Hypercyclic Linear Groups (HLG) Volume 60- Issue 5

Behnam Razzaghmaneshi*

• Assistant Professor of mathematics, Algebra, Islamic Azad University, Iran

Received: March 06, 2025; Published: March 12, 2025

*Corresponding author: Behnam Razzaghmaneshi, Assistant Professor of mathematics, Algebra, Islamic Azad University, Talesh Branch, Talesh, Iran

DOI: 10.26717/BJSTR.2025.60.009527

Abstract PDF

There exist no hypercyclic linear groups with paraheight w.

Keywords: Linear Groups; Paraheight Groups; Finite Groups

Let G be an arbitrary group A a G-module and H the split extension of A by G. We say that A is a hypercyclic G-module with paraheight y if A is H-hypercyclic (as a normal subgroup of H) with H-paraheight y.

Proposition

Let G be a finite group and A a hypercyclic G-module. Then A has paraheight at most W. If A is torsion-free as Z-module A has paraheight at most 1 G 1 and if A has jinite exponent e as Z-module and (e, 1 G 1) = 1 then A has paraheight at most 1 G 1 [logs e].

1. Suppose that A is Z-torsion-free. Embed A in V = Q 0~ A by identifying A and 1 @ A. V is a QG-module containing an ascending series of QG-submodules whose factors have Q-dimension at most 1. Also, V is completely reducible as QG-module (Extension of Maschke’s Theorem, Ref. [1-7], IV.8.j) and thus V is a direct sum of irreducible QG-modules of Q-dimension 1. Let VI, V, ,..., V, be the homogeneous components of V as QG-module and put Ai = A n Vi. Suppose that Ai i: (0). If g E G there exists a rational number s such that ag = sa for all a in Ai. Since A is hypercyclic there exists b E A, \(O) such that (b) is a G-submodule of Ai, that is 6g = nb for some integer n. Hence s = n is an integer (since A is torsion-free) and consequently every Z-submodule of Ai is a G-module. Thus, A has paraheight at most Y and clearly Y < / G j.

2. Suppose that A is an elementary abelian p-group where (p, / G I) = 1. By the extension of Maschke’s Theorem A is completely reducible as F, G-module. Let A, A, ,..., A, be the homogeneous components of A as lF,G-module. Since A is hypercyclic G acts on each Ai as a group of scalars and thus every Z-submodule of each Ai is a G-submodule. Therefore, A has paraheight at most Y < 1 G I. If A has finite exponent e then A contains a characteristic series of length at most [log, e] whose factors are elementary abelian. Hence by 2. if (e, 1G I) = 1, then A has paraheight at most I G / [log, e].

3. General Case. A contains a free abelian subgroup E such that A/E is a periodic abelian group. Then E1 = fiBEG Eg is a free abelian G-submodule of A and A/E, is still periodic as Z-module (since G is finite). Hence by 1. above we may assume that A is periodic as Z-module. A = 0, A, where A, is the p-primary component of A. If I3 is a G-submodule of A such that every Z-submodule of every primary component of B is G-invariant then every Z-submodule of B is G-invariant. Hence the paraheight of A is equal to the upper bound of the paraheights of the A, . Thus we may suppose that A is a p-group. Let A, be the G-submodule of A generated by all the irreducible G-sub-modules of A of order p and inductively define A,+,/A, = (A/A,), . Put A, = (JB A, . A,+JA, is elementary abelian and completely reducible as lF,G-module. There exists only a finite number (r say) of irreducible lF,G-modules of order p up to isomorphism. Let A,+,(j)/A, , j = 1, 2,..., r, be the homogeneous components A,+JA, as IF,G-module. As in 2. Above every Z-sub module of A,+,(j)/A, is G-invariant and thus A, has paraheight at most w. It remains to show that A, = A. If a E A, (aG) is a finite G-submodule of A (since G is finite and A is periodic). Now A is hypercyclic, so there exists a finite series of G-submodules, {0} = B, C B, C **. C B, = (aG) such that (B, : B,-J = p f or each i. Clearly Bi C Ai for each i and thus (aG) C A, . Therefore A = A, and 4.1 is proved. It follows from 4.1 and 3.3 that a locally supersoluble linear group has paraheight less than ~2 and that a locally supersoluble linear group over a finitely generated integral domain is parasoluble (use the existence of a nilpotent subgroup of finite index and for the second part [lo, 4.101). TO obtain the relative case and the bounds in terms of 11we have to a little more work that remarkably parallels the hypercentral situation ([lo] Chap. 8).

Lemma

Let V be a vector space of dimension n over the field F, G a subgroup of AutF(V) and W an FG-submodule of V of F-dimension d. If A is a G-hypercyclic normal subgroup of G stabilizing the series (0) C W 2 V then A has G-paraheight at most d (n - d). Proof. If a E A let a’ be the linear mapping of VI W into W given by v + W H v (u - 1). The map 4: a ++ a’ is a (group) monomorphism of A into the additive group of Hom, (V/ W, W). The latter is an FG-module, the G action being given byf 9: v + W H (vg-l + W) fg; f E Hom, (V/ W, W) and g E G. A is abelian and so is a G-module via conjugation. A simple check shows that $ is a G-module homomorphism. Let B be the F-subspace of Hom, (V/ W, W) spanned by A+. B is an FG- module and since A is G-hypercyclic contains an ascending series of FG-submodules whose factors have F-dimension at most 1. B has F-dimension at most d (n - d). Thus, there exists a series of FG-submodules (0) =B, C B, C ... C B, = B where each B, +lIB, has F-dimension 1 and r < d (n - d).

Supersoluble Linear Groups 53: Since A$ is G-hypercyclic and spans B there exists c in B, +, /B, \{O} such that (c) is a G-module. Now if g E G there exists 01 in F such that xg = OLXfor all x in B, +, /B, Hence EC = nc for some integer n and thus a=TZl,. Therefore every Z-submodule of B, +,/B, is a G-submodule and so B has paraheight (as G-module) at most Y. Since A sc A+ _CB the result follows.

Corollary

Let G be a subgroup of GL(n,F) and U a unipotent G-hypercyclic normal subgroup of G. Then U has G-paraheight at most +z(n - 1). Proof. Let V be the n-row vector space over F regarded as a G-module in the usual way. U is unitriangularizable over F [IO, I.211 and so there exists a non-trivial subspace of V on which U acts trivially. Let W = C,(U). Then d = dim, W 3 1 and W is G-invariant. By induction on II, UCo(V/W)/Co(V/ W) has G-paraheight at most +(n - d)(n - d - 1). By 4.2 U n C,( V/W) has G-paraheight at most d(n - d). Thus U has G-paraheight at most t(n - d)(n + d - 1) < +n(n -1) since I < d < n.

Lemma

If G is an irreducible subgroup of GL(n, F) then X(G) contains a diagonalizable subgroup A normal in G such that /\(G)/A is isomorphic to a subgroup of S, , the symmetric group on n letters. Proof. Let V be the n-row vector space over F regarded as G-module in the usual way and suppose that {V, , V, ,..., V,} is a minimal system of imprimitivity of V as FG-module. If H = NG(Vf), H acts primitively and irreducibly on Vi ([lo] 1.10). Let + be the induced homomorphism of H into AutF(VJ and put 2 = (H n h(G))+ n &(H$). H n A(G) is an H-hypercyclic subgroup of H. If (H n h(G))+ # 2 there exists an element h of (H n X(G))+\ 2 such that (2, h) is normal in H4. But clearly (2, h) is abelian and so lies in the centre of H$ by Blichfeldt’s theorem [ 10, 1.131. This contradiction shows that H n X(G) acts on Vi as a group of scalars. Let A = X(G) n h No(VJ.i=l Then A is a diagonalizable normal subgroup of G and /\ (G)/A is isomorphic to a subgroup of S, (since the elements of G permute the Vi among them-selves).

Main Theorem: If G is a subgroup of GL(n, F) G has paraheight at most w + [log, n!]. If G is a subgroup of GL(n, R) where R is a finitely generated integral domain then G has finite paraheight.

Proof: There exist irreducible representations pr , pa ,..., p,. of G and an x in GL(n, F) such that for all g in G. Let U = & ker pi and put H = {diag(gp, ,..., gp7) : g E G> C GL(n, F). U is a unipotent normal subgroup of G and G/U is isomorphic to the completely reducible subgroup H of GL(n,F). By 4.4 H contains a diagonalizable subgroup A that is normal in H such that h(H)/A is isomorphic to a subgroup of S, . Clearly A(H)/A is H-hypercyclic with H-paraheight at most[log, n!]. By [lo] 1.12 (H : C,(A)) is finite. Hence A has H-paraheight at most w (4.1) and thus H has paraheight at most w + [log, n!]. Since G/ U s H, h(G)/ U n h(G) has G-paraheight at most w + [log, n!] and by 4.3 U n A(G) has G-paraheight at most +z(n - 1). Thus G has paraheight at most w + [logsn!]. Suppose now that G _CGL(n, R). It follows from [8- 10] 4.10 that A is finitely generated. Therefore by 4.1 H, and thus G, has finite paraheight. We now consider what paraheights are possible for hypercyclic linear groups of low degree. For each positive integer n and each characteristic p denote by ~(n,p) the upper bound of the paraheights of hypercyclic linear groups of degree n over a field of characteristic p. We shall see that o(n, p) is never a limit ordinal.

Proposition

There exist no hypercyclic linear groups with para-height w. Proof. Let G be a hypercyclic subgroup of GL(n,F) with paraheight w. By induction on n and 4.3 G is irreducible. Hence G contains an abelian normal subgroup A of finite index. G contains a series of normal subgroups (1) = G,cG,C ... c G, = G of length w such that every subgroup of Gi+JG, is abelian and normal in G. Since (G : A) is finite there exists a finite r such that AG, = G. Then G/G, is abelian and so G has paraheight at most r + 1. This contradiction proves the proposition. However, there do exist linear groups with paraheight w. These may be constructed exactly as the example 8.7 of [lo] using examples (e.g. (C’s, 1 Cs) constructed below in place of C,, 1 C, .

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