ldots ,x_n)=f(\alpha _{1}(x_1),\alpha _2(x_2),\ldots ,\alpha _n(x_n)), \end{aligned}$$

(5)

then $T = (\alpha _1,\alpha _2,\ldots ,\alpha _{n},\delta )$ is called an autotopy or a weak automorphism of (G, f). The set of all autotopies of (G, f) is denoted by ${{\mathcal {A}}}(G,f)$. It is clear that ${{\mathcal {A}}}(G,f)$ is a group with respect to the composition of autotopies:

$$\begin{aligned} (\alpha _1,\alpha _2,\ldots ,\alpha _{n+1})\circ (\beta _1,\beta _2,\ldots ,\beta _{n+1})=(\alpha _1\beta _1,\alpha _2\beta _2,\ldots ,\alpha _{n+1}\beta _{n+1}). \end{aligned}$$

The group $\mathrm{Aut}(G,f)$ of all automorphisms of (G, f), considered as an $(n+1)$-tuples $(\varphi ,\ldots ,\varphi )$, is a normal subgroup of the group ${{\mathcal {A}}}(G,f)$.

We start with the following theorem proved by Belousov (cf. [1]) and give a short proof of this theorem.

Theorem 4.1

Groups of autotopies of isotopic n-groups are isomorphic.

Proof

Let $A=(\alpha _1,\alpha _2,\ldots ,\alpha _n,\alpha )$ and $B=(\beta _1,\beta _2,\ldots ,\beta _n,\beta )$ be autotopies of n-groups (G, f) and (H, g), respectively.

Consider the mapping $\beta :{{\mathcal {A}}}(G,f)\rightarrow {{\mathcal {A}}}(H,g)$ defined by $\beta (A)=T\circ A\circ T^{-1}$, where $T=(\delta _1,\delta _2,\ldots ,\delta _n,\delta )$ is an isotopy from (G, f) to (H, g). Then

$$\begin{aligned} T^{-1}:\,\delta ^{-1}g(x_1^n)\!= & {} \!f(\delta _1^{-1}(x_1),\ldots ,\delta _n^{-1}(x_n))\\ A:\,\alpha \delta ^{-1}\!g(x_1^n)\!= & {} \!\alpha f(\delta _1^{-1}(x_1),\ldots ,\delta _n^{-1}(x_n))\!=\!f(\alpha _1\delta _1^{-1}(x_1),\ldots ,\alpha _n\delta _n^{-1}(x_n))\\ T:\,\delta \alpha \delta ^{-1}g(x_1^n)= & {} \delta f(\alpha _1\delta _1^{-1}(x_1),\ldots ,\alpha _n\delta _n^{-1}(x_n))\\= & {} g(\delta _1\alpha _1\delta _1^{-1}(x_1),\ldots ,\delta _n\alpha _n\delta _n^{-1}(x_n)). \end{aligned}$$

Hence, $\beta (A)=(\delta _1\alpha _1\delta _1^{-1},\ldots ,\delta _n\alpha _n\delta _n^{-1},\delta \alpha \delta ^{-1})\in {\mathcal {A}}(H,g)$.

In a similar way, we can show that $\Phi (B)=T^{-1}\circ B\circ T\in {{\mathcal {A}}}(G,f)$. Thus $\beta$ is one-to-one and onto. Obviously $\beta (A_1\circ A_2)=\beta (A_1)\circ \beta (A_2)$. So, $\beta$ is an isomorphism. $\square$

Corollary 4.2

For any n-group (G, f) and any $a\in G$ the groups ${\mathcal {A}}(G,f)$ and ${{\mathcal {A}}}(\mathrm{der}(\mathrm{ret}_a(G,f)))$ are isomorphic. In particular,

$${{\mathcal {A}}}(\mathrm{der}_{\varphi ,b}(G,\cdot ))\cong {\mathcal {A}}(\mathrm{der}(G,\cdot )).$$

As a consequence of Theorem 3.7 and Corollary 3.8 we obtain

Corollary 4.3

Every autotopy T of an n-group $(G,f)=\mathrm{der}(G,\cdot )$ can be uniquely presented in the form (2), where $a_1,\ldots ,a_n\in G$, $c_i=a_1a_2\cdots a_i$, $I_c(x)=c^{-1}xc$ and $\beta \in \mathrm{Aut}(G,\cdot )$.

Corollary 4.4

Every autotopy T of a medial n-group $(G,f)=\mathrm{der}(G,\cdot )$ can be uniquely presented in the form (4), where $a_1,\ldots ,a_n\in G$, $c_n=a_1a_2\cdots a_n$ and $\beta \in \mathrm{Aut}(G,\cdot )$.

Theorem 4.5

The number of autotopies of an n-group (G, f) of order |G| is equal to $\,|G|^n\!\cdot \! |\mathrm{Aut}(G,\cdot )|$.

Proof

By Hosszú Theorem, any n-group (G, f) is $\langle \varphi ,b\rangle$-derived from certain group $(G,\cdot )$, i.e. $f(x_1^n)=x_1\varphi (x_2)\varphi ^2(x_2)\cdots \varphi ^{n-1}(x_n) b,$ where $b\in G$ is fixed and $\varphi$ is an automorphism of $(G,\cdot )$. This means that (G, f) is isotopic to the n-group $(G,g)=\mathrm{der}(G,\cdot )$. The isotopy has the form $T=(\varepsilon ,\varphi ,\varphi ^2,\ldots ,\varphi ^{n-2},R_b\varphi ^{n-1},\varepsilon )$. Hence, these two n-groups have the same number of autotopies (Theorem 4.1). i-central We will calculate the number of autotopies of the n-group $(G,g)=\mathrm{der}(G,\cdot )$. According to Corollary 4.3 each autotopy of (G, g) can be written in the form

$$\begin{aligned} T=T_1\circ T_2=(L_{a_1}I_{c_1},L_{a_2}I_{c_2},\ldots ,L_{a_n}I_{c_n},R_{c_n})\circ (\beta ,\beta ,\ldots ,\beta ). \end{aligned}$$

Since the autotopy $T_1$ is uniquely determined by $a_1,a_2,\ldots ,a_n\in G$, there are exactly $|G|^n$ autotopies. If the autotopy P belongs to the common part of $\{(L_{a_1}I_{c_1},L_{a_2}I_{c_2},\ldots ,L_{a_n}I_{c_n},R_{c_n})\,|\, a_1,\ldots , a_n\in G\}$ and $\{(\beta ,\ldots ,\beta )\,|\,\beta \in \mathrm{Aut}(G,\cdot )\}$, then all its components are equal to some automorphism $\beta \in \mathrm{Aut}(G,\cdot )$. In particular, $L_{a_1}I_{c_1}(x)=\beta (x)$, which for $x=e$ gives $a_1=e$. Consequently, $\beta =\varepsilon$. This means that all components of the autotopy P are equal. Therefore $|{{\mathcal {A}}}(G,g)|=|G|^n\cdot |\mathrm{Aut}(G,\cdot )|$. $\square$

The number of autotopies of a given n-group is very large. For example, for n-groups with p elements (p prime) it is equal to $p^n(p-1)$.

In the case when an n-group (G, f) has $p^m$ elements (p prime) we can determine the number of autotopies using the formula for the number of automorphisms of abelian p-groups (cf. [12]). Using this formula and Theorem 4.5 we obtain

Theorem 4.6

The number of autotopies of an n-group $\langle \varphi ,b\rangle$-derived from the group ${\mathbb {Z}}_{p^{\alpha _1}}\oplus {\mathbb {Z}}_{p^{\alpha _2}}\oplus \cdots \oplus {\mathbb {Z}}_{p^{\alpha _m}}$ is equal to

$$\begin{aligned} p^{n(\alpha _1+\cdots +\alpha _m)}\prod _{k=1}^m(p^{d_k}-p^{k-1})\prod _{j=1}^m(p^{\alpha _j})^{m-d_j}\prod _{i=1}^m(p^{\alpha _i-1})^{m+1-c_i} \;, \end{aligned}$$

where $d_k=\max \{l:\alpha _l=\alpha _k\}$ and $c_k=\min \{l:\alpha _l=\alpha _k\}$.

For medial n-groups we have a stronger result.

Theorem 4.7

If an n-group (G, f) is medial, then

${{\mathcal {A}}}(G,f)\cong \bigoplus \limits _{i=1}^n (G,\cdot )_i\ltimes \mathrm{Aut}(G,\cdot ),$

where $\bigoplus \limits _{i=1}^n (G,\cdot )_i$ is the direct sum of n copies of the group $(G,\cdot )=\mathrm{ret}_a(G,f)$.

Proof

An n-group (G, f) is $\langle \varphi ,b\rangle$-derived from the group $(G,\cdot )=\mathrm{ret}_a(G,f)$. All retracts of this n-group are isomorphic to $(G,\cdot )$. So, it is sufficient to prove our theorem for $(G,f)=\mathrm{der}(G,\cdot )$ (Theorem 4.1). Since (G, f) is medial, $(G,\cdot )$ is an abelian group. Hence, by Corollary 4.4, any autotopy T of $(G,f)=\mathrm{der}(G,\cdot )$ can be uniquely presented in the form

$$\begin{aligned} T=T_1\circ T_2=(L_{a_1},L_{a_2},\ldots ,L_{a_n},R_{a_1\cdots a_n})\circ (\beta ,\beta ,\ldots ,\beta ), \end{aligned}$$

where $a_1,a_2,\ldots ,a_n\in G$, $\beta \in \mathrm{Aut}(G,\cdot )$.

The set ${{\mathcal {T}}}_1=\{(L_{a_1},L_{a_2},\ldots ,L_{a_n},R_{a_1\cdots a_n})\,|\, a_1,\ldots ,a_n\in G\}$ forms a subgroup in ${\mathcal {A}}(G,f)$. Since for any autotopy $T=(L_{b_1}\beta ,\ldots ,L_{b_n}\beta ,R_{b_1\cdots b_n}\beta )$ its inverse has the form $T^{-1}=(L_{d_1}\beta ^{-1}\!,\ldots ,L_{d_n}\beta ^{-1}\!,R_{d}\beta ^{-1})$, where $d_i=\beta ^{-1}(b_i^{-1})$ and $d=(b_1\cdots b_n)^{-1}$. Then for all $T_1\in {{\mathcal {T}}}_1$ and $T\in {{\mathcal {A}}}(G,f)$

$$\begin{aligned} T^{-1}\circ T_1\circ T= & {} (\ldots ,L_{d_i}\beta ^{-1}L_{a_i}L_{b_i}\beta ,\ldots , R_d\beta ^{-1}R_{a_1\cdots a_n}R_{b_1\cdots b_n}\beta )\\= & {} (\ldots ,L_{\beta ^{-1}(a_i)},\ldots , R_{b_1\cdots b_na_1\cdots a_n(b_1\cdots b_n)^{-1}})\in {{\mathcal {T}}}_1. \end{aligned}$$

Thus ${{\mathcal {T}}}_1$ is a normal subgroup of ${{\mathcal {A}}}(G,f)$.

Clearly, the autotopy $T_1=(L_{a_1},L_{a_2},\ldots ,L_{a_n},R_{a_1\cdots a_n})$ is uniquely determined by $a_1,a_2,\ldots ,a_n\in G$. So, ${{\mathcal {T}}}_1$ can be identified with the direct sum of n copies of the group $(G,\cdot )$.

Elements of the set ${\mathcal {T}}_2=\{(\beta ,\beta ,\ldots ,\beta )\,|\,\beta \in \mathrm{Aut}(G,f)\}$ can be identified with $\beta$. Consequently, ${{\mathcal {T}}}_2$ can be identified with $\mathrm{Aut}(G,\cdot )$.

Since ${{\mathcal {T}}}_1\cap {\mathcal {T}}_2=\{(\varepsilon ,\varepsilon ,\ldots ,\varepsilon )\}$, ${\mathcal {A}}(G,f)$ is isomorphic to the semidirect product of $\bigoplus \limits _{i=1}^n (G,\cdot )_i$ and $\mathrm{Aut}(G,\cdot ).$ $\square$

5 Automorphisms

From the results obtained in the previous sections we deduce the following characterizations of automorphisms of n-groups.

From Proposition 3.6 we can deduce

Corollary 5.1

If $(G,\star )$ is a group with a neutral elements $e^{\star }$, then a mapping $h:G\rightarrow G$ such that $h(e^{\star })=e^{\star }$ is an automorphism of an n-group $(G,f)=\mathrm{der}_{\varphi ,a}(G,\star )$ if and only if $\,h\in \mathrm{Aut}(G,\star )$, $h(a)=a$ and $h\varphi =\varphi h$.

Lemma 3.9 gives

Corollary 5.2

A mapping $h:G\rightarrow G$ is an automorphism of an n-group $(G,f)=\mathrm{der}(G,\cdot )$ if and only if there is an automorphism of the group $(G,\cdot )$ such that $h=R_{h(e)}\varphi$ and $h(e)^n=h(e)$, where e is the identity of $(G,\cdot )$, belongs to the center of $(G,\cdot )$.

As a consequence of Theorem 3.10 and Corollary 3.11 we obtain

Proposition 5.3

A mapping $h:G\rightarrow G$ is an automorphism of an n-group $(G,f)=\mathrm{der}_{\varphi ,a}(G,\cdot )$ if and only if $h=R_c\beta$, where $\beta$ is an automorphism of $(G,\cdot )$ such that $\beta \varphi =\varphi \beta$

• (1) $\varphi (c)=c=c^n,$ $\,c\in Z(G,\cdot )$ and $\beta (a)=a$, or equivalently

• (2) $c\in Z(G,\cdot )$ and $\beta (a)=\varphi (c)\varphi ^2(c)\cdots \varphi ^{n-1}(c)a$.

Corollary 5.4

Let $(G,f)=\mathrm{der}_{\varphi ,a}(G,\cdot )$, $c\in G$ and $\beta \in \mathrm{Aut}(G,\cdot )$ be fixed. Then $R_c\beta \in \mathrm{Aut}(G,f)$ if and only if $R_c\in \mathrm{Aut}(G,f)$ and $\beta \in \mathrm{Aut}(G,f)$.

Proof

Indeed, if $h=R_c\beta$ is an automorphism of $(G,f)=\mathrm{der}_{\varphi ,a}(G,\cdot )$, then $\beta \varphi =\varphi \beta$ and $\beta (a)=a$. Therefore,

$$\begin{aligned} \begin{array}{rll} \beta f(x_1^n)=&{}\beta (x_1\varphi (x_2)\varphi ^2(x_3)\cdots \varphi ^{n-1}(x_n)a)\\ =&{}\beta (x_1)\varphi \beta (x_2)\varphi ^2\beta (x_3)\cdots \varphi ^{n-1}\beta (x_n)a\\ =&{}f(\beta (x_1),\beta (x_2),\ldots ,\beta (x_n)). \end{array} \end{aligned}$$

Hence $\beta \in \mathrm{Aut}(G,f)$. Consequently, also $R_c=h\beta ^{-1}\in \mathrm{Aut}(G,f)$.

The converse statement is obvious. $\square$

Theorem 5.5

If $(G,f)=\mathrm{der}_{\varphi ,a}(G,\cdot )$, then

$$\begin{aligned} \mathrm{Aut}(G,f)\cong {{\mathcal {R}}}_{\varphi }(G,f)\ltimes \mathrm{Aut}_{\varphi ,a}(G,\cdot ) \end{aligned}$$

where

$$\begin{aligned} {\mathcal {R}}_{\varphi }(G,f)= & {} \{R_c\,|\,\varphi (c)=c=c^n, \ \,c\in Z(G,\cdot )\}\,\, and\\ \mathrm{Aut}_{\varphi ,a}(G,\cdot )= & {} \{\beta \in \mathrm{Aut}(G,\cdot )\,|\,\beta \varphi =\varphi \beta , \ \beta (a)=a\}. \end{aligned}$$

Proof

${{\mathcal {R}}}_{\varphi }$ and $\mathrm{Aut}_{\varphi ,a}(G,\cdot )$ are subgroups of $\mathrm{Aut}(G,f)$ and $\mathrm{Aut}(G,\cdot )$, respectively.

By Proposition 5.3, for every $h=R_d\delta \in \mathrm{Aut}(G,f)$ and every $R_c\in {{\mathcal {R}}}_{\varphi }$ we have $h^{-1}R_ch=\delta ^{-1}R_{d^{-1}}R_cR_d\delta =R_{\delta ^{-1}(c)}\in {\mathcal {R}}_{\varphi }$. Hence ${{\mathcal {R}}}_{\varphi }$ is a normal subgroup of $\mathrm{Aut}(G,f)$. Moreover, if $\psi \in {\mathcal {R}}_{\varphi }(G,f)\cap \mathrm{Aut}_{\varphi ,a}(G,\cdot )$, then $\psi =R_c=\beta$. Thus, $R_c(a)=\beta (a)=a$, which implies $c=e$. Therefore, ${{\mathcal {R}}}_{\varphi }(G,f)\cap \mathrm{Aut}_{\varphi ,a}(G,\cdot )=\{\varepsilon \}$. Therefore $\mathrm{Aut}(G,f)\cong {{\mathcal {R}}}_{\varphi }(G,f)\ltimes \mathrm{Aut}_{\varphi ,a}(G,\cdot )$. $\square$

An idempotent of an n-group $(G,f)=\mathrm{der}_{\varphi ,a}(G,\cdot )$ is called quasi-central if it belongs to the center of the group $(G.\cdot )$. Quasi-central idempotents exist in certain n-groups. For example, in ternary group $(S_3,f)$ with the operation $f(x,y,z)=xyza$, where $a\in S_3$ is a fixed element of order 3, there are no such idempotents. The set of all quasi-central idempotents of an n-group is denoted by $CE_{(G,f)}$.

Proposition 5.6

Quasi-central elements of an n-group $(G,g)=\mathrm{der}_{\varphi }(G,\cdot )$ form a normal subgroup of $(G,\cdot )$ and a commutative n-subgroup of (G, g).

Proof

Let $(G,g)=\mathrm{der}_{\varphi }(G,\cdot )$. Then the neutral element e of the group $(G,\cdot )$ is a quasi-central idempotent of (G, g). Thus, the set $CE_{(G,g)}$ is nonempty.

Furthermore, for any $a\in CE_{(G,g)}$ we have $a=a\varphi (a)\varphi ^2(a)\cdots \varphi ^{n-1}(a)$ which implies $\varphi (a)\varphi ^2(a)\cdots \varphi ^{n-1}(a)=e=\varphi (a^{-1})\varphi ^2(a^{-1})\cdots \varphi ^{n-1}(a^{-1})$. Thus $a\in CE_{(G,g)}$ if and only if $a^{-1}\in CE_{(G,g)}$. Also $ac\in CE_{(G,g)}$ for $a,c\in CE_{(G,g)}$. Hence $CE_{(G,g)}$ is a normal subgroup of $(G,\cdot )$.

Consequently, for $a_1,\ldots ,a_n\in CE_{(G,g)}$ we have

$$\begin{aligned} \begin{array}{rlll} g(a_1^n)=&{}a_1\varphi (a_2)\varphi ^2(a_3)\cdots \varphi ^{n-1}(a_n)\\ =&{}a_1\varphi (a_1)\varphi ^2(a_1)\cdots \varphi ^{n-1}(a_1)\cdot \varphi \Big (a_2\varphi (a_2)\varphi ^2(a_2)\cdots \varphi ^{n-1}(a_2)\Big )\\ &{}\cdot \varphi ^2\big (a_3\varphi (a_3)\varphi ^2(a_3)\cdots \varphi ^{n-1}(a_3)\Big )\cdots \varphi ^{n-1}\Big (a_n\varphi (a_n)\varphi ^2(a_n)\cdots \varphi ^{n-1}(a_n)\Big )\\ =&{}a_1\varphi (a_2)\varphi ^2(a_3)\cdots \varphi ^{n-1}(a_n)\cdot \varphi \Big (a_1\varphi (a_2)\varphi ^2(a_3)\cdots \varphi ^{n-1}(a_n)\Big )\\ &{}\cdot \varphi ^2\Big (a_1\varphi (a_2)\varphi ^2(a_3)\cdots \varphi ^{n-1}(a_n)\Big )\cdots \varphi ^{n-1}\Big (a_1\varphi (a_2)\varphi ^2(a_3)\cdots \varphi ^{n-1}(a_n)\Big )\\ =&{}g(g(a_1^n),g(a_1^n),\ldots ,g(a_1^n)). \end{array} \end{aligned}$$

This shows that $(CE_{(G,g)},g)$ is an idempotent n-semigroup contained in (G, g). Since in (G, g) we have $g(x,a,\ldots ,a)=g(a,\ldots ,a,x)=x$ for all $a,x\in CE_{(G,g)}$, $(CE_{(G,g)},g)$ is an n-group (Theorem 2.1) contained in (G, g). $\square$

Theorem 5.7

If an n-group $(G,f)=\mathrm{der}_{\varphi ,a}(G,\cdot )$ has at least one quasi-central idempotent, then

$$\begin{aligned} \mathrm{Aut}(G,g)\cong {{\mathcal {R}}}_{E(G,g)}\ltimes Z(\varphi ) \end{aligned}$$

where ${{\mathcal {R}}}_{E(G,f)}=\{R_c\,|\,c\in CE_{(G,f)}\}$ and $Z(\varphi )=\{\beta \in \mathrm{Aut}(G,\cdot )\,|\,\varphi \beta =\beta \varphi \}.$

Proof

Let $(G,f)=\mathrm{der}_{\varphi ,a}(G,\cdot )$. We will show first that (G, f) is isomorphic to $(G,g)=\mathrm{der}_{\varphi }(G,\cdot )$.

Let $c\in CE_{(G,f)}$. Then $c=f(c,c,\ldots ,c)=c\varphi (c)\varphi ^2(c)\cdots \varphi ^{n-1}(c)a$. Thus, $\,ac^{-1}=c^{-1}\varphi (c^{-1})\varphi ^2(c^{-1})\cdots \varphi ^{n-1}(c^{-1})$. Hence

$$\begin{aligned} \begin{array}{rlll} R_{c^{-1}}f(x_1^n)=&{}x_1\varphi (x_2)\varphi ^2(x_3)\cdots \varphi ^{n-1}(x_n)ac^{-1}\\ =&{}x_1\varphi (x_2)\varphi ^2(x_3)\cdots \varphi ^{n-1}(x_n)c^{-1}\varphi (c^{-1})\varphi ^2(c^{-1})\cdots \varphi ^{n-1}(c^{-1})\\ =&{}x_1c^{-1}\,\varphi (x_2)\varphi (c^{-1})\,\varphi ^2(x_3)\varphi ^2(c^{-1})\,\cdots \,\varphi ^{n-1}(x_n)\varphi ^{n-1}(c^{-1})\\ =&{}x_1c^{-1}\,\varphi (x_2c^{-1})\,\varphi ^2(x_3c^{-1})\,\cdots \,\varphi ^{n-1}(x_nc^{-1})\\ =&{}R_{c^{-1}}(x_1)\,\varphi R_{c^{-1}}(x_2)\,\varphi ^2R_{c^{-1}}\,\cdots \,\varphi ^{n-1}R_{c^{-1}}(x_n)\\ =&{}g(R_{c^{-1}}(x_1),R_{c^{-1}}(x_2),\ldots ,R_{c^{-1}}(x_n)).\\ \end{array} \end{aligned}$$

Therefore $R_{c^{-1}}:(G,f)\rightarrow (G,g)$ is a homomorphism. Since it is one-to-one and onto, $(G,f)\cong (G,g)$. Then also $\mathrm{Aut}(G,f)\cong \mathrm{Aut}(G,g)$ and ${{\mathcal {R}}}_{E(G,f)}\!\cong \!{\mathcal {R}}_{E(G,g)}.$ So it is sufficient to prove our theorem for (G, g).

The neutral element of $(G,\cdot )$ is a quasi-central idempotent of (G, g). Thus the set ${{\mathcal {R}}}_{E(G,g)}$ is non-empty. Because $R_dR_c=R_{cd}$ and $cd\in CE_{(G,g)}$ for all $c,d\in CE_{(G,g)}$, ${{\mathcal {R}}}_{E(G,g)}$ is a subgroup of $\mathrm{Aut}(G,g)$. Moreover, $(R_b\beta )^{-1}\circ R_c\circ R_b\beta =R_{\beta ^{-1}(c)}$ for any $R_b\beta \in \mathrm{Aut}(G,g)$ and $c\in CE_{(G,g)}$.

Since, by Corollary 5.4, $\,\beta$ is an automorphism of (G, g), we have $\beta ^{-1}(c)=\beta ^{-1}g(c,c,\ldots ,c)=g(\beta ^{-1}(c),\beta ^{-1}(c),\ldots ,\beta ^{-1}(c))$. Thus $\beta ^{-1}(c)\in CE_{(G,g)}$. Consequently, $R_{\beta ^{-1}(c)}\in {{\mathcal {R}}}_{E(G,g)}$. Therefore, ${\mathcal {R}}_{E(G,g)}$ is a normal subgroup of $\mathrm{Aut}(G,g)$.

Clearly $Z(\varphi )$ also is a subgroup of $\mathrm{Aut}(G,g)$ and ${{\mathcal {R}}}_{E(G,g)}\cap Z(\varphi )=\{\varepsilon \}$. This means that $\mathrm{Aut}(G,g)\cong {{\mathcal {R}}}_{E(G,g)}\ltimes Z(\varphi )$. $\square$

Corollary 5.8

An n-group $(G,f)=\mathrm{der}_{\varphi ,a}(G,\cdot )$ containing at least one quasi-central idempotent has $|CE_{(G,f)}|\!\cdot \!|Z(\varphi )|$ automorphisms.

Corollary 5.9

If an n-group $(G,f)=\mathrm{der}_{\varphi ,a}(G,\cdot )$ has only one quasi-central idempotent, then

$$\begin{aligned} \mathrm{Aut}(G,f)\cong Z(\varphi ). \end{aligned}$$

Corollary 5.10

If m|n, then $\mathrm{Aut}({\mathbb {Z}}_m,f)\cong \mathrm{Aut}({\mathbb {Z}}_m,+)$ for an n-group $(G,f)=\mathrm{der}_{\varphi }({\mathbb {Z}}_m,+)$.

Corollary 5.11

If a medial n-group $(G,f)=\mathrm{der}_{a}(G,\cdot )$ has only one idempotent, then

$$\begin{aligned} \mathrm{Aut}(G,f)\cong \mathrm{Aut}(G,\cdot ). \end{aligned}$$

Corollary 5.12

A medial n-group $(G,f)=\mathrm{der}_{\varphi ,a}(G,\cdot )$ containing at least one idempotent has only the identity automorphism if and only if $(G,\cdot )$ is the direct product of cyclic groups of order 2.

Proof

In abelian group $\alpha (x)=x^{-1}$ is an automorphism commuting with all other automorphisms. If $|\mathrm{Aut}(G,f)|=1$, then also $|Z(\varphi )|=1$. Therefore must be $\alpha =\varepsilon$. Consequently, $x=\alpha (x)=x^{-1}$ for all $x\in G$. Hence $(G,\cdot )$, as a Boolean group, is the direct product of copies of the group ${\mathbb {Z}}_2$. $\square$

To calculate the number of automorphisms of a given n-group (G, f) let’s assume that $(G,f)=\mathrm{der}_{\varphi ,a}(G,\cdot )$ and $\delta (x)=\varphi (x)\varphi ^2(x)\cdots \varphi ^{n-1}(x)$. Then $\delta$ is an endomorphism of $(G,\cdot )$. Moreover let $K=\mathrm{Ker}\,\delta$, $P=\{\beta (a)a^{-1}\,|\,\beta \in Z(\varphi )\}\cap \delta (G)$, $D=\{(c,\beta )\in Z(G,\cdot )\times Z(\varphi )\,|\,\beta (a)=a\delta (c)\}$ and $S_a=\{\beta \in Z(\varphi )\,|\,\beta (a)=a\}$.

By Proposition 5.3, the mapping $h:G\rightarrow G$ is an automorphism of an n-group $(G,f)=\mathrm{der}_{\varphi ,a}(G,\cdot )$ if and only if there are $c\in Z(G,\cdot )$ and $\beta \in Z(\varphi )$ such that $h=R_c\beta$ and $\beta (a)=\delta (c)a$.

Thus, the enumeration of all automorphisms of an n-group $(G,f)=\mathrm{der}_{\varphi ,a}(G,\cdot )$ can be reduced to the enumeration of all pairs $(c,\beta )\in Z(G,\cdot )\times Z(\varphi )$ such that $\beta (a)=\delta (c)a$. Denote the set of such pairs by D.

Note that $(c,\beta )\in D$ implies $(cd,\beta )\in D$ for every $d\in K=\mathrm{Ker}\,\delta$. On the other hand, from $(c,\beta )=(b,\beta )$ it follows $cb^{-1}\in K$. Thus each automorphism $\beta \in Z(\varphi )$ determines exactly |K| pairs $(c,\beta )\in D$, i.e. exactly |K| elements in G. Thus $|D|=|K|\cdot t$, where t is the number of possible automorphisms $\beta \in Z(\varphi )$ for which there is a pair $(c,\beta )\in D$.

To determine the number of $\beta$ corresponding to the fixed $c\in P,$ note first that for each $c\in P$ there is at lease one $g_c\in G$ that that $c=\beta (a)a^{-1}=\delta (g_c)$. If also $\delta (g)=c$, then $\delta (gg_c^{-1})=e$, i.e. $gg_c^{-1}\in K=\mathrm{Ker}\,\delta$. Therefore, $g=g_cd$ for some $d\in K.$ This means that each element $c\in P$ corresponds to exactly |K| elements $g_c\in G$ such that $\delta (g_c)=c$. This element c corresponds to $|S_a|$ elements $\beta _c$ of the group $Z(\varphi )$ such that $\beta _c(a)a^{-1} = c.$ If $c,d\in P,$ $c\ne d$, then $\delta (g_c)\ne \delta (g_d)$ and $\beta _c\ne \beta _d$. Therefore $|Aut(Q, f)| = |K|\cdot |P|\cdot |S_a|$.

In this way we obtain

Theorem 5.13

If $(G,f)=\mathrm{der}_{\varphi ,a}(G,\cdot )$, then $|Aut(Q, f)| = |P|\cdot |K|\cdot |S_a|.$