S. A. Chunikhin

ON THE FACTORIZATION OF FINITE GROUPS BY MEANS OF INDEXIALS

(Presented by Academician I. M. Vinogradov, 22 VI 1964)

§ 1.

In the paper ((^{1})) we proposed a general method of factorization of finite groups, based on the use of the orders of the factor groups of a principal series of the group.

The question of the possibility of using, for the same purpose, “smaller” elements—the orders of subgroups of the indicated factor groups—has apparently not been raised until now. Indirectly, one such case was indicated by us in ((^{2})).

Theorems 1 and 2 of the present paper give a positive answer to this question in one further case. At the same time, these theorems are parallel also to the criterion for factorization of a group into two factors contained in Theorem 1 on groups of type Π-2 from ((^{2})), generalizing it in the particular case in which the normal series considered there is also invariant.

As is not difficult to see from a comparison of the definitions of groups of type Π-2 (see ((^{2}))) and of the indexial of a group used below, this generalization is achieved by dropping, in the hypotheses of the theorem on groups of type Π-2, the coprimeness of the orders of the factors and the solvability of one of them, while the requirement that certain subgroups be conjugate is weakened.

Theorems 3 and 4 given below generalize Theorem E1* from ((^{3})). Thus, in the present paper we continue the study begun by us in ((^{2})) of the connections between the factorization properties of a group and those of the factor groups of its normal and invariant series. In doing so we use the notation and results of the method of indexials set out in ((^{4})).

§ 2.

Let

[
\mathfrak{G}=\mathfrak{G}0 \supset \mathfrak{G}_1 \supset \cdots \supset \mathfrak{G}\lambda=\mathfrak{E}
\tag{R}
]

be some series of normal divisors of the finite group (\mathfrak{G}\ne \mathfrak{E}) (i.e. its invariant series). In the case (\mathfrak{G}=\mathfrak{E}), as the series ((R)) we shall consider the series (\mathfrak{E}, \mathfrak{E}, \ldots, \mathfrak{E}) ((\lambda+1) times).

Suppose that, for the factor groups of the series ((R)), the factorizations

[
\left.
\begin{gathered}
\mathfrak{G}{i-1}/\mathfrak{G}_i
=
[\mathfrak{F}}/\mathfrak{Gi]\,[\mathfrak{F}}/\mathfrak{Gi],
\quad i=1,2,\ldots,\lambda,\
\text{for which the subgroups}\
\mathfrak{F}}/\mathfrak{G1,\ \mathfrak{F}}/\mathfrak{G2,\ \ldots,\ \mathfrak{F}}/\mathfrak{G\lambda,\quad j=1,2,\
\text{are such that one can form (see ((^{4}))) the indexial}\
(\mathfrak{F}}/\mathfrak{G1)(\mathfrak{F}}/\mathfrak{G2)\cdots(\mathfrak{F}_\lambda)=(h_j)_R.}/\mathfrak{G
\end{gathered}
\right}
\tag{F}
]

are valid.

Definition 1. The system of factorizations ((F)) will be called an indexial system of the series ((R)) of the group (\mathfrak{G}), or, more briefly, an indexial system of the group (\mathfrak{G}).

Let us note that in the factorizations ((F)) we impose no restrictions on the orders of the factors. Taking one of them to be the identity subgroup (\mathfrak{G}_i/\mathfrak{G}_i), we see that every invariant series of any finite group has at least one indexial system.

By Theorem 6 of [4], the indexial ((h_j)_R,\ j=1,2,) has at least one regular Sylow expansion

[
(\mathfrak F_{1,j}/\mathfrak G_1)(\mathfrak C_{2,j}/\mathfrak G_2)\cdots(\mathfrak C_{\lambda,j}/\mathfrak G_\lambda)
=(c_jh_j)_R,\qquad j=1,2.
\tag{1}
]

Then, on the basis of the definition of a regular indexial [4], there exists at least one suitable subgroup for the indexial (1). Let (\mathfrak H_j,\ j=1,2,) be any one of them. By the property of a regular Sylow expansion of an indexial (see [4]), we have ((\mathfrak H_j)=c_jh_j) and (\Pi(c_j)\subset \Pi(h_j),\ j=1,2.) The subgroups (\mathfrak H_j) have a number of other properties as well, a list of which can be found in [4].

Definition 2. A pair of subgroups (\mathfrak H_1,\mathfrak H_2) will be called a suitable pair of subgroups of the indexial system (F) of the series (R) of the group (\mathfrak G).

It follows immediately from the preceding that

Theorem 1. For every indexial system of a group (\mathfrak G) there exists at least one pair of suitable subgroups.

Theorem 2. If (\mathfrak H_1) and (\mathfrak H_2) are some subgroups of a group (\mathfrak G), then for the existence of a factorization (\mathfrak G=\mathfrak H_1\mathfrak H_2) it is necessary and sufficient that the pair of subgroups (\mathfrak H_1,\mathfrak H_2) be a suitable pair of subgroups for some indexial system of the group (\mathfrak G).

Proof. Sufficiency. Suppose that the theorem is false. Then let (\mathfrak G) be one of the groups of least order for which the theorem does not hold. This means that there exists such a pair of suitable subgroups (\mathfrak H_1) and (\mathfrak H_2) of some indexial system (F) of the group (\mathfrak G) that (\mathfrak G\ne\mathfrak H_1\mathfrak H_2). Since, for (\mathfrak G=\mathfrak E), we would have (\mathfrak H_1=\mathfrak E,\ \mathfrak H_2=\mathfrak E), and (\mathfrak G=\mathfrak H_1\mathfrak H_2), it follows that (\mathfrak G\ne\mathfrak E).

The subgroups (\mathfrak H_1) and (\mathfrak H_2) correspond to the indexials ((h_1)_R) and ((h_2)_R) of the series (R). Consider the factorization

[
\mathfrak G/\mathfrak G_1=[\mathfrak F_{1,1}/\mathfrak G_1]\,[\mathfrak F_{1,2}/\mathfrak G_1].
\tag{2}
]

Let first (\lambda=1), i.e., let (\mathfrak G_1=\mathfrak E). Then (\mathfrak G=\mathfrak F_{1,1}\mathfrak F_{1,2}).

According to the definition of a suitable subgroup of an indexial and in view of (\mathfrak G_1=\mathfrak E), we have

[
\mathfrak F_{1,1}=[\mathfrak H_1\cap\mathfrak G_0]\mathfrak G_1=\mathfrak H_1,\qquad
\mathfrak F_{1,2}=[\mathfrak H_2\cap\mathfrak G_0]\mathfrak G_1=\mathfrak H_2.
]

Thus, (\mathfrak G=\mathfrak H_1\mathfrak H_2). We have obtained a contradiction.

Hence, (\lambda>1), i.e., (\mathfrak G_2) exists. It is now not difficult to verify that, since (1) is an indexial of the series (R), the product
((\mathfrak C_{2,j}/\mathfrak G_2)(\mathfrak C_{3,j}/\mathfrak G_3)\cdots(\mathfrak C_{\lambda,j}/\mathfrak G_\lambda),\ j=1,2,) will be an indexial for the series

[
\mathfrak G_1\supset \mathfrak G_2\supset\cdots\supset \mathfrak G_\lambda=\mathfrak E.
\tag{R}
]

We shall now show that

[
\left.
\begin{gathered}
\mathfrak H_j\cap\mathfrak G_1=\mathfrak H'j,\qquad j=1,2,\
\text{will be a suitable subgroup for the indexial}\
(\mathfrak G_1/\mathfrak G_1)(\mathfrak C/\mathfrak G_\lambda)\}/\mathfrak G_2)\cdots(\mathfrak C_{\lambda,j
\text{of the series (R) and for the indexial}\
(\mathfrak C_{2,j}/\mathfrak G_2)\cdots(\mathfrak C_{\lambda,j}/\mathfrak G_\lambda)\
\text{of the series }(R_1).
\end{gathered}
\right}
\tag{3}
]

Since (\mathfrak h_j' \subseteq \mathfrak G_1 \subset \mathfrak G_0=\mathfrak G), we have

[
\mathfrak G_1=[\mathfrak h_j'\cap \mathfrak G_0]\mathfrak G_1 .
\tag{4}
]

Further, taking into account that (\mathfrak h_j,\ j=1,2,) is a suitable subgroup for the indexial (1) and that (\mathfrak G_{i-1}\supset \mathfrak G_i,\ i=2,3,\ldots,\lambda), we have:

[
\mathfrak G_{i,j}=[\mathfrak h_j\cap \mathfrak G_{i-1}]\mathfrak G_i
=[\mathfrak h_j'\cap \mathfrak G_{i-1}]\mathfrak G_i .
\tag{5}
]

But (4), (5), and (\mathfrak h_j'\subseteq \mathfrak G_1), on the basis of the definition of a suitable subgroup (see (4)), prove (3).

We shall further distinguish the following two cases.

1) (\mathfrak G/\mathfrak G_1=\mathfrak F_{1,1}/\mathfrak G_1). In this case it is obvious that (\mathfrak F_{1,1}=\mathfrak G=\mathfrak h_1\mathfrak G_1) and that from (F), on the basis of the definition of an indexial (4), for (i=2,3,\ldots,\lambda) there is obtained an indexial system of the series ((\mathrm R_1)) of the group (\mathfrak G_1). Replacing in (F) the factors (\mathfrak F_{i,j}/\mathfrak G_i) by their extensions (\mathfrak C_{i,j}/\mathfrak G_i), we obtain the new factorization

[
\mathfrak G_{i-1}/\mathfrak G_i=[\mathfrak C_{i,1}/\mathfrak G_i]\,[\mathfrak C_{i,2}/\mathfrak G_i],
\quad i=2,3,\ldots,\lambda .
\tag{F'}
]

It is obvious that the equalities ((\mathrm F')) are also an indexial system of the series ((\mathrm R_1)) of the group (\mathfrak G_1).

Since the series ((\mathrm R)) has no repetitions, ((\mathfrak G_1)<(\mathfrak G)), and therefore the theorem will be true for (\mathfrak G_1). From this it follows, taking (3) into account, that the following factorization of the group (\mathfrak G_1) holds:

[
\mathfrak G_1=\mathfrak h_1\mathfrak h_2'
=[\mathfrak h_1\cap \mathfrak G_1]\,[\mathfrak h_2\cap \mathfrak G_1].
\tag{6}
]

As we established above, (\mathfrak G=\mathfrak h_1\mathfrak G_1). Hence, in view of (6), we have

[
\mathfrak G=\mathfrak h_1[\mathfrak h_1\cap \mathfrak G_1][\mathfrak h_2\cap \mathfrak G_1]
=\mathfrak h_1[\mathfrak h_2\cap \mathfrak G_1]\subseteq \mathfrak h_1\mathfrak h_2 .
]

But (\mathfrak h_1\mathfrak h_2\subseteq \mathfrak G). Consequently, (\mathfrak G=\mathfrak h_1\mathfrak h_2)—a contradiction.

2) (\mathfrak G/\mathfrak G_1\ne \mathfrak F_{1,1}/\mathfrak G_1). Consider the proper subgroup (\mathfrak F_{1,1}/\mathfrak G_1) of the group (\mathfrak G/\mathfrak G_1). If it were the case that (\mathfrak F_{1,1}=\mathfrak G_1), then, obviously, (\mathfrak G/\mathfrak G_1=[\mathfrak G_1/\mathfrak G_1]\times[\mathfrak F_{1,2}/\mathfrak G_1]=\mathfrak F_{1,2}/\mathfrak G_1), and we would again arrive at a contradiction, arguing as in case 1). Hence, taking into account the inequality (\lambda>1) established earlier, we conclude that there exists a series without repetitions

[
\mathfrak F_{1,1}\supset \mathfrak G_1\supset \mathfrak G_2\supset \cdots \supset \mathfrak G_\lambda=\mathfrak E,
\tag{R_2}
]

and moreover the factorizations

[
\mathfrak F_{1,1}/\mathfrak G_1=[\mathfrak F_{1,1}/\mathfrak G_1]\,[\mathfrak G_1/\mathfrak G_1],
]

[
\mathfrak G_{i-1}/\mathfrak G_i=[\mathfrak F_{i,1}/\mathfrak G_i]\,[\mathfrak F_{i,2}/\mathfrak G_i],
\quad i=2,3,\ldots,\lambda .
\tag{7}
]

Replacing in the last of the equalities (7) the groups (\mathfrak F_{i,j}/\mathfrak G_i) by the extensions (\mathfrak C_{i,j}/\mathfrak G_i), we arrive at the following factorizations

[
\mathfrak F_{1,1}/\mathfrak G_1=[\mathfrak F_{1,1}/\mathfrak G]\,[\mathfrak G_1/\mathfrak G_1],
]

[
\mathfrak G_{i-1}/\mathfrak G_i=[\mathfrak C_{i,1}/\mathfrak G_i]\,[\mathfrak C_{i,2}/\mathfrak G_i],
\quad i=2,3,\ldots,\lambda .
\tag{F''}
]

According to the definition of a suitable subgroup, we have: (\mathfrak F_{1,1}=\mathfrak h_1\mathfrak G_1) and (\mathfrak F_{1,2}=\mathfrak h_2\mathfrak G_1). Therefore (\mathfrak G/\mathfrak G_1=[\mathfrak F_{1,1}/\mathfrak G_1]\,[\mathfrak F_{1,2}/\mathfrak G_1]=[\mathfrak h_1\mathfrak G_1/\mathfrak G_1]\,[\mathfrak h_2\mathfrak G_1/\mathfrak G_1]). But then

[
\mathfrak G=\mathfrak h_1\mathfrak h_2\mathfrak G_1=\mathfrak F_{1,1}\mathfrak h_2 .
\tag{8}
]

It is not hard to verify that ((\mathfrak F_{1,1}/\mathfrak G_1)(\mathfrak C_{2,1}/\mathfrak G_2)\ldots(\mathfrak C_{\lambda,1}/\mathfrak G_\lambda)) and ((\mathfrak G_1/\mathfrak G_1)\times(\mathfrak C_{2,2}/\mathfrak G_2)\ldots(\mathfrak C_{\lambda,2}/\mathfrak G_\lambda)) will be indexials also for the group (\mathfrak F_{1,1}) with respect to the series ((\mathrm R_2)). This means that the equalities ((\mathrm F'')) form an indexial system of the series ((\mathrm R_2)) of the group (\mathfrak F_{1,1}).

By hypothesis, ((\mathfrak F_{1,1})<(\mathfrak G)). Therefore the theorem will be valid for (\mathfrak F_{1,1}). Hence, taking into consideration that from (\mathfrak F_{1,1}=\mathfrak h_1\mathfrak G_1) it follows that (\mathfrak h_1\subseteq \mathfrak F_{1,1}), and also that the replacement of ((\mathrm R)) by ((\mathrm R_2)) preserves (3), we see,

that for (\mathfrak{F}{1,1}) the factorization (\mathfrak{F}_1]) holds. Hence, on the basis of (8), we obtain:}=\mathfrak{H}_1[\mathfrak{H}_2\cap\mathfrak{G

[
\mathfrak{G}=\mathfrak{F}_{1,1}\mathfrak{H}_2
=\mathfrak{H}_1[\mathfrak{H}_2\cap\mathfrak{G}_1]\mathfrak{H}_2
=\mathfrak{H}_1\mathfrak{H}_2 .
]

Again we have obtained a contradiction.

Necessity. Let (\mathfrak{G}=\mathfrak{H}_1\mathfrak{H}_2), where (\mathfrak{H}_1) and (\mathfrak{H}_2) are some subgroups of (\mathfrak{G}). Then (\mathfrak{H}_1) and (\mathfrak{H}_2) will, obviously, be suitable subgroups of the indexials ((\mathfrak{H}_1/\mathfrak{G})) and ((\mathfrak{H}_2/\mathfrak{G})) of the series (\mathfrak{G}\supseteq\mathfrak{E}) of the group (\mathfrak{G}).

§ 3. Theorem 3. If (\mathfrak{G}) has a normal divisor (\mathfrak{G}_1), and a factorization

[
\mathfrak{G}/\mathfrak{G}_1=[\mathfrak{F}_1/\mathfrak{G}_1][\mathfrak{F}_2/\mathfrak{G}_1]\cdots[\mathfrak{F}_k/\mathfrak{G}_1]
]

holds, then the factorization (\mathfrak{G}=\overline{\mathfrak{F}}_1\overline{\mathfrak{F}}_2\cdots\overline{\mathfrak{F}}_k\mathfrak{G}_1) also holds, where (\overline{\mathfrak{F}}_i) is a subgroup of (\mathfrak{G}) for which (\overline{\mathfrak{F}}_i\cap\mathfrak{G}_1=\mathfrak{C}_i) is a special subgroup, and moreover (\overline{\mathfrak{F}}_i/\mathfrak{C}_i\simeq \mathfrak{F}_i/\mathfrak{G}_1) and (\Pi((\mathfrak{C}_i))\subseteq \Pi((\mathfrak{F}_i/\mathfrak{G}_1))), (i=1,2,\ldots,k).

Proof. Consider the indexial ((\mathfrak{F}_i/\mathfrak{G}_1)(\mathfrak{G}/\mathfrak{G})), (i=1,2,\ldots,k), of the series (\mathfrak{G}\supseteq\mathfrak{G}_1\supseteq\mathfrak{E}). By Theorem 6 ((^4)), this indexial has a regular s.a.z. extension ((\mathfrak{F}_i/\mathfrak{G}_1)(\mathfrak{C}_i/\mathfrak{G})), an arbitrary suitable subgroup of which we denote by (\overline{\mathfrak{F}}_i). By the definition of a suitable subgroup ((^4)), we have:

[
\mathfrak{F}_i=\overline{\mathfrak{F}}_i\mathfrak{G}_1,\qquad
\mathfrak{C}_i=[\overline{\mathfrak{F}}_i\cap\mathfrak{G}_1]\mathfrak{E}
=\overline{\mathfrak{F}}_i\cap\mathfrak{G}_1 .
\tag{9}
]

Hence (\overline{\mathfrak{F}}_i/\mathfrak{C}_i\simeq \mathfrak{F}_i\mathfrak{G}_1). According to the definition of a s.a.z. extension of an indexial ((^4)), the subgroup (\mathfrak{C}_i) is special and (\Pi((\mathfrak{C}_i))\subseteq\Pi((\mathfrak{F}_i/\mathfrak{G}_1))).

Further, from the conditions of the theorem and (9) it follows that

[
\mathfrak{G}=\mathfrak{F}_1\mathfrak{F}_2\cdots\mathfrak{F}_k
=\overline{\mathfrak{F}}_1\overline{\mathfrak{F}}_2\cdots\overline{\mathfrak{F}}_k\mathfrak{G}_1 .
]

All these properties of the subgroups (\overline{\mathfrak{F}}_1,\overline{\mathfrak{F}}_2,\ldots,\overline{\mathfrak{F}}_k) show that the theorem holds for (\mathfrak{G}).

Theorem 4. If (\mathfrak{G}) has a normal series (\mathfrak{G}=\mathfrak{G}0\supseteq\mathfrak{G}_1\supseteq\cdots\supseteq\mathfrak{G}\lambda=\mathfrak{E}), and factorizations

[
\mathfrak{G}{i-1}/\mathfrak{G}_i
=[\mathfrak{F}}/\mathfrak{Gi]\times[\mathfrak{F}}/\mathfrak{Gi]\cdots[\mathfrak{F}_i],}/\mathfrak{G
\quad i=1,2,\ldots,\lambda,\quad k_i\geq 1,
]

hold, then the factorization

[
\mathfrak{G}
=[\overline{\mathfrak{F}}{1,1}\overline{\mathfrak{F}}\cdots
\overline{\mathfrak{F}}{1,k_1}]
[\overline{\mathfrak{F}}}\overline{\mathfrak{F}{2,2}\cdots
\overline{\mathfrak{F}}]
\cdots
[\overline{\mathfrak{F}}{\lambda,1}\overline{\mathfrak{F}}\cdots
\overline{\mathfrak{F}}{\lambda,k\lambda}]
]

*also holds, where (\overline{\mathfrak{F}}{i,j}) is a subgroup of (\mathfrak{G}) for which (\overline{\mathfrak{F}}}\cap\mathfrak{Gi=\mathfrak{C}}) is a special subgroup, and moreover (\overline{\mathfrak{F}{i,j}/\mathfrak{C}}\simeq \mathfrak{F{i,j}/\mathfrak{G}_i) and (\Pi((\mathfrak{C}}))\subseteq\Pi((\mathfrak{F{i,j}/\mathfrak{G}_i))), and (\overline{\mathfrak{F}}) is a special group when (i}\cap\overline{\mathfrak{F}}_{i_1,j_1