Full Text
D. A. SUPRUNENKO
ON PERIODIC SUBGROUPS OF SOLVABLE MATRIX GROUPS
(Presented by Academician A. I. Mal'cev on 28 V 1962)
Let \(\Gamma\) be a maximal solvable subgroup of the full linear group over an algebraically closed field of arbitrary characteristic. In the present article the conjugacy of maximal periodic subgroups of the group \(\Gamma\) in \(\Gamma\) is proved.
By the letter \(\Delta\) we shall denote an algebraically closed field of arbitrary characteristic, and by \(E_n\) the identity matrix of order \(n\).
§ 1. Lemma 1. Let \(\Gamma\) be a maximal irreducible primitive solvable subgroup of the full linear group \(GL(n,\Delta)\). Then the set \(\Phi\) of all elements of finite order from \(\Gamma\) forms a normal divisor of the group \(\Gamma\).
Proof. Let \(\Delta^*\) be the multiplicative group of the field \(\Delta\), and let
\[
\Gamma_0=\Gamma\cap SL(n,\Delta),
\]
where \(SL(n,\Delta)\) is the special linear group. Then \(\Gamma_0\) is a finite group, and moreover
\[
\Gamma=\Gamma_0\Delta^*E_n \tag{1}
\]
An element \(g=\lambda g_0,\ \lambda\in\Delta^*,\ g_0\in\Gamma_0\), has finite order if and only if \(\lambda\) belongs to the periodic part \(P\) of the group \(\Delta^*\). Consequently,
\[
\Phi=\Gamma_0 P E_n.
\]
The lemma follows from this.
Lemma 2. Let \(\Gamma\) be a maximal irreducible solvable subgroup of \(GL(n,\Delta)\). Then all maximal periodic subgroups \(\Gamma\) are conjugate in \(\Gamma\).
Proof. Let \(\Delta^n\) be the linear space in which \(\Gamma\) acts. Write the complete decomposition of \(\Delta^n\) into the systems of imprimitivity of the group \(\Gamma\):
\[
\Delta^n=Q_1\dot{+}Q_2\dot{+}\cdots\dot{+}Q_k.
\]
Then, as shown in (2), \(\Delta^n\) has a basis
\[
u_1^1,\ldots,u_m^1,\ldots,u_1^k,\ldots,u_m^k, \tag{1}
\]
such that \(\Gamma=GT\), the group \(G\) consists of all matrices of the form
\[
g=
\begin{bmatrix}
A_1 & 0 & \cdots & 0\\
0 & A_2 & \cdots & 0\\
\cdot & \cdot & \cdots & \cdot\\
0 & 0 & \cdots & A_k
\end{bmatrix},
\]
where the \(A_j\), independently of one another, run through a maximal primitive solvable subgroup \(\mathfrak G\) of the group \(GL(m,\Delta)\), and the group \(T\) is isomorphic to a certain maximal solvable transitive subgroup of the symmetric group \(S_k\); for \(t\in T\),
\[
t(u_j^\nu)=u_j^\mu,\qquad \nu=1,\ldots,k;\ j=1,\ldots,m.
\]
We add also that \(u_1^\nu,\ldots,u_m^\nu\) is a basis of \(Q_\nu\).
Let \(\Phi\) be a maximal periodic subgroup of the group \(\mathfrak G\). Then
\[
\Psi=\Phi^k T,
\]
where \(\Phi^k\) consists of all matrices of the form
\[ \left[ \begin{array}{cccc} \varphi_1 & . & . & 0\\ . & . & . & .\\ 0 & . & . & \varphi_k \end{array} \right], \qquad \varphi_j\in\Phi, \]
is obviously a periodic subgroup of the group \(\Gamma\).
We shall show that any periodic subgroup of the group \(\Gamma\) is conjugate in \(\Gamma\) to some subgroup of the group \(\Psi\). Let \(H\) be an arbitrary periodic subgroup of the group \(\Gamma\); \(T_1\) the group of substitutions effected by \(H\) on \(Q_1,\ldots,Q_k\), and let \(Q_1,\ldots,Q_l\) be one of the systems of imprimitivity of the group \(T_1\). Then in \(H\) there are operators \(h_1,h_2,\ldots,h_l\) such that
\[ h_1=E_n,\qquad h_2Q_1=Q_2,\ldots,\qquad h_lQ_1=Q_l . \]
For \(h_\nu\) \((\nu=1,\ldots,l)\) one may write
\[ h_\nu=g_\nu t_\nu,\qquad t_\nu\in T,\qquad g_\nu\in G, \]
\[ t_\nu(u_j^1)=u_j^\nu,\qquad j=1,\ldots,m, \]
\[ g_\nu= \left[ \begin{array}{cccc} A_{1\nu} & . & . & 0\\ . & . & . & .\\ 0 & . & . & A_{k\nu} \end{array} \right], \qquad A_{i\nu}\in\mathfrak G . \]
We now choose a new basis of the space \(\Delta^n\):
\[ v_1^1,\ldots,v_m^1,\ v_1^2,\ldots,v_m^2,\ \ldots,\ v_1^k,\ldots,v_m^k, \tag{2} \]
where \(v_j^\nu=u_j^\nu\), if \(\nu=1\) or \(\nu>l\),
\[ v_j^\nu=h_\nu(u_j^1)=g_\nu(u_j^\nu),\qquad \text{if } 1<\nu\le l. \]
The matrix expressing the new basis (2) in terms of the old basis (1) has the form
\[ c= \left[ \begin{array}{ccccc} E_m & 0 & . & . & 1\\ 0 & A_{22} & . & . & 0\\ . & . & . & . & .\\ 0 & . & . & . & A_{ll}0\\ 0 & . & . & . & E_m \end{array} \right], \tag{3} \]
where \(A_{ij}\in\mathfrak G\). Consequently, \(c\in G\).
Let \(h\) be any operator from \(H\). Then \(hQ_\mu=Q_\nu\), \(1\le\mu,\nu\le l\). Obviously,
\[ \left[ \begin{array}{c} h(v_1^\mu)\\ .\\ .\\ .\\ h(v_m^\mu) \end{array} \right] = B_\mu \left[ \begin{array}{c} v_1^\nu\\ .\\ .\\ .\\ v_m^\nu \end{array} \right], \tag{4} \]
where \(B_\mu\) is an \(m\times m\)-matrix over \(\Delta\).
We shall prove that \(B_\mu\in\Phi\). For this, consider \(a=h_\mu h_\nu^{-1}h\). Obviously, \(a\in H\), \(aQ_\mu=Q_\mu\). Hence, from Lemma 1 it follows that the restriction of \(a\) to \(Q_\mu\) belongs to \(\Phi\). On the other hand, for \(a\), using (4), one can write
\[ \left[ \begin{array}{c} a(v_1^\mu)\\ .\\ .\\ .\\ a(v_m^\mu) \end{array} \right] = B_\mu \left[ \begin{array}{c} h_\mu h_\nu^{-1}(v_1^\nu)\\ .\\ .\\ .\\ h_\mu h_\nu^{-1}(v_m^\nu) \end{array} \right] = B_\mu \left[ \begin{array}{c} v_1^\mu\\ .\\ .\\ .\\ v_m^\mu \end{array} \right]. \]
Consequently, in the basis \(v_1^\mu,\ldots,v_m^\mu\) the matrix of the restriction of \(a\) to \(Q_\mu\) coincides with \(B_\mu\). In view of (3), in the basis \(u_1^\mu,\ldots,u_m^\mu\) the matrix of the restriction of \(a\) to \(Q_\mu\) is equal to \(AB_\mu A^{-1}\), where \(A\in\mathfrak G\). Thus \(AB_\mu A^{-1}\in\Phi\). Since \(\Phi\) is a normal divisor of \(\mathfrak G\), it follows that \(B_\mu\in\Phi\). We shall now regard the elements of \(H\) as matrices. Then, from the fact that \(B_\mu\in\Phi\), it follows that the restriction of \(c^{-1}Hc\) to
\(Q_1 \dotplus \cdots \dotplus Q_l = L\) is contained in the restriction of \(\Psi\) to \(L\) (see (3)). Repeating these arguments for each system of imprimitivity of the group \(T_1\), we obtain
\[ d^{-1}Hd \subseteq \Psi,\quad \text{where } d \in G. \]
The lemma follows from this.
§ 2. Theorem. Let \(\Gamma\) be a maximal solvable subgroup of \(GL(n,\Delta)\), where \(\Delta\) is an algebraically closed field. Then the maximal periodic subgroups of the group \(\Gamma\) are conjugate in \(\Gamma\).
Proof. First consider the case where the characteristic of the field \(\Delta\) is zero. In this case a maximal periodic subgroup of the group \(\Gamma\) is contained in some maximal completely reducible subgroup of the group \(\Gamma\). As was proved in \((^3)\), the maximal completely reducible subgroups of \(\Gamma\) are conjugate in \(\Gamma\). On the other hand, a maximal completely reducible subgroup \(D\) of the group \(\Gamma\) is isomorphic to the direct product of its irreducible components, each of which is a maximal solvable irreducible group \((^3)\). By virtue of Lemma 2, the maximal periodic subgroups of the group \(D\) are conjugate in \(D\). Consequently, for fields of characteristic zero the theorem is proved.
Let now the characteristic of the field \(\Delta\) be \(p>0\). Write the matrices of the group \(\Gamma\) in the form
\[ g= \begin{bmatrix} a_1(g) & * & \ldots & *\\ 0 & a_2(g) & \ldots & *\\ \cdots & \cdots & \cdots & \cdots\\ 0 & 0 & \ldots & a_k(g) \end{bmatrix}, \tag{5} \]
where \(a_j(g)\) runs through a maximal irreducible solvable subgroup \(\Gamma_j\) of the group \(GL(n_j,\Delta)\), \(n_1+n_2+\cdots+n_k=n\). Obviously,
\[ g=du, \tag{6} \]
where
\[ d= \begin{bmatrix} a_1(g) & \ldots & 0\\ \cdots & \cdots & \cdots\\ 0 & \ldots & a_k(g) \end{bmatrix}, \qquad u= \begin{bmatrix} E_{n_1} & * & \ldots & *\\ \cdots & \cdots & \cdots & \cdots\\ 0 & \ldots & E_{n_k} \end{bmatrix}. \]
All matrices \(u\) from (6) form an invariant \(p\)-subgroup \(U\) of the group \(\Gamma\), while the matrices \(d\) form a completely reducible subgroup \(D\). Obviously, the mapping \(g \mapsto d\) is a homomorphism \(\varphi\) of the group \(\Gamma\) onto \(D\). If \(H\) is a maximal periodic subgroup of the group \(\Gamma\), then \(\varphi(H)\) is a maximal periodic subgroup of \(D\). Consequently, any maximal periodic subgroup \(H\) of the group \(\Gamma\) is representable in the form
\[ H=VU, \]
where \(V\) is a maximal periodic subgroup of the group \(D\). Since the maximal periodic subgroups of the group \(D\) are conjugate in \(D\), it follows from (7) and from the invariance of \(u\) in \(\Gamma\) that the maximal periodic subgroups of the group \(\Gamma\) are conjugate in \(\Gamma\). The theorem is proved.
As is known, in \(GL(a,\Delta)\) the set of all maximal solvable subgroups splits into only a finite number of classes of conjugate subgroups. Hence, from the theorem just proved there follows
Corollary. The set of all maximal solvable periodic subgroups of \(GL(n,\Delta)\) splits into only a finite number of classes of subgroups conjugate in \(GL(n,\Delta)\).
Received
22 V 1962
REFERENCES
- H. Zassenhaus, Abh. math. Semin. Hansische Univ., 12, H. 3/4, 312 (1938).
- D. Suprunenko, Matem. sborn., 41 (82), 317 (1957).
- D. Suprunenko, Dokl. AN BSSR, 5, No. 9, 371 (1961).