UDC 519.44

MATHEMATICS

V. D. MAZUROV

ON FINITE GROUPS WITH A GIVEN SYLOW 2-SUBGROUP

(Presented by Academician A. I. Mal’cev on 18.IX.1965)

Gorenstein and Walter proved the following theorem ((2), Theorem 1).

Let \(G\) be a finite group containing a subgroup of order four that coincides with its centralizer in \(G\). Let \(K\) be the largest normal divisor of odd order of the group \(G\). Then one of the following assertions is true:

I. In the group \(G\) there is no subgroup of index 2, and a Sylow 2-subgroup \(S\) of the group \(G\) has the form

\[ S=\langle \alpha,\beta\rangle,\qquad \alpha^{2^{n+1}}=\beta^2=1,\quad \beta\alpha\beta=\alpha^{-1+2^n},\quad n\geqslant 2. \tag{*} \]

II. The group \(G\) has an invariant series \(G\supset G_0\supset K\supset \{1\}\), where \([G:G_0]\leqslant 2\), \(G_0/K\simeq PSL(2,q), PGL(2,q), SL(2,q)\) (\(q\) odd) or \(A_7\).

III. The factor group \(G/K\) is isomorphic to a Sylow 2-subgroup \(S\) of the group \(G\).

In the present paper it is established that, in the case where assertion I of this theorem holds, \(G/K\simeq PSL(3,3)\) or \(M_{11}\), the quadruple transitive Mathieu group of degree 11.

The structure is also established of finite groups whose Sylow 2-subgroups are of type (*) and in which the centralizer of the element \(\tau=\alpha^{2^n}\) in the group \(G\) is soluble.

Theorem 1. Let \(G\) be a finite group and let a Sylow 2-subgroup \(S\) of the group \(G\) have the form (). Let \(K\) be the largest normal divisor of odd order of the group \(G\). If the centralizer \(C_G(\tau)\) of the element \(\tau=\alpha^{2^n}\) in the group \(G\) is soluble, then one of the following assertions is true:*

1) \(G\) has a normal series \(G\supset G_0\supset T\supset K\), where \([T:K]\leqslant 2\), \([G:G_0]=2\), and \(G_0/T\) is isomorphic to a subgroup of \(P\Gamma L(2,q)\) containing \(PSL(2,q)\) (\(q\) odd). Here \(P\Gamma L(2,q)\) is the group of all nonsingular projective semilinear transformations of the vector space of dimension 2 over a finite field of \(q\) elements.

2) \(G/K\simeq PSL(3,3)\) or \(M_{11}\).

3) \(G/K\) is isomorphic to a Sylow 2-subgroup \(S\) of the group \(G\).

Proof. Let \(G/K\) not be isomorphic to \(S\). If \(G\) contains a subgroup \(G_0\) of index 2, then a Sylow 2-subgroup \(S_0\) of the group \(G_0\) is a dihedral or quaternion group. If \(S_0\) is a dihedral group, then, by (3), \(G_0/K\) is isomorphic to a subgroup of \(P\Gamma L(2,q)\) containing \(PSL(2,q)\), or to the alternating group \(A_7\). Since the automorphism group of \(A_7\) is isomorphic to \(S_7\), whose Sylow 2-subgroup is not of type (), assertion 1) holds with \(T=K\). If \(S_0\) is a quaternion group, then (1) shows that \(\overline{G}_0=G_0/K\) has center \(\overline{T}\) of order 2 and in \(\overline{G}_0/\overline{T}\) the Sylow 2-subgroups are dihedral groups. In this case, by (3*), assertion 1) holds, where \([T:K]=2\).

Suppose that in the group \(G\) there is no subgroup of index 2. By (4), \(H=C_G(\tau)\) has the following structure: \(H\) contains a subgroup \(H_0\) of index 2, in which there is no subgroup of index 2 and whose Sylow 2-subgroup is a quaternion group. We shall show that if \(A\) is the largest normal divisor of odd order of the group \(H\), then \(H/A\simeq GL(2,3)\). We may assume that \(A=\{1\}\) and that in \(H\) there are no invariant subgroups of odd order. Consider-

we consider \(\bar H_0=H_0/\{\tau\}\). In \(\bar H_0\) a Sylow \(2\)-subgroup is a dihedral group and \(\bar H_0\) has no nonidentity invariant subgroups of odd order, for otherwise \(H_0\), and with it also \(H\), would have such subgroups. From (3) and the solvability of \(H\) it follows that \(H_0 \cong PGL(2,3)\), whence it is easy to obtain that \(H \cong GL(2,3)\). Thus, in \(G\) there is no subgroup of index \(2\) and \(H/A \cong GL(2,3)\). In (4) such a situation was considered under the assumption that \(A\) is abelian. In this case \(G/K \cong SL(3,3)\) or \(M_{11}\). However, by slightly modifying the proof in (4), one can see that the requirement that \(A\) be commutative can be removed, which completes the proof of Theorem 1.

Theorem 2. Let \(G\) be a finite group containing a subgroup of order four that coincides with its centralizer in \(G\). Suppose \(G\) contains no subgroup of index \(2\). If a Sylow \(2\)-subgroup \(S\) of \(G\) is of type \((*)\), then \(G/K \cong M_{11}\) or \(PSL(3,3)\). Here \(K\) is the largest normal divisor of odd order of the group \(G\).

Proof. By (4), \(H=C_G(\tau)\), where \(\tau=a^{2^n}\), contains a subgroup \(H_0\) of index \(2\), in which there is no subgroup of index \(2\) and a Sylow \(2\)-subgroup of which is a quaternion group. Theorem 1 from (2) shows that if \(A\) is the largest normal divisor of odd order of the group \(H\), then \(H_0/A\) is isomorphic to \(SL(2,q)\), where \(q\) is odd. The element \(\beta\) induces in \(A\) a regular automorphism; otherwise \(C(\{\tau,\beta\})\ne \{\tau,\beta\}\), which contradicts the condition. Therefore \(A\) is an abelian subgroup, and for any element \(\xi\in A\),
\[ \beta \xi \beta=\xi^{-1}. \]
The element \(\beta\) induces in \(\bar H_0=H_0/A\) an automorphism of second order. Suppose that this automorphism leaves fixed an element \(\eta A\) of the factor group \(H_0/A\). Then \(\beta\eta\beta=\eta\xi\), where \(\xi\in A\). Since the order of \(A\) is odd, there exists a natural number \(k\) such that \(\xi^{2k}=\xi\). Then
\[ \beta(\eta\xi^k)\beta=\eta\xi\xi^{-k}=\eta\xi^{2k}\xi^{-k}=\eta\xi^k, \]
and the element \(\eta\xi^k\in C_H(\beta)\). Therefore \(\eta\xi^k\in\{\tau\}\), and \(\eta A=\{\tau A\}\). Thus \(\beta\) induces in \(\bar H_0=H_0/A\) an automorphism that fixes only the elements of the subgroup \(\{\tau A\}\). We may assume \(A=\{1\}\), otherwise all arguments can be carried out for \(H/A\). We shall identify \(H_0\) with \(SL(2,q)\); \(H=H_0\{\beta\}\), and \(\beta\) induces in \(SL(2,q)\) an automorphism fixing only the elements of the cyclic subgroup
\[ \left\{\begin{pmatrix}-1&0\\0&-1\end{pmatrix}\right\}. \]
We first show that the order of a Sylow \(2\)-subgroup \(S\) of the group \(H\) is \(16\). A Sylow \(2\)-subgroup \(S_0\) of the group \(H_0\) is a maximal quaternion subgroup of some Sylow \(2\)-subgroup \(S\) of the group \(H\). It is easy to see that any element of \(S\setminus S_0\) has order \(2\) or \(2^{n+1}\); therefore any \(2\)-element of \(H\setminus H_0\) has order \(2\) or \(2^{n+1}\). In \(SL(2,q)\) there exist cyclic subgroups \(X\) and \(Y\) of orders \(q-1\) and \(q+1\). Let \(\{\rho\}\) be one of these subgroups. The subgroup \(\beta\{\rho\}\beta\) belongs to \(SL(2,q)\), and, since all cyclic subgroups of the same order are conjugate in \(SL(2,q)\), there exists an element \(\eta\in SL(2,q)\) such that
\[ \beta\{\rho\}\beta=\eta\{\rho\}\eta^{-1} \]
and \(\beta\eta\in N_H(\{\rho\})\). It is obvious that \(\beta\eta\notin H_0\). Let \(\beta\eta=\eta_1\eta_2=\eta_2\eta_1\), where \(\eta_2\) is a \(2\)-element and the order of \(\eta_1\) is odd. Since \([H:H_0]=2\), \(\eta_1\in H_0\), and \(\eta_2\in H\setminus H_0\). Since \(\eta_2\) is some power of the element \(\beta\eta\), it follows that \(\eta_2\in N_H(\{\rho\})\). The order of \(\eta_2\) is \(2^{n+1}\) or \(2\).

Suppose the order of \(\eta_2\) is \(2^{n+1}\). Then \(\eta_2^2\in N_{H_0}(\{\rho\})\), and from the properties of \(SL(2,q)\) it follows that
\[ \eta_2^4=(\eta_2^2)^2\in \{\rho\}, \]
and in this case the order of \(\{\rho\}\) is divisible by \(2^{n-1}\).

Suppose the order of \(\eta_2\) is \(2^{n+1}\). Since \(\eta_2^2\in N_{H_0}\{\rho\}\), and from the properties of \(SL(2,q)\) are conjugate in \(H\) to \(\beta\), we may assume that \(\eta_2=\beta\). Let \(\rho=\rho_1\rho_2=\rho_2\rho_1\), where \(\rho_2\) is a \(2\)-element and the order of \(\rho_1\) is odd. Then \(\beta\in N_H(\{\rho_1\})\), and, since the element \(\beta\) commutes with no nonidentity element of \(\{\rho_1\}\),
\[ \beta\rho_1\beta^{-1}=\rho_1^{-1}. \]
\(\{\rho_2\}\{\beta\}\) belongs to some Sylow \(2\)-subgroup of \(H\) and does not coincide with it. Therefore \(\{\rho_2\}\{\beta\}\) is a dihedral group and
\[ \beta\rho_2\beta=\rho_2^{-1}. \]
Hence
\[ \beta\rho\beta^{-1}=\rho^{-1}. \]
But in \(SL(2,q)\) there exists an element \(\zeta\) such that
\[ \zeta\rho\zeta^{-1}=\rho^{-1} \]
and \(\zeta\beta\in C_G(\rho)\). Let \(\zeta\beta=\xi_1\xi_2=\xi_2\xi_1\), where \(\xi_2\) is a \(2\)-element and the order of \(\xi_1\) is odd. \(\xi_2\in H\setminus H_0\) and \(\xi_2\rho=\rho\xi_2\). If the order of \(\xi_2\) is \(2\), then \(\xi_2\) is conjugate to \(\beta\), and we obtain that \(\{\rho\}=\{\tau\}\), while since the order of \(\rho\) ...

is equal to \(q-1\) or \(q+1\), then \(q=3\), and the order of a Sylow \(2\)-subgroup \(S\) of the group \(H\) is \(16\). If, however, the order of \(\xi_2\) is \(2^{n+1}\), then, as we have already shown, \(2^{n-1}\) divides the order of \(\langle \rho\rangle\). Thus, if \(q\ne 3\), \(2^{n-1}\) divides both \(q-1\) and \(q+1\), and hence their greatest common divisor is equal to \(2\), whence \(n-1\le 1\), \(n\le 2\); but, by hypothesis, \(n\ge 2\), and therefore in every case the order of \(S\) is \(16\).

The order of the group \(H\) is \(2q(q-1)(q+1)\). If \(q=r^2\), where \(r\) is an integer, then \(q-1=(r-1)(r+1)\) is divisible by \(8\), whence \(32\) divides the order of \(H\). Therefore \(q\) is not a square, and the field \(GF(q)\) has no automorphisms of order two. Every automorphism of \(SL(2,q)\) is induced by a mapping \(X\to AX^\sigma A^{-1}\), where \(A\in GL(2,q)\), \(X\in SL(2,q)\), and \(\sigma\) is an automorphism of the field \(GF(q)\). Since \(\beta^2=1\), for the corresponding automorphism we have \(\sigma^2=1\); but, since \(q\) is not a square, \(\sigma=1\). The automorphism induced in \(SL(2,q)\) by \(\beta\) is obtained by conjugating \(SL(2,q)\) by a matrix \(A\in GL(2,q)\) for which \(A^2\) lies in the center of \(GL(2,q)\). If \(q\equiv -1\pmod 4\), then as \(A\) one may choose the matrix \(\begin{pmatrix}0&1\\ i&0\end{pmatrix}\), where \(i^2=-1\), \(i\in GF(q)\), but it commutes with matrices of the form \(\begin{pmatrix}x&y\\ iy&x\end{pmatrix}\), while the equation \(x^2-iy^2=1\) has in \(GF(q)\) solutions other than \(x=\pm 1,\ y=0\). Therefore \(q\equiv 1\pmod 4\). As \(A\) one may choose the matrix \(\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}\), and it commutes with matrices in \(SL(2,q)\) of the form \(\begin{pmatrix}v&0\\ 0&v^{-1}\end{pmatrix}\), where \(v\in GF(q)\), \(v\ne 0\). The number of such matrices is \(q-1\). Therefore \(q-1=2\) and \(q=3\). We have shown that the centralizer of the element \(\tau\) is solvable. Now Theorem 1 gives \(G/K\simeq PSL(3,3)\) or \(M_{11}\).

Sverdlovsk Branch
of the V. A. Steklov Mathematical Institute
Academy of Sciences of the USSR

Received
9 IX 1965

References

1. R. Brauer, M. Suzuki, Proc. Nat. Acad. Sci. U.S.A., 45, 1757 (1959).
2. D. Gorenstein, J. Walter, Ill. J. Math., 6, 553 (1962).
3. D. Gorenstein, J. Walter, J. Algebra, 2, 85 (1965).
4. W. J. Wong, J. Austr. Math. Soc., 4, 1, 90 (1964).