UDC 519.44

MATHEMATICS

L. A. SHEMETKOV

ON THE EXISTENCE OF \(\Pi\)-COMPLEMENTS TO NORMAL SUBGROUPS OF FINITE GROUPS

(Presented by Academician V. M. Glushkov on 6 IV 1970)

In a survey lecture at the Edinburgh Mathematical Congress \((^{1})\), Wielandt drew attention to the importance of eliminating the commutativity condition on the normal subgroup in the following theorem of Gaschütz \((^{2})\): a normal abelian subgroup \(K\) of a group \(G\) has a complement in \(G\), if for every prime \(p\) there exists a complement to \(G_p \cap K\) in \(G_p\). In the present paper, continuing our investigations on the theory of complements \((^{3})\), we obtain a solution of this question (Theorem 2).

We consider only finite groups. Notation: \(\Pi\) is some set of primes; \(G\) is a finite group of order \(|G|\); \(O^{\Pi}(G)\) is the subgroup of the group \(G\) generated by all those elements of \(G\) whose orders are not divisible by numbers from \(\Pi\); \(G_p\) is one of the Sylow \(p\)-subgroups of the group \(G\), \(p\) is prime.

We shall say that a subgroup \(K\) of a group \(G\) has a \(\Pi\)-complement \(H\) in \(G\) if \(HK=G\) and \(|H\cap K|\) is not divisible by numbers from \(\Pi\); a subgroup \(H\) with this property will be called a \(\Pi\)-complement to (for) \(K\) in \(G\). If \(\Pi\) is the set of all prime numbers, then a \(\Pi\)-complement becomes an ordinary complement.

Theorem 1. A normal subgroup \(K\) of a group \(G\) has a \(\Pi\)-complement in \(G\), if for every \(p \in \Pi\) the subgroup \(G_p \cap K\) is abelian and complemented in \(G_p\).

Before proceeding to the proof of Theorem 1, we note some of its applications.

Theorem 2. A normal subgroup \(K\) of a group \(G\) has a complement in \(G\), if for every prime divisor \(p\) of the index \(|G:K|\) the subgroup \(G_p \cap K\) is abelian and complemented in \(G_p\).

Proof. Let \(\Pi\) be the set of all prime divisors of the index \(|G:K|\) of the subgroup \(K\). By Theorem 1, there exists a \(\Pi\)-complement \(H\) to the subgroup \(K\) in \(G\). Since \(|H\cap K|\) is not divisible by numbers from \(\Pi\), while \(H/H\cap K\) is a \(\Pi\)-group, there exists a complement to \(H\cap K\) in \(H\), which is the required complement for \(K\).

Theorem 2 contains both Gaschütz’s theorem and the Schur–Zassenhaus theorem on the existence of a complement to a Hall normal subgroup. The following theorem also generalizes Gaschütz’s result.

Theorem 3. A normal subgroup \(K\) of a group \(G\) has a complement in \(G\), if for every prime \(p\) a Sylow \(p\)-subgroup of \(K\) is abelian and complemented in a Sylow \(p\)-subgroup of \(G\) containing it.

Theorem 4. A normal subgroup \(K\) of a group \(G\) has a \(\Pi\)-complement in \(G\), if for every \(p \in \Pi\) a Sylow \(p\)-subgroup of \(K\) is abelian and \(O^{p}(K)=K\).

Proof. Let \(P\) be a Sylow \(p\)-subgroup of the group \(G\), \(p \in \Pi\). According to a theorem of Gaschütz \(((^{5}), p. 426)\), \(P\cap K\) has a complement in \(P\). It remains to apply Theorem 1.

For \(K=O^{\Pi}(G)\), Theorem 4 yields a result giving an answer to the question posed in \((^{3})\) (see also \((^{4})\), problem 3.59).

Theorem 5. If, for every \(p \in \Pi\), the Sylow \(p\)-subgroup of \(O^{\Pi}(G)\) is abelian, then \(O^{\Pi}(G)\) has a complement in \(G\).

An extension of a group \(G\) is a group \(\Gamma\) containing a normal subgroup \(N\) isomorphic to \(G\). The extension \(\Gamma\) is split if \(N\) has a complement in \(\Gamma\). If \(\Gamma/N\) is a \(\Pi\)-group, then \(\Gamma\) is called an extension of \(G\) by means of a \(\Pi\)-group.

Theorem 6. Let the group \(G\) coincide with its commutator subgroup and have an abelian Sylow \(p\)-subgroup for every \(p \in \Pi\). Then every extension of the group \(G\) by means of a \(\Pi\)-group is split.

We proceed to the proof of Theorem 1. Suppose that there exist groups for which Theorem 1 is not fulfilled. Choose among them a group \(G\) of least order. Thus, there will be a set \(\Pi\) of primes and a normal subgroup \(K\) of the group \(G\) such that there is no \(\Pi\)-complement to \(K\) in \(G\). The set \(\Pi\) is nonempty, since in the case of empty \(\Pi\) the required \(\Pi\)-complement could be \(G\) itself. For the same reason \(\Pi\) contains at least one prime divisor of the order of \(K\). For convenience we shall assume that every number in \(\Pi\) divides the order of \(G\).

Now take such a subset \(\omega\) of the set \(\Pi\) for which the following conditions are satisfied: 1) there is no \(\omega\)-complement to the subgroup \(K\) in \(G\); 2) the subgroup \(K\) has an \(\omega_1\)-complement in \(G\) for every nontrivial subset \(\omega_1\) of the set \(\omega\). The set \(\omega\), obviously, is nonempty and contains at least one prime divisor of \(|K|\). Let \(p\) be some element of \(\omega\) dividing \(|K|\). Denote by \(\tau\) the set obtained from \(\omega\) by deleting the element \(p\). Then \(K\) has a \(\tau\)-complement \(H\) in the group \(G\).

Since \(HK=G\), in \(H\) there is a subgroup \(M\) such that \(MK=G\), but \(M_1K \ne G\) for every nontrivial subgroup \(M_1\) of \(M\) (such a subgroup \(M\) was called in [3] an addition to \(K\) in \(G\)). It is not difficult to see that \(M \cap K\) lies in the Frattini subgroup of the group \(M\), and hence is nilpotent. Since \(M \cap K\) is contained in \(H \cap K\), the order \(|M \cap K|\) is not divisible by the numbers in \(\tau\), i.e. \(M\) is a \(\tau\)-complement to \(K\) in \(G\). Note also that \(p\) divides \(|M \cap K|\), since \(K\) has no \(\omega\)-complements.

In what follows \(M_p\) denotes some (fixed) Sylow \(p\)-subgroup of \(M\), and \(G_p\) one of the Sylow \(p\)-subgroups of the group \(G\) that contains \(M_p\). Obviously, \(P=G_p \cap K\) is a Sylow \(p\)-subgroup in \(K\). Moreover, \(G_p=M_pP\), since \(|MK:M_pP|\) is not divisible by \(p\). Denote by \(P_1\) the intersection
\[ M_p \cap (M \cap K)=M_p \cap K=M_p \cap G_p \cap K=M_p \cap P. \]
It is clear that \(P_1\) is a Sylow \(p\)-subgroup of \(M \cap K\). The order of \(M \cap K\) is divisible by \(p\), therefore \(P_1\) is distinct from the identity. Consider the normalizer \(N\) of the subgroup \(P_1\) in the group \(G\). We shall distinguish two cases.

First case. \(N \ne G\). The subgroup \(N\) contains \(M\), since \(M \cap K\) is nilpotent and normal in \(M\). Since \(P\) is abelian by hypothesis, \(P\) is also contained in \(N\). Thus \(G_p=M_pP\) is contained in \(N\), and \(P\), being Sylow in \(K\) and in \(N \cap K\), has a complement in \(G_p\) by hypothesis. Let \(q \in \omega\), \(q \ne p\). Take a Sylow \(q\)-subgroup \(N_q\) of \(N\) containing a Sylow \(q\)-subgroup \(M_q\) of \(M\). Then \(N_q=M_qQ\), where \(Q=N_q \cap K=N_q \cap (N \cap K)\) is a Sylow \(q\)-subgroup of \(N \cap K\). The intersection \(M_q \cap Q\) is equal to the identity, since \(M_q \cap Q\) is contained in the subgroup \(M \cap K\), whose order is not divisible by \(q\). Thus we see that for \(N\), \(N \cap K\), and the set \(\omega\) all the conditions are fulfilled. Therefore, in view of \(|N|<|G|\), \(N \cap K\) has an \(\omega\)-complement \(S\) in \(N\). From \(S(N \cap K)=N\) it follows that \(SK=G\), since \(MK=G\) and \(N \supseteq M\). The order of \(S \cap K\) is not divisible by the numbers in \(\omega\), since \(S \cap (N \cap K)=(S \cap K)\cap N=S \cap K\). Hence \(S\) is an \(\omega\)-complement to \(K\) in \(G\). We have arrived at a contradiction.

Second case. \(N=G\), i.e. the subgroup \(P_1\) is normal in \(G\). Denote by \(R\) the product of all those normal subgroups of the group \(K\) each of which has no composition factors of order \(p\). Obviously, the subgroup \(R\) is characteristic in \(K\) and also does not ...

has composition factors of order \(p\). Since the Sylow \(p\)-subgroup \(P\) of \(K\) is abelian by assumption, applying Theorem 3.3 of [6] and the theorem of P. Hall and G. Higman ([5], p. 691), we see that \(PR/R\) is a normal Sylow \(p\)-subgroup of the group \(K/R\). By assumption \(G_p=P\overline P\), \(P\cap \overline P=1\). Since \(G_p\cap R=P\cap R\) is a Sylow subgroup in \(R\), it follows that \(\overline P\cap R=1\), and hence \(PR/R\cap \overline PR/R=R/R\). Thus \(\overline PR/R\) is a complement to \(PR/R\) in \(G_pR/R\). By Gaschütz’s theorem ([5], p. 121), there exists a complement \(U/R\) to the subgroup \(PR/R\) in the group \(MPR/R\).

Let \(q\in\omega,\ q\ne p\). Since \(|M\cap PR|\) is not divisible by \(q\), a Sylow \(q\)-subgroup \(Q\) of \(MPR\) can be represented in the form \(Q=M_qQ_1\), where \(Q_1\) is a Sylow \(q\)-subgroup of \(R\), intersecting \(M_q\) trivially. But the Sylow \(q\)-subgroups of \(MPR\) and \(U\) have the same orders, and hence \(Q^x\le U\) for some \(x\in MPR\). Thus the Sylow \(q\)-subgroup of \(R\) has a complement in a Sylow \(q\)-subgroup of \(U\) containing it. Further, \(O^p(R)=R\), since, by construction, \(R\) has no composition factors of order \(p\). Therefore, by Gaschütz’s theorem ([5], p. 426), a Sylow \(p\)-subgroup of \(R\) has a complement in a Sylow \(p\)-subgroup of \(U\) containing it. Thus the hypothesis of the theorem is satisfied for the set \(\omega\), the group \(U\), and its normal subgroup \(R\). Since \(|U|<|G|\), the theorem is true for \(U\). Hence \(R\) has an \(\omega\)-complement \(L\) in the group \(U\). The subgroup \(L\) will be an \(\omega\)-complement to \(K\) in \(G\). Indeed, from the three equalities \(LR=U\), \(UPR=MPR\), \(MK=G\), it follows that \(LK=G\). Moreover, it is not difficult to note that for any \(q\in\omega\) the order of a Sylow \(q\)-subgroup of \(L\) is equal to the order of a Sylow \(q\)-subgroup of the factor group \(G/K\). Therefore \(|L\cap K|\) is not divisible by numbers from \(\omega\). We have again arrived at a contradiction.

Theorem 1 is proved.

Gomel Laboratory
of the Institute of Mathematics
of the Academy of Sciences of the BSSR

Received
3 III 1970

REFERENCES

1. International Mathematical Congress in Edinburgh, 1958 (survey reports), Moscow, 1962.
2. W. Gaschütz, J. reine und angew. Math., 190, 93 (1952).
3. L. A. Shemetkov, Dokl. Akad. Nauk SSSR, 178, no. 3, 559 (1968).
4. Kourovka Notebook (unsolved problems in group theory), Novosibirsk, 1969.
5. B. Huppert, Endliche Gruppen. I, Berlin—Heidelberg—New York, 1967.
6. B. Huppert, Acta scient. math. Szeged, 22, no. 1–2, 46 (1961).