UDC 519.44

MATHEMATICS

S. A. RUSAKOV

A GENERALIZATION OF POST’S THEOREM

(Presented by Academician V. M. Glushkov on 29 I 1970)

§ 1. As is known, Lagrange’s theorem for finite \(n\)-groups is in general irreversible already for \(n=2\). Therefore it is very important to find those classes of finite \(n\)-groups for which the converse of Lagrange’s theorem holds at least for a certain kind of divisor of the order of the \(n\)-group.

In the present paper we study the “saturation” of Abelian \(n\)-groups (see Definition 3) by subgroups; the results obtained by us generalize a well-known result of E. Post \(\left({}^{2}\right.\), p. 284) on the existence of subgroups in cyclic \(n\)-groups, expressed by the following theorem:

Let \(\mathfrak{G}\) be a cyclic \(n\)-group of order \(g=rs\), where \(r\) is the largest divisor of \(g\) relatively prime to \(n-1\). Then \(\mathfrak{G}\) has at least one element and exactly one subgroup of those and only those orders \(\gamma\) for which \(\gamma=\delta s\), where \(\delta\) is an arbitrary divisor of \(r\).

§ 2. We give the definitions and notation used in the paper.

Let \(\mathfrak{G}\) be a set on which an \(n\)-ary operation \(\sigma_n\) is defined \(\left({}^{1}\right.\), p. 5), where \(n \ge 2\) is the arity of the operation, and let \(\sigma_n(x_1x_2 \ldots x_n)\) be the value of the \(n\)-ary operation \(\sigma_n\) applied to the elements \(x_1,x_2,\ldots,x_n\) of \(\mathfrak{G}\). Then we have

Definition 1 (cf. \(\left({}^{3}\right.\), p. 158; \(\left({}^{2}\right.\), p. 213). The set \(\mathfrak{G}\) is called an \(n\)-group if the following postulates are satisfied:

1. The operation \(\sigma_n\) is associative, i.e., for any elements \(x_1,x_2,\ldots,x_n,x_{n+1},\ldots,x_{2n-1}\) of \(\mathfrak{G}\) the equality
   \[ \sigma_n(\sigma_n(x_1x_2 \ldots x_n)x_{n+1}\ldots x_{2n-1}) = \sigma_n(x_1x_2 \ldots x_j\sigma_n(x_{j+1}\ldots x_{j+n})\ldots x_{2n-1}), \]
   where \(j=1,2,\ldots,n-1\), holds.

2. The law of single-valued and unrestricted invertibility holds, i.e., for any elements \(a_1,a_2,\ldots,a_{i-1},a_{i+1},\ldots,a_n,a\) belonging to \(\mathfrak{G}\), each of the equations
   \[ \sigma_n(a_1a_2 \ldots a_{i-1}x_i a_{i+1}\ldots a_n)=a \quad (i=1,2,\ldots,n) \]
   is always solvable in \(\mathfrak{G}\) with respect to \(x_i\), and uniquely.

The concept of a subgroup for \(n\)-groups with \(n>2\) is introduced in the same way as for \(n=2\).

We shall denote the cardinality of any set \(\mathfrak{S}\) (in particular, of an \(n\)-group \(\mathfrak{G}\)) by \(|\mathfrak{S}|\). If \(\mathfrak{S}=\mathfrak{G}\), then \(|\mathfrak{G}|\) will be called the order of the \(n\)-group \(\mathfrak{G}\). If \(|\mathfrak{G}|\) is finite, the \(n\)-group \(\mathfrak{G}\) is also called finite.

Let \(\Gamma\) be the set of all nonempty subsets composed of elements of the \(n\)-group \(\mathfrak{G}\). On this set we define an \(n\)-ary operation \(\omega_n\) in the following way. Let \(\mathfrak{M}_i \in \Gamma\) \((i=1,2,\ldots,n)\). Then by
\[ \omega_n(\mathfrak{M}_1\mathfrak{M}_2 \ldots \mathfrak{M}_n) \]
we shall understand the set of all elements of \(\mathfrak{G}\), each of which is equal to \(\sigma_n(m_1m_2 \ldots m_n)\), where \(m_i \in \mathfrak{M}_i\) and \(m_i\) takes an arbitrary value from \(\mathfrak{M}_i\). It is clear that some of \(\mathfrak{M}_1,\mathfrak{M}_2,\ldots,\mathfrak{M}_n\), or possibly all of them, may consist of a single element. If \(\mathfrak{M}_i=\{m_i\}\) \((i=1,2,\ldots,n)\), then obviously
\[ \omega_n(m_1m_2 \ldots m_n)=\sigma_n(m_1m_2 \ldots m_n). \]

Definition 2 \(\left(\left({}^{3}\right.\right.\), p. 165). A subgroup \(\mathfrak{H}\) of an \(n\)-group \(\mathfrak{G}\) is called invariant in \(\mathfrak{G}\) if, for any element \(x \in \mathfrak{G}\), the equality
\[ \omega_n(x\mathfrak{H}\ldots \mathfrak{H}) = \omega_n(\underbrace{\mathfrak{H}\ldots \mathfrak{H}}_{i-1}x\underbrace{\mathfrak{H}\ldots \mathfrak{H}}_{n-i}) \quad (i=2,\ldots,n) \]
holds. If
\[ \omega_n(x\mathfrak{H}\ldots \mathfrak{H}) = \omega_n(\mathfrak{H}\ldots \mathfrak{H}x), \]
then \(\mathfrak{H}\) is called a semi-invariant subgroup of the \(n\)-group \(\mathfrak{G}\).

Definition 3 (cf. (²), p. 217). An \(n\)-group \(\mathfrak G\) will be called Abelian if, for any elements \(x_1, x_2, \ldots, x_n\) of \(\mathfrak G\), the value \(\sigma_n(x_1x_2\ldots x_n)\) does not change under any permutation of these elements.

§ 3. We shall also need the following theorems.

Theorem 1 (cf. (³), p. 163). Let \(\mathfrak H\) be some subgroup of an arbitrary \(n\)-group \(\mathfrak G\), and let \(a_1, a_2, \ldots, a_{k-1}, a_{k+l}, \ldots, a_n\) be fixed elements of the \(n\)-group \(\mathfrak G\), where \(k \geqslant 1\), \(l \geqslant 1\). If in

\[ \omega_n(a_1a_2 \ldots a_{k-1}\underbrace{\mathfrak H \ldots \mathfrak H}_{l}a_{k+l}\ldots a_n) \]

one of the elements \(a_i\) \((i=1,2,\ldots,k-1,k+l,\ldots,n)\) is replaced by the variable element \(x\), then, for distinct \(x\), two such obtained subsets of the \(n\)-group \(\mathfrak G\) either coincide or have not a single common element; such subsets have the same cardinality and, taken together, exhaust the entire \(n\)-group \(\mathfrak G\). If, moreover, \(\mathfrak G\) is finite, then the cardinality of each such subset is equal to \(|\mathfrak H|\).

Theorem 2 ((³), p. 165). If \(\mathfrak H\) is a semi-invariant subgroup for an \(n\)-group \(\mathfrak G\), then all subsets of the \(n\)-group \(\mathfrak G\) of the form \(\omega_n(x\mathfrak H\ldots\mathfrak H)\) form an \(n\)-group with respect to the operation \(\omega_n\).

We shall call such an \(n\)-group the factor group for \(\mathfrak G\) with respect to \(\mathfrak H\), and denote it by \(\mathfrak G/\mathfrak H\). In what follows we shall consider only finite \(n\)-groups.

§ 4. We now state the results obtained by us.

Theorem 3. A nonempty subset \(\mathfrak H\) of a finite \(n\)-group \(\mathfrak G\) is an \(n\)-subgroup if and only if \(\mathfrak H\) is a subset on which the \(n\)-ary operation \(\sigma_n\) is defined.

The proof is carried out by the same method as for \(n=2\).

Theorem 4. Let the factor group \(\mathfrak G/\mathfrak N\) for a finite \(n\)-group \(\mathfrak G\) with respect to \(\mathfrak N\) possess some subgroup \(\mathfrak B\). Then \(\mathfrak G\) contains such a subgroup \(\mathfrak B\) that \(|\mathfrak B|=|\mathfrak N|\,|\overline{\mathfrak B}|\).

Proof. On the basis of Theorem 1, the \(n\)-group \(\mathfrak G\) can be represented as

\[ \mathfrak G=\omega_n(x_1\mathfrak N\ldots\mathfrak N)+\omega_n(x_2\mathfrak N\ldots\mathfrak N)+\ldots+\omega_n(x_\rho\mathfrak N\ldots\mathfrak N), \tag{1} \]

where \(|\omega_n(x_i\mathfrak N\ldots\mathfrak N)|=|\mathfrak N|\), \(i=1,2,\ldots,\rho\). Now considering in (1) each summand as a separate element, we obtain, by Theorem 2, the factor group \(\mathfrak G/\mathfrak N\). Since \(\overline{\mathfrak B}\subseteq \mathfrak G/\mathfrak N\), we have

\[ \overline{\mathfrak B}=\omega_n(y_1\mathfrak N\ldots\mathfrak N)+\omega_n(y_2\mathfrak N\ldots\mathfrak N)+\ldots+\omega_n(y_\tau\mathfrak N\ldots\mathfrak N), \tag{2} \]

where \(y_1,y_2,\ldots,y_\tau\) are among \(x_1,x_2,\ldots,x_\rho\), and \(\tau=|\overline{\mathfrak B}|\).

Let now \(\mathfrak B\) be the collection of all elements of the \(n\)-group \(\mathfrak G\) of the form \(\sigma_n(y_jv_1\ldots v_{n-1})\), where \(j=1,2,\ldots,\tau\) and \(v_1,\ldots,v_{n-1}\) are arbitrary elements of \(\mathfrak N\). Then it is obvious that

\[ \sigma_n(y_jv_1\ldots v_{n-1}) \in \omega_n(y_j\mathfrak N\ldots\mathfrak N). \tag{3} \]

Let us show that \(\mathfrak B\) is a subgroup of the \(n\)-group \(\mathfrak G\). Indeed, let
\[ \sigma_n(z_1v'_1\ldots v'_{n-1}),\quad \sigma_n(z_2v''_1\ldots v''_{n-1}),\quad \ldots,\quad \sigma_n(z_nv^{(n)}_1\ldots v^{(n)}_{n-1}) \]
be arbitrary elements of \(\mathfrak B\), where \(z_1,z_2,\ldots,z_n\) are among the elements \(y_1,y_2,\ldots,y_\tau\). Then, taking equality (3) into account, we obtain

\[ \sigma_n\bigl(\sigma_n(z_1v'_1\ldots v'_{n-1}),\sigma_n(z_2v''_1\ldots v''_{n-1}),\ldots,\sigma_n(z_nv^{(n)}_1\ldots v^{(n)}_{n-1})\bigr)=z\in\mathfrak N= \]
\[ =\omega_n\bigl(\omega_n(z_1\mathfrak N\ldots\mathfrak N)\omega_n(z_2\mathfrak N\ldots\mathfrak N)\ldots\omega_n(z_n\mathfrak N\ldots\mathfrak N)\bigr). \]

Since \(\overline{\mathfrak B}\) is a subgroup of the factor group \(\mathfrak G/\mathfrak N\), it follows that \(\mathfrak N\in\overline{\mathfrak B}\), and, consequently, \(\mathfrak N=\omega_n(y_\lambda\mathfrak N\ldots\mathfrak N)\) \((1\leqslant\lambda\leqslant\tau)\). Hence \(z\in\mathfrak B\), and by Theorem 3, \(\mathfrak B\) will be a subgroup of the \(n\)-group \(\mathfrak G\). Further, since each summand in (2) has \(|\mathfrak N|\) elements from \(\mathfrak G\) and these summands have no common elements, we have \(|\mathfrak B|=|\mathfrak N|\tau=|\mathfrak N|\,|\overline{\mathfrak B}|\). The theorem is proved.

The definitions and notation we use, relating to the degree and order of an element of an \(n\)-group, can be found in \((^2)\), p. 282.

Theorem 5. An Abelian \(n\)-group \(\mathfrak G\) of order \(g=rs\), where \((r,s)=1\) and \((r,n-1)=1\), has a subgroup of order \(\delta s\), where \(\delta\) is an arbitrary divisor of \(r\).

Proof. Suppose that the theorem is false. Then, among all Abelian \(n\)-groups satisfying the condition of the theorem, choose an \(n\)-group \(\mathfrak G\) of least order \(g\) for which the theorem does not hold. Since for \(g=1\) the theorem holds, we have \(g>1\).

We shall subsequently consider the following possibilities:

1. In \(\mathfrak G\) there is at least one element of first order. Let \(a\) be an element of first order of the \(n\)-group \(\mathfrak G\), i.e. \(\sigma_n(aa\ldots a)=a\). Then the \((n-1)\)-term sequence \(\{a,a,\ldots,a\}\) is the identity of the \(n\)-group \(\mathfrak G\) (see \((^2)\), p. 214). On the set \(\mathfrak G\) we define a binary operation \(\sigma_2\) as follows:

\[ \sigma_2(x_1x_2)=x_1x_2=\sigma_n(x_1x_2a\ldots a), \tag{4} \]

where \(x_1\) and \(x_2\) are arbitrary elements of \(\mathfrak G\).

We shall show that with respect to this operation \(\mathfrak G\) is a 2-group. Indeed, taking (4) into account, we have
\((x_1x_2)x_3=\sigma_n(\sigma_n(x_1x_2a\ldots a)x_3a\ldots a)\) and
\(x_1(x_2x_3)=\sigma_n(x_1\sigma_n(x_2x_3a\ldots a)a\ldots a)\), where \(x_1\), \(x_2\), and \(x_3\) are arbitrary elements of \(\mathfrak G\).

By postulate 1 of Definition 1 and taking into account that \(\mathfrak G\) is an Abelian \(n\)-group, we obtain that
\[ \sigma_n(\sigma_n(x_1x_2a\ldots a)x_3a\ldots a) = \sigma_n(x_1\sigma_n(x_2x_3a\ldots a)a\ldots a). \]
Therefore \((x_1x_2)x_3=x_1(x_2x_3)\), i.e. associativity holds for the binary operation.

Since the equation
\(\sigma_n(xb_1a\ldots a)=b\) (\(b_1\) and \(b\) are arbitrary elements of the \(n\)-group \(\mathfrak G\)) is always uniquely solvable in \(\mathfrak G\) with respect to \(x\), the equation \(xb_1=b\) is also uniquely solvable in \(\mathfrak G\) with respect to \(x\). The same assertion is valid for the equation \(b_1y=b\). Consequently, \(\mathfrak G\) is a 2-group.

We now show that for any elements \(x_1,x_2,\ldots,x_n\) of \(\mathfrak G\) the equality

\[ x_1x_2\ldots x_n=\sigma_n(x_1x_2\ldots x_n) \tag{5} \]

holds.

Indeed, from equality (4) it follows that

\[ \begin{aligned} x_1x_2x_3\ldots x_n &=(\ldots((x_1x_2)x_3)\ldots)x_{n-1})x_n \\ &=\sigma_n\bigl(\sigma_n(\ldots(\sigma_n(\sigma_n(x_1x_2a\ldots a)x_3a\ldots a)\ldots)x_{n-1}a\ldots a)x_na\ldots a\bigr) \\ &=\sigma_n(x_1x_2\underbrace{a\ldots a}_{n-2}x_3\underbrace{a\ldots a}_{n-2}\ldots x_{n-1}\underbrace{a\ldots a}_{n-2}x_n\underbrace{a\ldots a}_{n-2}), \end{aligned} \]

and, as is easy to show, the number of all \(a\)’s occurring under the sign \(\sigma_n\) is \((n-1)(n-2)\). Since \(\mathfrak G\) is an Abelian \(n\)-group, we have

\[ x_1x_2\ldots x_n = \sigma_n\bigl(x_1x_2\ldots x_n\underbrace{a\ldots a}_{(n-1)(n-2)}\bigr) = \sigma_n\bigl(x_1x_2\ldots x_na\ldots a\ldots a\ldots a\bigr). \]

Here the last displayed occurrence consists of two blocks of \(a\)’s, each containing \(n-1\) terms:

\[ \sigma_n\bigl(x_1x_2\ldots x_n \underbrace{a\ldots a}_{n-1} \ldots \underbrace{a\ldots a}_{n-1} \bigr). \]

Since the \((n-1)\)-term sequence \(\{a,a,\ldots,a\}\) is the identity of the \(n\)-group \(\mathfrak G\), it follows that \(x_1x_2\ldots x_n=\sigma_n(x_1x_2\ldots x_n)\).

It is not difficult to show that the 2-group \(\mathfrak G\) is also Abelian. Therefore \(\mathfrak G\), as an Abelian 2-group of order \(g=rs\), has a subgroup \(\mathfrak H\) of order \(\delta s\), where \(\delta\) is an arbitrary divisor of \(r\). We shall show that \(\mathfrak H\) is a subgroup of the \(n\)-group \(\mathfrak G\). Indeed, let \(h_1,h_2,\ldots,h_n\) be arbitrary elements of \(\mathfrak H\). Then \(h_1h_2\ldots h_n\in\mathfrak H\). Hence, from equality (5) we conclude that \(\sigma_n(h_1h_2\ldots h_n)\in\mathfrak H\), i.e. by Theorem 3, \(\mathfrak H\) is a subgroup of the \(n\)-group \(\mathfrak G\). We have a contradiction.

1. The order of any element of the \(n\)-group \(\mathfrak G\) is different from 1.

Let \(b\) be an arbitrary element of the \(n\)-group \(\mathfrak G\), and let \(\mathfrak B\) be the cyclic subgroup generated by this element. By Lagrange’s theorem for \(n\)-groups \(\left({}^{2},\text{ p. }222\right)\), \(g_1=|\mathfrak B|\) is a divisor of \(g\). Since \((r,s)=1\), we may write \(g_1\) as follows: \(g_1=r_1s_1\), where \(r_1\) and \(s_1\) divide \(r\) and \(s\), respectively. Then \((r_1,s_1)=1\). Therefore \(\mathfrak B\) contains an element, and consequently also a subgroup, of order \(s\). Indeed, let us require that \((b^{[\theta]})^{[s_1]}=b^{[\theta]}\), where \(\theta\) is, for the moment, an unknown number. On the basis of relation 2 (see \(\left({}^{2},\text{ p. }282\right)\)) we have
\[ b^{[(n-1)\theta s_1+\theta+s_1]}=b^{[\theta]}. \]
Since the order of the element \(b\) is the number \(g_1\), it follows, according to E. Post’s assertion \(\left({}^{2},\text{ p. }283\right)\), that the congruence
\[ (n-1)\theta s_1+\theta+s_1-\theta\equiv 0\pmod {g_1} \]
holds. Hence it follows that
\[ s_1(n-1)\theta\equiv -s_1\pmod {g_1}. \tag{6} \]

According to the hypothesis of the theorem, \((r,n-1)=1\). Therefore \((s_1(n-1),g_1)=s_1\), and hence the congruence (6) has in all \(s_1\) distinct solutions
\[ t,\ t+r_1,\ldots,\ t+(s_1-1)r_1, \]
where \(t\) is a solution of the congruence
\[ (n-1)\theta\equiv -1\pmod {r_1}. \]

Now consider the class of numbers congruent to \(t\) modulo \(r_1\). Let \(c\) be an arbitrary positive number belonging to this class. Then
\[ c=g_1q+g_2,\quad \text{where } 0\geq g_2<g_1, \]
and therefore
\[ (b^{[g_1q+g_2]})^{[s_1]}=b^{[g_1q+g_2]}. \]
Hence, and from the fact that \(g_1\) is the order of the element \(b\), follows the equality
\[ (b^{[g_2]})^{[s_1]}=b^{[g_2]}, \]
i.e., in \(\mathfrak B\) there exists an element \(b_1=b^{[g_2]}\), and consequently also a subgroup \(\mathfrak S\) of order \(s_1\). If \(s_1=1\), then in \(\mathfrak G\) there would exist an element \(b_1\) of first order, contrary to the case under consideration. Therefore we shall assume \(s_1>1\), i.e. \(|\mathfrak S|>1\). In view of the abelianness of the \(n\)-group \(\mathfrak G\), we conclude that \(\mathfrak S\) is an invariant subgroup.

Consider the factor group \(\mathfrak G/\mathfrak S\). It is easy to show that \(\mathfrak G/\mathfrak S\) is an abelian \(n\)-group. Since
\[ |\mathfrak G/\mathfrak S|=r\frac{s}{s_1}<g \]
and
\[ \left(r,\frac{s}{s_1}\right)=1, \]
the theorem is true for \(\mathfrak G/\mathfrak S\), i.e. \(\mathfrak G/\mathfrak S\) contains a subgroup \(\mathfrak H\) of order
\[ \delta\frac{s}{s_1}, \]
where \(\delta\) is an arbitrary divisor of \(r\). Then, on the basis of Theorem 4, we conclude that \(\mathfrak G\) contains a subgroup \(\mathfrak H\) of order \(\delta s\). We have again obtained a contradiction. Thus the theorem is completely proved.

From Theorem 5 follows the following

Corollary 1. An abelian \(n\)-group \(\mathfrak G\) of order \(g=rs\), where \(r\) is the greatest divisor of \(g\) relatively prime to \(n-1\), possesses a subgroup of order \(\delta s\), where \(\delta\) is an arbitrary divisor of \(r\).

Proof. Since \(r\) is the greatest divisor of \(g\) relatively prime to \((n-1)\), we have \((r,s)=1\), and on the basis of Theorem 5 we conclude that \(\mathfrak G\) possesses a subgroup of order \(\delta s\), where \(\delta\) is an arbitrary divisor of \(r\).

Corollary 1 generalizes E. Post’s theorem \(\left({}^{2},\text{ p. }284\right)\) on the existence of subgroups in cyclic \(n\)-groups.

Gomel Laboratory
Institute of Mathematics
Academy of Sciences of the BSSR

Received
12 XII 1969

References

\(^{1}\) V. D. Belousov, Foundations of the Theory of Quasigroups and Loops, Moscow, 1967.
\(^{2}\) E. L. Post, Trans. Am. Math. Soc., 48, No. 2, 208 (1940).
\(^{3}\) A. K. Sushkevich, Theory of Generalized Groups, Kharkov—Kiev, 1937.