MATHEMATICS

S. A. RUSAKOV

ON THE EMBEDDING OF SUBGROUPS

(Presented by Academician A. I. Mal’cev, 1 VI 1962)

§ 1. In the present note a number of theorems obtained by us on the embedding of subgroups are given; as special cases, Theorems 2, 3, 5 of our paper \((^3)\) follow from them. Only finite groups are considered. We use the notation and definitions of our previous papers \((^{2,3})\), as well as the following definition.

Definition. Let \(\sigma\) and \(\tau\) be such (empty or not) sets of prime numbers that \(\sigma \cap \tau\) is empty. Then we shall say that the strong \(N_\sigma^\tau\)-Sylow theorem holds for a group \(\mathfrak G\), if for the normalizer of every \(\tau\)-subgroup in the group \(\mathfrak G\) the strong \(\sigma\)-Sylow theorem holds.

Since the identity subgroup \(\mathfrak E\) of the group \(\mathfrak G\) is a \(\tau\)-subgroup, it follows from the definition that the strong \(\sigma\)-Sylow theorem also holds for the group \(\mathfrak G\).

§ 2. Let \(\mathfrak G\) possess a subgroup \(\mathfrak H=\mathfrak H_\sigma \times \mathfrak G_\tau\), where \(\mathfrak H_\sigma\) is a \(\sigma\)-Hall subgroup of \(\mathfrak H\), and \(\mathfrak G_\tau\) is some \(\tau\)-Hall subgroup of \(\mathfrak G\). Then the following theorems hold:

Theorem 1. If the strong \(N_\sigma^\tau\)-Sylow theorem and the \(\tau\)-Sylow theorem hold for \(\mathfrak G\), then for any \(\sigma\)-solvable subgroup \(\mathfrak M\) whose order divides \((\mathfrak H)\), there exists an element \(G \in \mathfrak G\) such that \(\mathfrak M^G \subseteq \mathfrak H\).

Proof. Suppose that the theorem is false for the group \(\mathfrak G\). Then, among all \(\sigma\)-solvable subgroups of the group \(\mathfrak G\) for which the assertion of the theorem does not hold, choose a subgroup \(\mathfrak M\) of least order. \((\mathfrak M)=(\mathfrak M)_\sigma(\mathfrak M)_\tau\). Obviously, \((\mathfrak M)>1\).

Consider the following possibilities:

1. \(\mathfrak M\) contains an invariant subgroup \(\mathfrak M_\tau\) of order \((\mathfrak M)_\tau\). On the basis of Schur’s theorem, \(\mathfrak M\) contains a subgroup \(\mathfrak M_\sigma\) of order \((\mathfrak M)_\sigma\). According to the hypothesis of the theorem, \(\mathfrak M_\tau^*=\mathfrak M_\tau^{G_1}\subseteq \mathfrak G_\tau\), where \(G_1\in\mathfrak G\). Since \(\mathfrak M_\tau^*\) is invariant in \(\mathfrak M^*=\mathfrak M_\sigma^*\mathfrak M_\tau^*\), where \(\mathfrak M_\sigma^*=\mathfrak M_\sigma^{G_1}\), the normalizer \(\mathfrak N_{\mathfrak M_\tau^*}^{\mathfrak G}\) contains \(\mathfrak M_\sigma^*\). It is also clear that \(\mathfrak N_{\mathfrak M_\tau^*}^{\mathfrak G}\) contains \(\mathfrak H_\sigma\). On the basis of the hypothesis of the theorem we can assert that \(\mathfrak M_\sigma^{*N}\subseteq\mathfrak H_\sigma\), where \(N\in \mathfrak N_{\mathfrak M_\tau^*}^{\mathfrak G}\). Then

\[ \mathfrak M^{*N}=\mathfrak M_\sigma^{*N}\mathfrak M_\tau^{*N} =\mathfrak M_\sigma^{*N}\mathfrak M_\tau^* \subseteq \mathfrak H_\sigma\times \mathfrak G_\tau=\mathfrak H. \]

Consequently, \(\mathfrak M^G\subseteq\mathfrak H\), where \(G=G_1N\). Contradiction.

1. \(\mathfrak M\) is any \(\sigma\)-solvable subgroup of the group \(\mathfrak G\). Then each index of a certain composition series
   \[ \mathfrak M \supset \mathfrak M^{(1)} \supset \cdots \supset \mathfrak E \]
   of the group \(\mathfrak M\) either is not divisible by any prime number from \(\sigma\), or is some prime number belonging to \(\sigma\).

Consider the following cases:

a) The index of \(\mathfrak M^{(1)}\) in \(\mathfrak M\) is not divisible by any prime number from \(\sigma\), i.e.
\[ (\mathfrak M^{(1)})=(\mathfrak M)_\sigma(\mathfrak M^{(1)})_\tau. \]
Since \(\mathfrak M^{(1)}\) is \(\sigma\)-solvable and \((\mathfrak M^{(1)})<(\mathfrak M)\), it follows that
\[ \mathfrak M^{(1)G_1}\subseteq \mathfrak H, \]
where \(G_1\in\mathfrak G\). Consequently, \(\mathfrak M^{(1)}\) contains an invariant solvable sub-

the group \(\mathfrak M_\sigma^{(1)}=\mathfrak M_\sigma\), which, as is easy to see, will also be invariant in \(\mathfrak M\). By Schur’s theorem, in \(\mathfrak M\) there exists a subgroup \(\mathfrak M_\tau\) of order \((\mathfrak M)_\tau\).

Suppose first that \((\mathfrak M)_\tau\) is divisible by only one prime number from \(\tau\). Since the strong \(\sigma\)-Sylow theorem holds for the group \(\mathfrak G\), there exists an element \(G_2\) in the group \(\mathfrak G\) such that
\[ \mathfrak M_\sigma^*=\mathfrak M_\sigma^{G_2}\subseteq \mathfrak H_\sigma . \]
It is clear that the normalizer \(N_{\mathfrak G}^{\mathfrak M_\sigma^*}\) contains \(\mathfrak M_\tau^*=\mathfrak M_\tau^{G_2}\) and \(\mathfrak G_\sigma\). By Sylow’s theorem,
\[ \mathfrak M_\tau^{*N}\subseteq \mathfrak G_\tau, \]
where
\[ N\in N_{\mathfrak G}^{\mathfrak M_\sigma^*}. \]
Consequently,
\[ \mathfrak M^{*N}=\mathfrak M_\sigma^{*N}\mathfrak M_\tau^{*N} =\mathfrak M_\sigma^*\mathfrak M_\sigma^{*N} \subseteq \mathfrak H_\sigma\times \mathfrak G_\tau=\mathfrak H, \]
or \(\mathfrak M^G\subseteq \mathfrak H\), where \(G=G_2N\). We have a contradiction. Now suppose that \((\mathfrak M)_\tau\) is divisible by more than one prime number from \(\tau\). Then it is easy to show that \(\mathfrak M_\tau\) is invariant in \(\mathfrak M\). Indeed, \(\mathfrak M_\sigma \mathfrak M_{q_i}\), where \(q_i\) is an arbitrary divisor of \((\mathfrak G)\) from \(\tau\) and \(\mathfrak M_{q_i}\subset \mathfrak M_\tau\), is a subgroup of the group \(\mathfrak M\). From the case just considered it follows that
\[ \mathfrak M_\sigma \mathfrak M_{q_i} = \mathfrak M_\sigma\times \mathfrak M_{q_i}. \]
From this it is not difficult to see that \(\mathfrak M_\sigma\) and \(\mathfrak M_\tau\) are elementwise permutable. Consequently, \(\mathfrak M_\tau\) is invariant in \(\mathfrak M\), and, on the basis of case 1,
\[ \mathfrak M^G\subseteq \mathfrak H. \]
A contradiction.

b) The index of \(\mathfrak M^{(1)}\) in \(\mathfrak M\) is some prime number from \(\sigma\). According to the induction hypothesis, it is easy to show that \(\mathfrak M\) contains an invariant subgroup \(\mathfrak M_\tau\). Then from case 1 it follows that
\[ \mathfrak M^G\subseteq \mathfrak H. \]
Again we have a contradiction.

Thus the theorem is proved completely.

From the theorem obtained and the theorem of S. A. Chunikhin and I. K. Chunikhina \((^1)\) it follows:

Theorem 2. If \(\sigma=\{p\}\) and the strong \(N_\sigma^\tau\)-Sylow theorem and the \(\tau\)-Sylow theorem hold for \(\mathfrak G\), then for every subgroup \(\mathfrak M\), whose order divides \((\mathfrak H)\), there exists an element \(G\in\mathfrak G\) such that
\[ \mathfrak M^G\subseteq \mathfrak H. \]

Applying Theorem 4 of our paper \((^2)\), it is easy to prove the following assertion:

Theorem 3. If the strong \(N_\sigma^\tau\)-Sylow theorem and the \(\tau\)-Sylow theorem hold for \(\mathfrak G\), then for every \(\sigma\Delta\)-subgroup \(\mathfrak M\), whose order divides \((\mathfrak H)\), there exists an element \(G\in\mathfrak G\) such that
\[ \mathfrak M^G\subseteq \mathfrak H. \]

Gomel Branch
of the Institute of Mathematics and Computer Technology
of the Academy of Sciences of the BSSR

Received
20 V 1962

REFERENCES

1. S. A. Chunikhin, I. K. Chunikhina, Mat. sborn., 15 (57), No. 2, 325 (1944).
2. S. A. Rusakov, DAN, 127, No. 2, 270 (1959).
3. S. A. Rusakov, DAN, 141, No. 2, 320 (1961).