Abstract
Full Text
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:
-
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. -
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}})
\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\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'),\quad
\sigma_n(z_2v''1\ldots v''),\quad \ldots,\quad
\sigma_n(z_nv^{(n)}1\ldots v^{(n)})
]
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'),\sigma_n(z_2v''1\ldots v''}),\ldots,\sigma_n(z_nv^{(n)1\ldots v^{(n)})\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:
- 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}}\ldots x_{n-1}\underbrace{a\ldots a{n-2}x_n\underbrace{a\ldots a}),
\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}
\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.
- 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_21), 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}