PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 126, Number 2, February 1998, Pages 343–348 S 0002-9939(98)04126-4

SUBNORMAL SUBGROUPS OF GROUP RING UNITS

ZBIGNIEW S. MARCINIAK AND SUDARSHAN K. SEHGAL

(Communicated by Ronald M. Solomon)

Abstract. Let G be an arbitrary group. If a ∈ ℤG satisfies a^{2} = 0, a ≠ 0, then the units 1+a, 1+a^{∗} generate a nonabelian free subgroup of units. As an application we show that if G is contained in an almost subnormal subgroup V of units in ℤG then either V contains a nonabelian free subgroup or all finite subgroups of G are normal. This was known before to be true for finite groups G only.

0. Introduction

Let ℤG denote the integral group ring of a group G. Let U_{1}ℤG be the group of invertible elements in this ring which are of augmentation one: if u = ∑ u_{g}g then ∑ u_{g} = 1.

Let G be an arbitrary group. Clearly G ≤ U_{1}ℤG. If not all finite subgroups of G are normal in G then we are able to construct a nontrivial unit u ∈ U_{1}ℤG, u ∉ G, in the following way. Choose x, y ∈ G so that o(x) = n < ∞ and y⟨x⟩y^{−1} ≠ ⟨x⟩. For a = (1 − x)y(1 + x + ⋯ + x^{n−1}) ∈ ℤG we have a ≠ 0 and a^{2} = 0. Hence u = 1 + a is a unit and u^{−1} = 1 − a. Units obtained in this way are called bicyclic units.

It is well known that U_{1}ℤG quite often contains a nonabelian free subgroup. In fact, the following theorem was proved by Sehgal [13, p. 200] and also by Hartley– Pickel [4] (see also [14, p. 19]).

0.1. Theorem. Let G be a solvable group which has a non normal finite subgroup. Then U_{1}ℤG contains a nonabelian free subgroup.

Also, all nonabelian finite groups G, except Hamiltonian 2-groups, have a free subgroup inside U_{1}ℤG.

Free subgroups in U_{1}ℤG were also studied by Jespers who characterized in [5] all finite groups G which have a free complement in U_{1}ℤG. Jespers, Leal and del Río classified in [6] all finite nilpotent groups G such that U_{1}ℤG has a subgroup of finite index which is a direct product of noncyclic free groups. Recently Leal and del Río [7] removed the nilpotent condition and classified all finite groups with the same property.

These results were obtained by a careful study of rational representations of finite groups. Except a few examples, these were really existence results.

\begin{matrix}c \\ ◯ \end{matrix}1998 American Mathematical Society

Received by the editors August 11, 1996.

1991 Mathematics Subject Classification. Primary 16S34, 16U60.

This research was supported by Canadian NSERC Grant A-5300 and Polish Scientific Grant 2P30101007.

Recently, we have found an elementary and explicit construction of a free subgroup inside U_{1}ℤG, also for infinite groups. Namely, we proved in [10] that if u ∈ ℤG is a bicyclic unit then ⟨u, u^{∗}⟩ is a nonabelian free subgroup of U_{1}ℤG, where (∑ a_{g}g)^{∗} = ∑ a_{g}g^{−1}.

Also assuming the existence of bicyclic units, Goncalves and Passman gave in [3] a construction of a free subgroup inside U_{1}KG when K is a field, charK = p > 0, tr.deg.K > 0 and G has no p-torsion.

In the present paper we extend the main result of [10] in the following way.

0.2. Theorem. Let a ∈ ℤG be an arbitrary element satisfying a ≠ 0, a^{2} = 0. Set u = 1 + a. Then the subgroup ⟨u, u^{∗}⟩ < U_{1}ℤG is free and nonabelian.

We apply this theorem to generalize the main result of [2]. Let us say that a subgroup V < U_{1}ℤG is almost subnormal if V is subnormal in a finite index subgroup of U_{1}ℤG.

0.3. Theorem (Goncalves-Ritter-Sehgal, [2]). Let G be a finite group. If V < U_{1}ℤG is an almost subnormal subgroup containing G then either

(i) V has a nonabelian free subgroup, or (ii) G is abelian or a Hamiltonian 2–group.

The original proof depended on the classification of finite groups all of whose proper subgroups and factor groups are Hamiltonian 2–groups or abelian, followed by a careful investigation of their rational representations.

Notice that all subgroups of groups mentioned in (ii) are normal. Keeping this in mind, we offer the following generalization:

0.4. Theorem. Let G be an arbitrary group. If V < U_{1}ℤG is an almost subnormal subgroup containing G then either

(i) V has a nonabelian free subgroup, or (ii) all finite subgroups of G are normal.

1. Free subgroups in U_{1}ℤG

We construct here some free subgroups inside U_{1}ℤG. In fact, it is equally easy to do this in a more general context.

Let A be a ℂ-algebra with an involution denoted by ∗. Following [12, p. 302] we say that a Hermitian inner product (-, -): A × A ⟶ ℂ is compatible with ∗ if (a^{∗}b, c) = (b, ac) holds for all a, b, c ∈ A.

1.1. Examples. (a) Let A = ℂG, the group algebra of an arbitrary group G with the standard involution (∑ a_{g}g)^{∗} = ∑ \overline{a}_{g}g^{−1}. The Hermitian inner product compatible with ∗ is defined by (a, b) = tr(a^{∗}b) where tr is the group ring trace: tr(∑ a_{g}g) = a_{1} (compare [11, p.33]).

(b) Let A = M_{n}(ℂ), the matrix ring with the involution a^{∗} = \overline{a}^{T}. The Hermitian inner product compatible with ∗ is given by (a, b) = Tr(a^{∗}b) where Tr is the usual matrix trace.

(c) Let A be any H^{∗}-algebra. It is a Banach ∗-algebra whose norm is a Hilbert space norm such that (a, bc^{∗}) = (ac, b) = (c, a^{∗}b) for all a, b, c ∈ A (see [1]).

We can equip any such algebra with a norm ∥.∥: A ⟶ ℝ defined by ∥a∥ = (a, a)^{1/2}.

The main result of this section is

1.2. Theorem. Let A be a ℂ-algebra with an involution ∗ and with a compatible Hermitian inner product (-, -). If an element a ∈ A satisfies a^{2} = 0 and ∥a^{∗}a∥ ≥ 4 · ∥1∥ then ⟨1 + a, 1 + a^{∗}⟩ is a free nonabelian subgroup of units in A.

For the proof we consider the sequence T = {1, c, c^{2}, c^{3}, … } of elements in A, where c = a^{∗}a ∈ A. In what follows we will assume that ∥c∥ ≥ 4 · ∥1∥. Our argument splits into two cases depending on the linear dependence of T.

1.3. Lemma. If the sequence T is linearly independent then

(i) the submonoid S of (A, ·) generated by {a, a^{∗}} is isomorphic to ⟨s, t|s^{2} = 0 = t^{2}⟩,

(ii) the linear span of S in A is isomorphic to the (reduced) semigroup algebra ℂS.

Proof. Let Σ be the monoid given by the presentation ⟨s, t|s^{2} = 0, t^{2} = 0⟩. Consider the monoid map ϕ: Σ ⟶ S given by ϕ(s) = a^{∗}, ϕ(t) = a.

To prove (i) it is enough to show that ϕ is injective. To this end we decompose Σ into five pairwise disjoint pieces: Σ = {0} ∪ T ∪ tT ∪ T s ∪ tT s, where T = { (st)^{k} | k ≥ 0}. Notice that ϕ is injective on T ⊂ Σ as, by assumption, all powers of c are distinct.

Suppose that ϕ(x) = ϕ(y) for two distinct elements x, y ∈ Σ \ T . If x, y belong to the same part of Σ, we can multiply them both on the left by s and/or on the right by t to obtain two distinct elements of T mapping to the same element of S. This is a contradiction with the last paragraph. On the other hand, if x, y belong to different parts of Σ then a simultaneous left and/or right multiplication gives us a new pair x′ = 0, y′ ∈ T . But then 0 = ϕ(x′) = ϕ(y′) ∈ T – a contradiction, as T is linearly independent. This proves (i).

To prove (ii) it is enough to show that S \ {0} is linearly independent in A. Consider any linear combination of elements from S \ {0}. We can write it in the form α + β + γ + δ = 0 where the four summands are combinations of elements from T, aT, T a^{∗} and aT a^{∗} respectively. Note that these sets cover S \ {0} and are pairwise disjoint by part (i).

If α ≢ 0 then we multiply our combination on both sides by a^{∗}a. We obtain a non-trivial combination (a^{∗}a)α(a^{∗}a) = 0 of elements from T, a contradiction. Hence α ≡ 0 and so β + γ + δ = 0.

We can continue in this way to show that also β ≡ γ ≡ δ ≡ 0. For details look in the proof of Lemma 6 in [10].

1.4. Lemma. If the sequence T is linearly dependent then the linear map f : A ⟶ A given by f(x) = cx has an eigenvalue λ with |λ| ≥ 4.

Proof. Choose a maximal positive integer d so that the set T_{d} = {1, c, …, c^{d−1}} is linearly independent. Let W be the ℂ-linear span of T_{d} in A. Then c^{d} ∈ W and hence the finite dimensional subspace W is invariant under f.

Notice that c^{∗} = c. Then for v, w ∈ W we have (v, f(w)) = (v, cw) = (c^{∗}v, w) = (cv, w) = (f(v), w), i.e. f: W ⟶ W is a Hermitian operator with respect to the Hermitian inner product (-, -)_{|W}. From the Spectral Theorem (see [9, Ch.XIV, §12]) it follows that W has an orthonormal basis v_{1}, …, v_{d} consisting of eigenvectors of f.

Let f(v_{i}) = λ_{i}v_{i} for i = 1, …, d. We prove that we must have |λ_{i}| ≥ 4 for at least one i. To this end write 1 ∈ W ⊂ A as a combination of the eigenvectors: 1 = ∑ α_{i}v_{i}, α_{i} ∈ ℂ. Then c = c · 1 = ∑ α_{i}cv_{i} = ∑ α_{i}f(v_{i}) = ∑ α_{i}λ_{i}v_{i}. Hence ∥c∥^{2} = (c, c) = ∑ |α_{i}λ_{i}|^{2}, by the Pythagorean Theorem.

Suppose that |λ_{i}| < 4 holds for i = 1, …, d. Then ∥c∥^{2} < 16 · ∑ |α_{i}|^{2} = 16 · ∥1∥^{2} and so ∥c∥ < 4 · ∥1∥ – a contradiction with the assumption that ∥c∥ ≥ 4 · ∥1∥.

Proof of Theorem 1.2. We look at the sequence T. If T is linearly independent then, by Lemma 1.3, the subalgebra R ⊂ A generated by {a^{∗}, a} is isomorphic to the semigroup algebra ℂΣ. Consider the monoid map ρ: Σ ⟶ M_{2}(ℂ) given by ρ(s) = \binom{0 2}{0 0}, ρ(t) = \binom{0 0}{2 0}. We extend ρ by linearity to an algebra homomorphism \overline{ρ}: R ≈ ℂΣ ⟶ M_{2}(ℂ). Then \overline{ρ}(1 + a^{∗}) = \binom{1 2}{0 1}, \overline{ρ}(1 + a) = \binom{1 0}{2 1} and hence 1 + a^{∗}, 1 + a generate together a nonabelian free subgroup, by a well known theorem of Sanov:

1.5. Theorem (Sanov, see [8, p. 92]). For any complex number z satisfying |z| ≥ 2 the matrices \binom{1 z}{0 1}, \binom{1 0}{z 1} generate a nonabelian free subgroup of GL_{2}(ℂ).

We are thus left with the case when T is linearly dependent. By Lemma 1.4 there exists an element v ∈ A, v ≠ 0, such that a^{∗}a · v = λv and |λ| ≥ 4. Choose μ ∈ ℂ so that μ^{2} = λ and define w = (1/μ)a · v ∈ A.

The subspace V = span_{ℂ}(v, w) ⊂ A is invariant under left multiplication by a^{∗} and by a. In fact,

\begin{matrix}a^{∗} · v = (1/λ)(a^{∗})^{2}av = 0, & a · v = μw, \\ a^{∗} · w = (1/μ)a^{∗}av = μv, & a · w = (1/μ)a^{2}v = 0. \end{matrix}

From the above equalities it follows that the vectors v, w are linearly independent and hence they form a basis of V.

Obviously, left multiplications by 1 + a^{∗} and 1 + a also preserve the subspace V. With respect to the basis {v, w} they are represented by the matrices

1 + a^{∗} ↔ (\begin{matrix}1 & μ \\ 0 & 1 \end{matrix}) , 1 + a ↔ (\begin{matrix}1 & 0 \\ μ & 1 \end{matrix}) .

However, we have |λ| ≥ 4 and hence |μ| ≥ 2. Therefore the above pair of matrices generates a free group, by Theorem 1.5. Hence the units 1 + a^{∗}, 1 + a ∈ A do the same.

Now we specialize to group algebras.

1.6. Theorem. If an element a ∈ ℂG satisfies a^{2} = 0 and ∥a∥ ≥ 2 then ⟨1 + a, 1 + a^{∗}⟩ is a free nonabelian subgroup of units in ℂG.

Proof. Let c = a^{∗}a. Then we have ∥c∥^{2} = (c, c) = tr(c^{∗}c) = ∑ |c_{g}|^{2} ≥ |c_{1}|^{2}. It follows that ∥c∥ ≥ |c_{1}| = |tr(c)| = |tr(a^{∗}a)| = |(a, a)| = ∥a∥^{2} ≥ 4. Hence ∥a^{∗}a∥ ≥ 4 = 4 · ∥1∥ and we may apply Theorem 1.2.

1.7. Theorem. If an element a ∈ ℤG satisfies a^{2} = 0 and a ≠ 0 then ⟨1+a, 1+a^{∗}⟩ is a free nonabelian subgroup of units in ℤG.

Proof. By the previous theorem it is enough to show that ∥a∥ ≥ 2. Suppose then that ∥a∥ < 2, i.e. 4 > ∥a∥^{2} = ∑ a_{g}^{2}. Hence all a_{g} belong to {−1, 0, 1} and a has at most three elements in its support. From a^{2} = 0 it follows that a has augmentation zero and hence a = g − h for some g, h ∈ G. Then from 0 = a^{2} = g^{2} + h^{2} − gh − hg it easily follows that g = h and a = 0 – a contradiction.

1.8. Corollary. If ℤG contains a nonzero nilpotent element then the unit group of ℤG contains a nonabelian free subgroup.

1.9. Corollary. Suppose that the group ring ℤG is not a domain. If G has no nonidentity finite normal subgroups then the group of units of ℤG contains a nonabelian free subgroup.

Proof. Let x, y ∈ ℤG be nonzero elements such that xy = 0. From Connell’s Theorem [11, Thm. 4.2.10] it follows that the algebra ℚG is prime and hence yℚGx ≠ 0. Pick an element z ∈ ℤG so that a = yzx ≠ 0. Then a^{2} = 0 and Theorem 1.7 applies.

2. An application

Here we prove Theorem 0.4. We start with

2.1. Lemma. Suppose that an almost subnormal subgroup V < U_{1}ℤG contains G. If ℤG has a bicyclic unit then V contains a nonabelian free subgroup.

Proof. Write U = U_{1}ℤG for short. We have

G < V = V_{r} ◁ V_{r−1} ◁ ⋯ ◁ V_{1} < U

and |U : V_{1}| < ∞. It follows there exists a normal subgroup N ◁U such that N < V_{1} and |U : N| < ∞. Consider N^{∗} = {u ∈ U|u^{∗} ∈ N}. Then also |U : N^{∗}| < ∞ and N^{∗} ◁ U. Set Ñ = N ∩ N^{∗}. Then we have Ñ^{∗} = Ñ, |U : Ñ| < ∞, Ñ ◁ U and Ñ < V_{1}.

Let u be a bicyclic unit and let k be the order of the image of u in U/Ñ. Then u^{k} ∈ Ñ and (u^{∗})^{k} = (u^{k})^{∗} ∈ Ñ^{∗} = Ñ.

By construction u = 1 − a where a = (x − 1)y(1 + x + ⋯ + x^{n−1}) for suitable x, y ∈ G. Consider the group ring elements a_{s} = k(x − 1)^{s}y(1 + x + ⋯ + x^{n−1}) for s ≥ 1. Clearly a_{s}^{2} = 0. Hence we have new units u_{s} = 1 − a_{s} ∈ U.

Notice that xu_{s}x^{−1} = 1 − xa_{s}. It follows that u_{s} = [x, u_{s−1}] for all s ≥ 2. In fact,

\begin{aligned}[x, u_{s−1}] & = x · (1 − a_{s−1}) · x^{−1} · (1 + a_{s−1}) = (1 − xa_{s−1})(1 + a_{s−1}) \\ & = 1 + a_{s−1} − xa_{s−1} = 1 − (x − 1)a_{s−1} = 1 − a_{s} = u_{s}. \end{aligned}

Finally, we show that u_{s}, u_{s}^{∗} ∈ V_{s} for all s ≥ 1. Clearly u_{1} = u^{k} and hence u_{1}, u_{1}^{∗} ∈ Ñ < V_{1}. Assume that u_{s−1} ∈ V_{s−1}. Then u_{s−1}x^{−1}u_{s−1}^{−1} ∈ V_{s} as x^{−1} ∈ G < V_{s} ◁ V_{s−1}. Then also u_{s} = x · u_{s−1}x^{−1}u_{s−1}^{−1} ∈ V_{s}. The proof that u_{s}^{∗} ∈ V_{s} is similar.

Now, by Theorem 1.7, ⟨u_{r}, u_{r}^{∗}⟩ = ⟨1+a_{r}, 1+a_{r}^{∗}⟩ < V_{r} = V is a free subgroup.

Proof of Theorem 0.4. If G has a finite subgroup which is not normal then U_{1}ℤG contains a bicyclic unit. From Lemma 2.1 it then follows that the subgroup V contains a nonabelian free subgroup.

2.2. Corollary. Let G be an arbitrary group. If U_{1}ℤG contains a subgroup V such that

(i) V is subnormal in a finite index subgroup of U_{1}ℤG, (ii) G ≤ V, (iii) V is nilpotent then all finite subgroups of G are normal.

2.3. Remark. Clearly, we could change nilpotent in the last corollary to solvable or any other nontrivial variety of groups. The proof remains the same.

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Institute of Mathematics, Warsaw University, ul. Banacha 2, 02-097 Warszawa, Poland

E-mail address: zbimar@mimuw.edu.pl

Department of Mathematical Sciences, University of Alberta, Edmonton, Canada T6G 2G1

E-mail address: S.Sehgal@ualberta.ca