UDC 519.4

MATHEMATICS

Ya. G. BERKOVICH

A GENERALIZATION OF THE THEOREMS OF CARTER AND WIELANDT

(Presented by Academician A. I. Mal'tsev on 1 II 1966)

§ 1. The purpose of this note is to generalize two well-known theorems of the theory of finite groups—the theorem of Carter \((^1)\) and the theorem of Wielandt \((^5)\). We precede each theorem by several simple auxiliary propositions, which makes it possible to avoid repetitions in the proofs.

We use the following notation and definitions. \(N(H)\) and \(C(H)\) are, respectively, the normalizer and the centralizer of the complex \(H\) in the group \(G\); \(\pi\) and \(\pi'\) are complementary sets of primes. A group is called \(\pi\)-separable \((^2)\) if its \(\pi\)-Hall subgroup is nilpotent and is a direct factor. A nilpotent subgroup that coincides with its normalizer (in \(G\)) will be called a Carter subgroup of the group \(G\). We shall say that a group \(G\) has the Carter property if it contains exactly one conjugacy class of Carter subgroups. For example, the simple group of order 168 has the Carter property.

§ 2. Lemma 1. Let \(H\) be a Carter subgroup of the group \(G\), and let the subgroup \(F\) containing \(H\) have the Carter property. Then \(N(F)=F\).

Indeed, applying the Frattini argument gives

\[ F\cdot [N(H)\cap N(F)] = N(F). \]

But the left-hand side of this equality, in view of \(N(H)=H\), coincides with \(F\).

Lemma 2. Let the condition of Lemma 1 be satisfied for \(F=HM\), where \(H\) is a Carter subgroup in \(G\), and \(M\) is such a normal divisor of the group \(G\) that the factor group \(G/M\) is nilpotent. Then \(HM=G\).

This is an immediate consequence of Lemma 1 and the normalizer property of nilpotent groups.

Lemma 3. If all subgroups of the group \(G\) (including \(G\) itself) have the Carter property, then the homomorphic images of any of its subgroups have this same property.

Proof. Obviously, it is enough to show that all homomorphic images of the group \(G\) have the Carter property. Let \(H\) be a Carter subgroup in \(G\), and let \(M\) be a normal divisor in \(G\). By Lemma 1, the subgroup \(HM/M\) is a Carter subgroup in \(G/M\). Let \(L/M\) be another Carter subgroup in \(G/M\). By assumption, \(L\) contains a Carter subgroup \(F\); \(L=FM\) by Lemma 2. Further, from the inclusion \(N(F)\subseteq N(FM)=L\) it follows that \(N(F)=F\). Therefore in \(G\) there is an element \(x\) such that \(H^{x}=F\). But then \((HM)^{x}=H^{x}M=FM=L\), i.e. \(HM\) and \(L\) are conjugate in \(G\). All the more so, \(HM/M\) and \(L/M\) are conjugate in \(G/M\), as was required.

Lemma 3a. Let \(M\) be a normal divisor in \(G\); let \(L/M\) be a Carter subgroup in \(G/M\); and let \(H\) be a Carter subgroup in \(L\). If \(HM\) has the Carter property, then \(N(H)=H\).

Indeed, by Lemma 2 we have \(HM=L\). From the inclusion \(N(H)\subseteq N(HM)=L\) it follows that \(N(H)=H\), as required.

For the proof of Theorem 1 the following special case of it is useful:

Lemma 4. Let \(R\) be a nilpotent \(\pi\)-Hall subgroup of the group \(G\), and let the \(\pi'\)-Hall subgroup be invariant in \(G\). If all \(\pi'\)-sub-

subgroups of the group \(G\) have the Carter property, then \(G\) also has the Carter property.

Proof. By hypothesis \(N(R)=R\times M\). Let \(F\) be a Carter subgroup in the \(\pi'\)-subgroup \(M\). Put \(H=R\times F\). We have \(N(H)\cap N(R)=H\). Since \(N(H)\subseteq N(R)\), it follows that \(N(H)=H\). Now let \(H_1\) be another Carter subgroup in \(G\). Using Lemma 2, we may, without loss of generality, assume that \(R\subseteq H\cap H_1\). Then \(N(R)\supseteq \{H,H_1\}\). The subgroups \(H/R\) and \(H_1/R\) coincide with their normalizers in \(N(R)/R\); therefore they are conjugate there. Hence \(H\) and \(H_1\) are conjugate in \(G\). Theorem 1 shows that a result analogous to Carter’s theorem holds for a broader class of groups than that for which Carter’s theorem was proved.

Theorem 1. If all \(\pi'\)-subgroups of a \(\pi\)-solvable group \(G\) have the Carter property, then the group \(G\) itself has this property.

Proof. We may assume that the theorem has already been proved for all proper subgroups of the group \(G\). By Lemma 3, all nontrivial homomorphic images of the group \(G\) then have the Carter property (here induction is also used). We may assume that the set \(\pi\) is nonempty (otherwise the conclusion is weaker than the hypotheses). Then \(G\) is not simple. A minimal normal divisor \(M\) of the group \(G\) is an \(\omega\)-subgroup, where \(\omega\) is one of the sets \(\pi,\pi'\).

Let \(A/M\) be a Carter subgroup in \(G/M\). The subgroup \(A\), by Lemma 4, contains a Carter subgroup \(H\), and moreover \(A\), by the same lemma, has the Carter property. Now by Lemma 3a we have \(N(H)=H\).

Now let \(H_1\) and \(H_2\) be distinct Carter subgroups of the group \(G\). By Lemmas 4 and 1, then \(N(H_iM)=H_iM,\ i=1,2\); hence \(H_1M/M\) and \(H_2M/M\) are Carter subgroups in \(G/M\), and by induction they are conjugate there. Therefore \(H_1M\) and \(H_2M\) are conjugate in \(G\). Thus \(H_1^x\subseteq H_2M\), \(x\) from \(G\). But then \(H_1^x\) and \(H_2\) are conjugate in \(H_2M\), by Lemma 4. Hence \(H_1\) and \(H_2\) are conjugate in \(G\), as required.

Example. Let, in a \(\pi\)-solvable group \(G\), its \(\pi'\)-Hall subgroup be isomorphic to \(LF(2,7)\). Then \(G\), by Theorem 1, has the Carter property.

Carter’s theorem is, evidently, a special case of Theorem 1. Lemma 5 is, of course, known, but is included for convenience of reading.

Lemma 5. Let \(A_0\) and \(B\) be subgroups of the group \(G\), with \(B\) invariant in \(G\). If \(A=A_0B\), then \(N(A/B)=N(A)/B\).

Proof. Let \(N(A/B)=N/B\). Then \(A\) is invariant in \(N\), i.e. \(N(A)\supseteq N\), and \(N(A)/B\supseteq N/B=N(A/B)\). Since \(A\) is invariant in \(N(A)\), the subgroup \(A/B\) is invariant in \(N(A)/B\). But then \(N(A/B)\supseteq N(A)/B\), and this completes the proof.

Proposition 1. Suppose \(H\) contains a Carter subgroup of \(F\), where \(F\) is some \(\pi\)-Hall subgroup of a \(\pi\)-solvable group \(G\). Then \(H\) is a \(\pi\)-Hall subgroup in \(N(H)\), if \(H\subseteq F\).

Proof. Let \(M\) be a minimal normal divisor in \(G\). Consider the following two possibilities for \(M\).

1) \(M\) is a \(\pi\)-subgroup.

Then \(M\subseteq F\) and \(N(HM)\cap F=HM\), by Lemma 1. Hence \(HM/M\) contains a Carter subgroup of \(F/M\). By induction, \(HM/M\) is a \(\pi\)-Hall subgroup in \(N(HM/M)\). By Lemma 5, \(N(HM/M)=N(HM)/M\), i.e. \(HM\) is a \(\pi\)-Hall subgroup in \(N(HM)\). But from \(HM\subseteq F\) it follows that \(N(H)\cap HM=H\). Therefore \(N(H)\cap N(HM)=N(H)\) contains \(H\) as a \(\pi\)-Hall subgroup.

2) \(M\) is a \(\pi'\)-subgroup.

By induction, \(HM/M\) is a \(\pi\)-Hall subgroup in \(N(HM/M)=N(HM)/M\). Therefore \(H\) is a \(\pi\)-Hall subgroup in \(N(HM)\). Since \(N(H)\subseteq N(HM)\), \(H\) is a \(\pi\)-Hall subgroup in \(N(H)\). The proposition is proved.

Corollary 1. Let a solvable group \(G = H_1H_2\), where \(H_1\) and \(H_2\) are subgroups of relatively prime orders. If \(F_i\) is a Carter subgroup in \(H_i\), \(i=1,2\), and \(\{F_1,F_2\}=F_1\times F_2\), then \(F=F_1\times F_2\) is a Carter subgroup in \(G\).

Indeed, the result follows from the relation \(N(F)\subseteq N(F_1)\cap N(F_2)\) and Proposition 1.

Corollary 2. Let \(H\) be a Carter subgroup of a \(\pi\)-Hall subgroup of a solvable group \(G\). If \(N(H)\) is a \(\pi\)-decomposable group, then \(N(H)\) contains a Carter subgroup of the group \(G\).

The proof is no different from the proof of Lemma 4.

Proposition 2 generalizes the main result of the note \((^3)\).

Following \((^1)\), we shall call a subgroup \(H\) abnormal in \(G\) if \(g\in \{H,H^g\}\) for all \(g\) in \(G\). \(H\) is abnormal in \(G\) if and only if \(H\) does not lie in two distinct conjugate subgroups of the group \(G\), and all subgroups containing \(H\) coincide with their normalizers in \(G\). Lemma 6 makes it possible to shorten the proof of Proposition 2.

Lemma 6. Let the subgroup \(F/M\) be abnormal in the group \(G/M\). If \(H\) is such an abnormal subgroup in \(F\) that \(HM=F\), then \(H\) is abnormal in \(G\).

Proof. Let the subgroup \(D\) contain \(H\). Then

\[ N(D)\subseteq N(DM)=N(DF)=DF. \]

From \(N(D)=D\cdot [F\cap N(D)]\) and the invariance of \(D\cap [F\cap N(D)]=D\cap F\) in \(F\cap N(D)\), it follows that
\[ N(D)=D\cdot [F\cap N(D)]=D\cdot (D\cap F)=D. \]
Now let \(H\subseteq K\cap K^g\). Then \(F=HM\subseteq KM\cap K^gM\), and therefore, from the abnormality of \(F\) in \(G\), we obtain \(K^gM=KM\) and \(g\) is contained in \(KM\). Then \(g=xy\) with \(x\) from \(K\) and \(y\) from \(M\). Further, \(H\subseteq K\cap K^{xy}=K\cap K^y\) and \(H^{y^{-1}}\subseteq K\cap K^{y^{-1}}\). Therefore \(y^{-1}\in \{H,H^{y^{-1}}\}\subseteq K\) and \(K^g=K^{xy}=K\). This proves the abnormality of \(H\) in \(G\).

Proposition 2. A \(\pi\)-solvable group \(G\) contains an abnormal \(\pi\)-decomposable subgroup.

Proof. \(M\) is a minimal normal divisor in \(G\), and is an \(\omega\)-subgroup, where \(\omega\) is one of the sets \(\pi,\pi'\). By induction, \(G/M\) contains a \(\pi\)-decomposable abnormal subgroup \(F/M\). Let \(F_1\) be a \(\pi\)-Hall subgroup and \(F_2\) a \(\pi'\)-Hall subgroup in \(F\). If \(\omega=\pi'\), then put \(H=N(F_1)\cap F\). If, however, \(\omega=\pi\), then put \(H=H_1\times F_2\), where \(H_1\) is a Carter subgroup in \(N(F_2)\cap F_1\). From the \(\pi\)-solvability of \(F\) it follows that \(H\) is abnormal in \(F\). In the first case, \(HM=F\) is obtained as a result of applying the Frattini argument. In the second case the same equality follows from
\[ N(HM/M)\cap F/M=HM/M, \]
the \(\pi\)-decomposability of \(F/M\), and the inclusion \(F_2\subseteq HM\). Now the abnormality of \(H\) in \(G\) is a consequence of Lemma 6.

§ 3. In this paragraph we generalize Wielandt’s theorem on the solvability of the product of pairwise permutable groups of relatively prime orders to the case in which the factors are nilpotent \((^5)\).

Lemma 7. Let \(N\) be a normal divisor of the group \(G=AB\), \((|A|,|B|)=1\). Then the subgroup \(N\) is factorable.

Recall that a subgroup \(N\) of the group \(G=AB\) is factorable if
\[ N=(N\cap A)\cdot (N\cap B). \]
Now the lemma follows from the equality
\[ |N|=|N\cap A|\cdot |N\cap B|. \]

Lemma 8. Let \(G=AB\), \((|A|,|B|)=1\), and let \(A_0\) and \(B_0\) be normal divisors in \(A\) and \(B\). Then the generated subgroup \(H=\{A_0,B_0\}\) is factorable.

By Lemma 7, from \((^5)\) the subgroup \(F=N(H)\) is factorable, i.e.
\[ F=(F\cap A)\cdot (F\cap B). \]
But then \(H\), by Lemma 7, is factorable in \(F\), i.e.
\[ H=(H\cap F\cap A)\cdot (H\cap F\cap B)=(H\cap A)\cdot (H\cap B). \]

Lemma 9. The group \(G=\{a,b\mid a^2=b^2=1\}\) is dihedral.

Indeed,
\[ G=\{a,ab\} \]
and
\[ a\cdot ab\cdot a=ba=(ab)^{-1}. \]

Theorem 2. Let \(G=AB\), \((|A|,|B|)=1\); \(A=P\times L\), where \(P\) is a Sylow \(2\)-subgroup in \(G\), and \(B\) is nilpotent. Then the group \(G\) is solvable.

Proof. Suppose that \(G\) is a counterexample of minimal order. Since all homomorphic images are factorized in the same way as \(G\), application of Lemma 7 and induction shows that: (a) the group \(G\) is simple. Using (a) and considering the indices \(|G:C(a)|\) and \(|G:C(b)|\), where \(a\) and \(b\) are nonidentity elements of \(Z(A)\) and \(Z(B)\), respectively, we conclude, by the well-known theorem of Burnside \({}^{6}\), that: (b) \(L\ne 1\) and the subgroup \(B\) is not primary. From the theorem on the solvability of groups of odd order it follows that (c) \(P\ne 1\).

Let \(u\) be an involution in \(Z(P)\). Suppose that \(C(u)\ne A\). Since \(C(u)\) is factorized, there is in \(B\cap C(u)\) a nonidentity element \(b\) of prime order. By (b), in \(B\) there exists a Sylow subgroup \(Q\) whose order is relatively prime to the order of \(b\). Then \(H=\{u,Q\}\subseteq C(b)\ne G\). By Lemma 8, the subgroup \(H\) is factorized; hence \(H\), by induction, is solvable. Let \(R\) be a nonidentity Sylow subgroup in the Fitting subgroup of the group \(H\). If \(R\subseteq H\cap A\), then \(F=\{L,Q\}\subseteq N(R)\ne G\). By Lemma 8, the subgroup \(F\) is factorized; hence it is, by induction, solvable. Then, by Lemma 10 of (5), we have \(L^x\cdot Q^y=Q^y\cdot L^x\) for all \(x\) and \(y\) from \(G\). Since \(LQ_0\ne G\), \(G\), by Theorem 3 of (7), is not simple, which contradicts (a). If, however, \(R\subseteq H\cap B=Q\), then all elements conjugate to \(u\) in \(G\) lie in \(\{u,B\}\subseteq N(R)\ne G\), which, by (a), is impossible. Thus: (d) if \(u\) is an involution from \(Z(P)\), then \(C(u)=A\).

At the same time we have proved that: (e) if \(u\) is an involution from \(Z(P)\), and \(Q\) is a nonidentity Sylow subgroup in \(B\), then \(\{u,Q\}=G\).

Let \(u\) be chosen as in (d), and let \(v\) be an involution of \(G\) distinct from \(u\). Suppose that the dihedral group (Lemma 9) \(D=\{u,v\}\) is not a 2-subgroup. Then it follows from (d) that \(u\) does not lie in the center of \(D\); therefore \(u\) inverts all elements of odd order in \(D\). Let \(T\) be a nonidentity primary subgroup of odd order of \(D\). From \(D\subseteq N(T)\) it follows that \(\{u\}\cdot T\) is a subgroup and, moreover, as we showed above, a nonnilpotent one. By the well-known \(D\)-theorem of Wielandt \({}^{8}\), the number \(2\cdot |T|\) does not divide the number \(|A|\); hence the number \(|T|\) divides the number \(|B|\). Without loss of generality, we may assume that \(T\) lies in \(B\) (otherwise we would replace \(B\) by a suitable conjugate subgroup). Let \(Q\) be such a nonidentity Sylow subgroup in \(B\) that \((|T|,|Q|)=1\) (see (b)). Then \(\{u,Q\}\subseteq N(T)\ne G\), which contradicts (e). Thus: (f) if \(u\) is an involution in \(Z(P)\), and \(v\) is an arbitrary involution in \(G\), then their generated subgroup \(\{u,v\}\) is a 2-subgroup.

Let \(u\) and \(v\) be the same involutions as in (f). By (f) and Sylow’s theorem, in \(G\) there is a Sylow 2-subgroup \(P^g\) that contains the subgroup \(\{u,v\}\). Then \(\{u,v\}\subseteq A^g\) and \(L^g\subseteq C(u)=A\). This means that \(L^g=L\) and \(v\in N(L)\). Thus an arbitrary involution \(v\) of the group \(G\) lies in \(N(L)\ne G\), which contradicts the simplicity of the group \(G\)—a contradiction that proves the theorem completely. With the aid of induction it is easy to prove

Lemma 10. Let
\[ G=A_1\cdot A_2\cdot \ldots \cdot A_n, \]
where \((|A_i|,|A_j|)=1\) and the products \(A_iA_j=A_jA_i\) are solvable for all \(i\) and \(j\). Then \(G\) is also solvable.

A consequence of Theorem 2 and Lemma 10 is

Theorem 3. Let
\[ G=A_0\cdot A_1\cdot A_2\cdot \ldots \cdot A_n, \]
where \((|A_i|,|A_j|)=1\), \(A_iA_j=A_jA_i\) for all \(i\) and \(j\). If \(A_0\) is 2-solvable, and the \(A_i\), \(i>0\), are nilpotent of odd orders, then the group \(G\) is solvable.

There remains unresolved the interesting question of whether \(G\) will be solvable (\(G\) from Lemma 10) if the requirement that the numbers \(|A_i|\) be relatively prime is dropped. The following is easily proved.

Theorem 4. Let \(G=AB\), where \(A\) is \(p\)-solvable, and \(B\) is a Hamiltonian group of order not divisible by \(p\). Then \(G\) is \(p\)-solvable.

Institute of Mathematics
Academy of Sciences of the BSSR

Received
20 I 1966

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