UDC 519.44

MATHEMATICS

A. I. STAROSTIN

FINITE GROUPS CLOSE TO SPLIT GROUPS

(Presented by Academician A. I. Mal’cev on 6 VI 1966)

A finite group in which the condition (QP) is satisfied: the intersection of any two distinct nonprimary maximal nilpotent subgroups of even order is the identity subgroup, will be called a QP-group, and a nonprimary maximal nilpotent subgroup of even order a μ₂-subgroup. The condition (QP) is hereditary for subgroups and factor groups, as follows from the properties given below and from the centralizer criterion for QP-groups. Throughout, only finite groups are considered.

Lemma 1. Let \(G\) be a group of even order with nontrivial center \(Z(G)\). Then, in order that \(G\) be a QP-group, it is necessary and sufficient that \(G\) be a group of one of the following types:

1. A nilpotent group.

2. \(G = A \lambda T\), where \(T\) is a Sylow 2-subgroup of \(G\), and \(T\) has a \(G\)-invariant subgroup \(T_0 \ne E\) such that \(G/T_0\) is a Frobenius group with invariant set isomorphic to \(A\).

3. \(G = T \lambda G_1\), where \(T\) is a Sylow 2-subgroup of \(G\), and \(G_1\) contains a nilpotent \(G\)-invariant subgroup \(G_0\) such that \(G/G_0\) is a Frobenius group with invariant set isomorphic to \(T\).

Theorem 1. A group is a QP-group if and only if the centralizer of each of its nonidentity elements has at most one μ₂-subgroup.

Corollary. A group is a QP-group if and only if the centralizer of each of its nonidentity elements either has odd order or is a group of one of types 1–3 of Lemma 1.

The condition (QP) is clearly satisfied if the group has only one μ₂-subgroup. It turns out that such groups are solvable and are exhausted by the following types:

1–3. As in Lemma 1.

1. \(G = S \lambda A \lambda \{b\}\), where \(S\) and \(\{b\}\) are 2-groups, \(S \lambda A\) is of type 3, and \(A \lambda \{b\}\) is a Frobenius group with invariant set \(A\).

Lemma 2. Let in a QP-group \(G\) one of the following conditions be satisfied:

(a) in \(G\) there are nonprimary elements of even order and a \(G\)-invariant 2-subgroup distinct from \(E\);

(b) in \(G\) there are elements of order \(2p\) (\(p\) an odd prime number) and a \(G\)-invariant \(p\)-subgroup distinct from \(E\).

Then in \(G\) there exists only one μ₂-subgroup.

Corollary. The solvable radical of a nonsolvable QP-group is trivial, unless \(G\) is a CJT-group.

A CJT-group is a group in which the centralizer of any involution is a 2-group \((^1)\). It is clear that every CJT-group is a QP-group. Suppose that \(G\) is a solvable QP-, but not a CJT-group. Let the order of \(G\) be even. If \(G\) has a \(G\)-invariant 2-subgroup distinct from \(E\), then by Lemma 2 \(G\) has exactly one μ₂-subgroup; consequently, \(G\) is a group of one of types 1–4. Suppose that in \(G\) there is more than one-

$\mu_2$-subgroup. Then $G$ has an invariant 2-complement, since its Sylow 2-subgroup is either cyclic or the quaternion group. Hence, applying induction on the order of the group $G$, we obtain the following theorem.

Theorem 2. In order that a solvable group $G$ of even order be a $QP$-group, it is necessary and sufficient that $G$ be a group of one of the following types:

1—4. As above.

5. A Frobenius group whose complement is a group of even order of one of types 1, 2.

Next we consider nonsolvable $QP$-groups. Let $G$ be a nonsolvable $QP$-, but not a $CJT$-group. Then in $G$ there exist $\mu_2$-subgroups, and all of them are distinct from their normalizers, by the well-known theorem of Frobenius. Let $H$ be one of these normalizers. By Theorem 2, $H$ is a group of one of types 2—4. Suppose that $H$ is a group of type 3. Then a Sylow 2-subgroup of $H$ is also a Sylow 2-subgroup in $G$. From condition (QP) it follows that the Sylow 2-subgroups in $G$ are pairwise mutually prime. From the description of such groups (2) it follows that $G$ must be isomorphic to $SL(2,q)$ or $Sz(q)$, where $q$ is a power of 2. However, the latter groups are $CJT$-groups. Thus, $H$ is a group of one of types 2, 4.

Introduce the following notation: $\mathfrak M_1$ is the collection of normalizers of $\mu_2$-subgroups of $G$, each of which is of type 4; $\mathfrak M_2$ is the collection of normalizers of $\mu_2$-subgroups of $G$, each of which is a Hall subgroup in $G$ of type 2; $\mathfrak M_3$ is the collection of normalizers of $\mu_2$-subgroups of $G$, each of which is a group of type 2, but is not a Hall subgroup in $G$.

Consider three cases.

1st case: $\mathfrak M_1 \ne \varnothing$. If $K=S \lambda A \lambda \{b\}\in \mathfrak M_1$, then $T=S\{b\}$ coincides with its normalizer in $G$, and therefore is a Sylow 2-subgroup in $G$, and all subgroups from $\mathfrak M_1$ are conjugate. The normalizer $N(A)$ of the subgroup $A$ in $G$ is a Frobenius group $N(A)=A\lambda B$, where the complement $B$ is a subgroup of the centralizer of a certain involution. Hence it follows that $K$ is a Hall subgroup in $G$. If in the group $G$ $\mathfrak M_2\cup\mathfrak M_3\ne\varnothing$, and $U$ is the Fitting radical of $K$, then $K/U\approx S_3$, and the order of the Sylow 2-subgroup of the Fitting radical of any subgroup $X$ from $\mathfrak M_2\cup\mathfrak M_3$ is equal to 2. Using this assertion, one can prove that if $N(A)\ne A\lambda\{b\}$, then in $\mathfrak M_1$ there is a subgroup $K_1$ such that $K\cap K_1=A\lambda\{b\}$. The same conclusion is reached by the assumption that there exists in $\mathfrak M_1$ a subgroup $K_2$, distinct from $K$ and such that $K\cap K_2$ contains elements of odd order. All subgroups of $G$ whose order is equal to the order of $A$ are conjugate. Comparing the number of these subgroups in $G$, in $K$, and the cardinality of the set $\mathfrak M_1$ shows that if $N(A)=A\lambda\{b\}$, then the intersection of any two distinct subgroups from $\mathfrak M_1$ is a 2-group. Suppose that the intersection of any two distinct subgroups from $\mathfrak M_1$ is a 2-group, and denote by $D$ a maximal one of these intersections. $D$ is a maximal intersection of Sylow 2-subgroups of $G$. Therefore $N(D)$ is not 2-closed. If $D$ is a cyclic group, then $N(D)\in\mathfrak M_2\cup\mathfrak M_3$, and, as shown above, $|D|=2$. If $D$ is noncyclic, then $D$ is an elementary abelian group of rank 2, $N(D)$ is a $CJT$-group of type 4, and $N(D)/D\approx S_3$. After these preliminary considerations one can prove the following lemma.

Lemma 3. In a nonsolvable $QP$-, but not $CJT$-group $G$, a Sylow 2-subgroup is a dihedral group if $\mathfrak M_1\ne\varnothing$.

Proof. Suppose the contrary. From the definition of groups of type 4 and from the structure of 2-groups containing a self-centralizing subgroup of order 4, it follows that in $G$ there are no self-centralizing subgroups of order 4. Suppose that $\mathfrak M_2\cup\mathfrak M_3\ne\varnothing$. If in $\mathfrak M_1$ there were two distinct subgroups whose intersection is not a 2-group, then the order of the subgroup $A$ would be 9. Considering various—

possibilities for \(C(b)\), it is not hard to verify that \(N(A)=A\lambda\{b\}\). This is impossible, as was shown above. Suppose that the maximal intersection \(D\) of subgroups from \(\mathfrak M_1\) is noncyclic. Then \(|D|=4\), and \(N(D)\approx S_4\). This again contradicts the assumption. Thus, \(|D|=2\), if \(\mathfrak M_2\cup\mathfrak M_3\ne\varnothing\). It turns out that the latter condition can be omitted. Moreover, \(N(D)\in\mathfrak M_2\cup\mathfrak M_3\).

Applying the first theorem of Grün, we find in \(G\) a subgroup \(G_0\) of index 2. Since \(K\cap G_0=S\lambda A\), the Sylow 2-subgroups in \(G_0\) are pairwise mutually disjoint. From (2) it follows that \(G_0\) is isomorphic to \(SL(2,q)\) or \(Sz(q)\), where \(q\) is a power of 2. But since the rank of the center \(S\) is equal to 2, the order of \(S\) must be 4. Therefore \(T\) is a dihedral group, which contradicts the assumption. The lemma is proved.

2nd case. \(\mathfrak M_1=\varnothing,\ \mathfrak M_2\ne\varnothing\). Denote by \(D\) the maximal intersection of any two distinct subgroups from \(\mathfrak M_2\). Since in groups (of type 2) from \(\mathfrak M_2\) all elements outside the Fitting radical are 2-elements, \(D\) is a 2-group isomorphic to a subgroup of a direct product of two 2-groups, each of which has a unique involution. In particular, the rank of the center of \(D\) is not greater than 2. Suppose that \(D\) is a Sylow 2-subgroup in \(G\). Since the lower layer \(D_0\) of the center of \(D\) is weakly closed in \(D\) with respect to \(G\), the greatest 2-factor group of the group \(G\) is isomorphic to the same factor group for \(N(D_0)\). From \(C(D_0)=D\) we have \(N(D_0)=D\lambda\{b\}\), a Frobenius group, where \(b^3=e\). It is easy to check that if a subgroup of a direct product of 2-groups, each of which has a unique involution, admits a regular automorphism of order 3, then this subgroup is a direct product of isomorphic cyclic groups. From the nonsolvability of \(G\) it must then follow, as Brauer showed in (3), that \(|D|=4\).

Suppose next that in \(G\) there are no self-centralizing subgroups of order 4. Thus, \(D\) cannot be a Sylow 2-subgroup in \(G\), and a Sylow 2-subgroup \(T\) of \(G\) coincides with its normalizer. Now it is easy to verify that \(D\) is a maximal intersection of Sylow 2-subgroups of the group \(G\), and either \(N(D)\in\mathfrak M_3\) (in this case \(D\) has a unique involution), or \(N(D)=N(D_0)\) is a \(CJT\)-group of type 4, \(N(D)/D=S_3\). Hence the center of a Sylow 2-subgroup of \(G\) is cyclic, and \(G\) is not 2-normal. Therefore there exists such a \(D\) that \(N(D)\notin\mathfrak M_3\). A Sylow 2-subgroup of \(N(D)\) is a wreath product of two cyclic groups: of order \(2^m,\ m>1\), and of order 2. It coincides with its normalizer, and therefore is a Sylow 2-subgroup in \(G\). By the first theorem of Grün, \(G\) has a subgroup \(G_0\) of index 2. The Sylow 2-subgroup in \(G_0\) is \(D\). From (3) it follows that \(G_0\), and hence also \(G\), is solvable. This contradicts the assumption. Taking Lemma 3 into account, we summarize the preceding arguments. We shall call a 2-group \(T\) semidihedral if

\[ T=\{a,b\},\quad a^{2^{n+1}}=b^2=e,\quad b^{-1}ab=a^{-1+2^n},\quad n\geqslant 2. \]

Lemma 4. A Sylow 2-subgroup of a nonsolvable \(QP\)-, but not \(CJT\)-group is dihedral or semidihedral if \(\mathfrak M_1\cup\mathfrak M_2\ne\varnothing\).

3rd case: \(\mathfrak M_1=\mathfrak M_2=\varnothing,\ \mathfrak M_3\ne\varnothing\). Suppose also that in \(G\) there are no self-centralizing subgroups of order 4. Let \(M\in\mathfrak M_3\), and let \(D\) be the maximal intersection of \(M\) with subgroups from \(\mathfrak M_3\) distinct from \(M\). \(D\) is a Sylow 2-subgroup in \(M\), and the number of involutions in \(D\) is 3. The center of a Sylow 2-subgroup \(T\) of \(G\) is therefore cyclic, \(T\) coincides with its normalizer, and \(G\) is not 2-normal. Note also that the number of involutions in \(T\) is greater than 3. Denote by \(D_0\) the lower layer of the center of \(D\). \(N(D_0)=N(D)\) and is a 2-subgroup, \(|N(D):D|=2\), and the number of involutions in \(N(D)\) is greater than three. Hence it follows that the involutions from \(M\) split into two classes of involutions conjugate in \(G\): the class of involutions each of which is contained in the center of some Sylow 2-subgroup of \(G\), and the class of the remaining involutions. \(N(D)\), in turn,

is distinct from its normalizer, and the latter is also a 2-subgroup. Therefore a $\Phi$-subgroup of $N(D)$ has only one involution. It is now easy to check that $M$ has the following structure: $M=\{s\}\times(A\lambda P)$, where $s^2=e$, $A\lambda P$ is a Frobenius group whose complement $P$ is a 2-group, and whose invariant set $A$ is an abelian Hall subgroup of $G$. Since, by assumption, a Sylow 2-subgroup $T$ of $G$ is not a group of maximal class, $T$ has an invariant noncyclic subgroup $V_T$ of order 4 ($^4$). Using the notion of a Thompson family of 2-subgroups and the method of ($^1$), one can show that every involution in $V_T$ is contained in the center of some Sylow 2-subgroup of $G$. Hence it follows that $C(V_T)$ is a maximal subgroup in $T$, the rank of the center of $C(V_T)$ is 2, $N(C(V_T))=C(V_T)\lambda\{b\}\lambda\{s\}$ is a group of type 4, where $b^3=s^2=e$, $C(s)\in\mathfrak M_3$. In particular, the nilpotency class of $C(V_T)$ is at most 2. Now one can apply the first theorem of Grün. We note that $C(V_T)$ does not contain those involutions $s$ for which $C(s)\in\mathfrak M_3$. Thus all the groups referred to in Grün’s theorem are contained in $C(V_T)$. Hence $G$ has a subgroup of index 2. From this fact, Lemma 4, and the principal results of ($^{1,5–7}$), the main result follows:

Theorem 3. In order that an insoluble group $G$ be a $QP$-group, it is necessary and sufficient that $G$ either be a $CNP$-group, or be isomorphic to $A_7$ or $H(q)$.

A $CNP$-group is a group in which the centralizer of every nonidentity element has a nilpotent splitting. Such groups are described in ($^7$). $A_7$ is the alternating group of degree 7. By $H(q)$, following ($^8$), is denoted a split extension of $PSL(2,q)$ by a group of order 2 whose Sylow 2-subgroup is semidihedral.

A $QNP$-group is a group in which the following condition is satisfied:

$(\mathrm{QNP}).$ The intersection of any two distinct non-invariant maximal nilpotent subgroups is equal to the identity subgroup.

Corollary. An insoluble group is a $QP$-group if and only if it is a $QNP$-group.

The description of the $QNP$-groups obtained by us will be published in another paper. The proofs, the outlines of which have been given here, largely repeat the arguments used in the description of $QNP$-groups.

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