A NOTE ON SUBNORMAL SUBGROUPS IN DIVISION RINGS CONTAINING SOLVABLE SUBGROUPS

Abstract

Let D be a division ring and N be a subnormal subgroup of the multiplicative group $D^*$. We show that if N contains a nonabelian solvable subgroup, then N contains a nonabelian free subgroup.

1 Introduction

Let D be a division ring and $D^*$ denote the multiplicative group of D. Recall that a subgroup N of $D^*$ is subnormal in $D^*$ if there is a finite sequence of subgroups

$$ \begin{align*}N=N_r\trianglelefteq N_{r-1}\trianglelefteq \cdots N_1 \trianglelefteq N_0=D^*.\end{align*} $$

The motivation of this paper is a result of Tits for linear groups. Tits showed that if N is a finitely generated subgroup of the general linear group $\mathrm {GL}_n(F)$ over a field F, then N contains either a nonabelian free subgroup or a nilpotent normal subgroup H of finite index in N [13]. This famous result of Tits is now referred to as the Tits Alternative. For general linear groups over division rings, Lichtman [10] proved that there exists a division ring D such that $D^*$ contains a finitely generated group which is not solvable-by-finite. Therefore, the Tits Alternative fails even for matrices of degree one over $D^*$ . Also in [10], Lichtman remarked that it was not known whether the multiplicative group of a noncommutative division ring contains a nonabelian free subgroup. One year later, Lichtman showed in [11] that if N is a normal subgroup of $D^*$ such that N contains a nonabelian nilpotent-by-finite subgroup, then N contains a nonabelian free subgroup. After that, Gonçalves and Mandel proposed the following conjecture.

Conjecture 1.1 [5, Conjecture 2].

For a division ring D with centre F and a subnormal subgroup N of $D^*$ , if $N\not \subseteq F$ , then N contains a nonabelian free subgroup.

We give an affirmative answer to this conjecture in the case when N contains a nonabelian solvable subgroup. This result can be seen as a generalisation of Lichtman’s result in [11].

Theorem 1.2 (Main Theorem).

Let D be a division ring and N a subnormal subgroup of the multiplicative group $D^*$ . If N contains a nonabelian solvable subgroup, then N contains a nonabelian free subgroup.

We note that many special cases of the Main Theorem have been studied by many authors. For example, a subgroup N of $D^*$ will contain nonabelian free subgroups if N satisfies any of the following assumptions:

• • N is normal in $D^*$ and N contains a nonabelian nilpotent-by-finite subgroup (Lichtman [11]);

• • N is normal in $D^*$ and N contains a nonabelian locally solvable-by-locally-finite subgroup (Bell and Gonçalves [1]);

• • N is subnormal in $D^*$ , N contains a nonabelian locally solvable subgroup and the transcendence degree of the centre of D over its prime subfield is infinite (Bien and Hai [2]).

For convenience, throughout this paper, the phrase ‘free subgroup’ means ‘nonabelian free subgroup’. The idea of the proof of the Main Theorem comes from [1]. We can sketch the proof as follows. Assume that N is a subnormal subgroup of $D^*$ containing a nonabelian solvable subgroup. First, we will study the existence of free subgroups in subnormal subgroups of the division ring of fractions $K(x,\sigma )$ of the skew polynomial ring $K[x,\sigma ]$ in a single-variable x twisted by a nonidentity automorphism $\sigma $ over a field K. Namely, we will prove that every subnormal subgroup of $K(x,\sigma )^*$ containing x contains a free subgroup. After that, we shall build a subring of D which is isomorphic to $K[x,\sigma ]$ , and based on the result on $K(x,\sigma )$ , the Main Theorem will follow.

Some special cases for the existence of free subgroups have also been considered. For example, the following conjecture was proposed by Hai and Thin [8].

Conjecture 1.3 [8, Conjecture 1].

Every locally solvable subnormal subgroup of $D^*$ is central.

In [7, Theorem 2.2], Hai and Thin proved that Conjecture 1.3 is true in the case of a locally nilpotent subnormal subgroup in an arbitrary division ring D and, in [7, Theorem 2.4], for the case of a division ring which is algebraic over its centre. After that, in [8], Conjecture 1.3 holds if D is a weakly locally finite division ring, that is, D satisfies the condition: for every finite subset S of D, the division subring $P(S)$ generated by S and a prime subfield P of D is finite-dimensional over the centre $\mathrm {Z}(P(S))$ . Recently, Conjecture 1.3 was fully confirmed in [3, Theorem 1]. By applying the Main Theorem, we can also show that Conjecture 1.3 is true.

2 Proof of the main theorem

The fundamental result used in proving the Main Theorem is Theorem 2.6. Our approach to the proof of Theorem 2.6 comes from [1]. However, thanks to insights from [2], our proof is more straightforward than that in [1].

Let K be a field and $\sigma $ be a field automorphism of K. Let us denote by $K[x,\sigma ]$ the skew polynomial ring in an indeterminate x twisted by $\sigma $ over K. Then $K[x,\sigma ]$ is a principal left ideal domain and also a principal right ideal domain, that is, every one-sided ideal of $K[x,\sigma ]$ can be generated by one element (see [6, Theorem 2.8]). The division ring of fractions of $K[x,\sigma ]$ is denoted by $K(x,\sigma )$ . Every element of $K(x,\sigma )$ is of the form $f(x)\cdot g(x)^{-1}$ , where $f(x)$ and $g(x)$ are in $K[x,\sigma ]$ . Additionally, we denote by $K((x,\sigma ))$ the skew Laurent series ring in the indeterminate x twisted by $\sigma $ over K. It is well known that $K((x,\sigma ))$ is a division ring (see [9, Example 1.8]). Clearly, $K[x,\sigma ]$ can be embedded into $K((x,\sigma ))$ , and so $K(x,\sigma )$ can be seen as a division subring of $K((x,\sigma ))$ .

Let D be a division ring with the centre F. We can regard D as an F-algebra. If $\mathrm {dim}_F(D)<\infty $ , then D is called centrally finite. From [9, Proposition 14.2], $K((x,\sigma ))$ is centrally finite if and only if the order of $\sigma $ is finite. As a corollary, if the order of $\sigma $ is finite, then $K(x,\sigma )$ is centrally finite. We give some results about the existence of free subgroups in $K(x,\sigma )$ . We need the following property.

Lemma 2.1 [4, Theorem 2.1].

Let D be a noncommutative centrally finite division ring and N a noncentral subnormal subgroup of $D^*$ . Then N contains a free subgroup.

We immediately have the following result.

Corollary 2.2. Let $K(x,\sigma )$ be the division ring of fractions of $K[x,\sigma ]$ and N a noncentral subnormal subgroup of $K(x,\sigma )^*$ . If $\sigma $ has a finite order, then N contains a free subgroup.

For an arbitrary automorphism $\sigma $ of K, whether a subnormal subgroup N of $K(x,\sigma )^*$ contains a free subgroup is an open question. In this section, we shall answer the question in case $\sigma $ is a nonidentity and $x\in N$ .

Lemma 2.3 [2].

Let K be a field and $\sigma $ a nonidentity automorphism of K with fixed subfield $k=\{a\in K:\sigma (a)=a\}$ . Assume that there exists an element $a\in K\setminus k$ such that $k(a,\sigma (a),\sigma ^2(a),\ldots )$ is not a finitely generated extension of k. If N is a subnormal subgroup of $K(x,\sigma )^*$ containing x, then N contains a free subgroup.

Proof. This is the assertion (i) of [2, Theorem 2.5].

With the notation as in Lemma 2.3, we shall now consider the remaining case, that is, $k(a,\sigma (a),\sigma ^2(a),\ldots )$ is a finitely generated extension of k. The next lemma is the key lemma to resolve this case.

Lemma 2.4. Let K be a field and $\sigma $ an automorphism of K with fixed subfield $k=\{a\in K: \sigma (a)=a\}$ . Assume that there exists an element $a\in K\setminus k$ such that $E=k(a,\sigma (a), \sigma ^2(a),\ldots )$ is a finitely generated extension of k. Then, the following hold.

1. (i) $E=k(\sigma ^{n_1}(a), \sigma ^{n_2}(a),\ldots , \sigma ^{n_r}(a))$ , where the $n_i$ are integers with $0\leq n_1<n_2<\cdots <n_r$ .

2. (ii) The restriction of $\sigma $ to the subfield E is an automorphism of E of finite order.

Proof. (i) Since $E/k$ is a finitely generated extension, $E=k(a_1,a_2,\ldots ,a_s)$ for some $a_1, a_2, \ldots , a_s$ in E. For each $i\in \overline {1,s}$ , there exist $f_i$ and $g_i$ in $k[a,\sigma (a), \sigma ^2(a),\ldots ]$ such that $a_i=f_ig_i^{-1}$ . We can select a finite number of elements $\sigma ^{n_{i,1}}(a), \sigma ^{n_{i,2}}(a),\ldots ,\sigma ^{n_{i,\ell _i}}(a)$ such that $f_i$ and $g_i$ are in $k[\sigma ^{n_{i,1}}(a), \sigma ^{n_{i,2}}(a),\ldots ,\sigma ^{n_{i,\ell _i}}(a)]$ . It follows that $a_i\in k(\sigma ^{n_{i,1}}(a), \sigma ^{n_{i,2}}(a),\ldots ,\sigma ^{n_{i,\ell _i}}(a))$ . We have

$$ \begin{align*}E=k(a_i\mid 1\leq i\leq s)\subseteq k(\sigma^{n_{i,j}}(a)\mid 1\leq i\leq s, 1\leq j\leq \ell_i)\subseteq k(a,\sigma(a), \sigma^2(a),\ldots)=E.\end{align*} $$

This leads to $E=k(\sigma ^{n_{i,j}}(a)\mid 1\leq i\leq s, 1\leq j\leq \ell _i)$ . Hence, we can write $E=k(\sigma ^{n_1}(a), \sigma ^{n_2}(a),\ldots , \sigma ^{n_r}(a))$ for $0\leq n_1<n_2<\cdots <n_r$ .

(ii) We have $\sigma (E)\subseteq E$ because $\sigma (\sigma ^{n}(a))=\sigma ^{n+1}(a)\in E$ . From assertion (i), $E=k(\sigma ^{n_1}(a), \sigma ^{n_2}(a),\ldots , \sigma ^{n_r}(a))$ for $0\leq n_1<n_2<\cdots <n_r$ . Put $t_i=\sigma ^{n_i}(a)$ , and so $E=k(t_1,t_2,\ldots ,t_r)$ . Remark that

$$ \begin{align*}\sigma^{n_i-n_j}(t_j)=\sigma^{n_i-n_j}(\sigma^{n_j}(a))=\sigma^{n_i}(a)=t_i \quad \text{for } 1\leq i,j\leq r.\end{align*} $$

Put $n=(n_1-n_2)(n_2-n_3)\ldots (n_{r-1}-n_r)(n_r-n_1)$ . For each $i=\overline {1,r}$ ,

$$ \begin{align*}n=(n_i-n_{i+1})(n_{i+1}-n_{i+2})\ldots(n_{r-1}-n_r)(n_r-n_1)(n_1-n_2)\ldots(n_{i-1}-n_{i}),\end{align*} $$

and so

$$ \begin{align*} \sigma^n(t_i) & = \sigma^{(n_i-n_{i+1})(n_{i+1}-n_{i+2})\cdots(n_{r-1}-n_r)(n_r-n_1) (n_1-n_2)\cdots(n_{i-1}-n_{i})}(t_i)\\ & =\sigma^{n_i-n_{i+1}}(\sigma^{n_{i+1}-n_{i+2}}(\cdots(\sigma^{n_{r-1}-n_r}(\sigma^{n_r-n_1}(\sigma^{n_1-n_2}(\ldots(\sigma^{n_{i-1}-n_{i}}(t_i)))\cdots{\kern-2pt}) =t_i. \end{align*} $$

Thus, the restriction $\sigma |_{E}$ has finite order. Suppose that the order of $\sigma |_{E}$ is $d>0$ . Since $\sigma (E)\subseteq E$ , we have

$$ \begin{align*}E=\sigma^d(E)\subseteq \sigma^{d-1}(E)\subseteq \cdots \subseteq\sigma^{1}(E).\end{align*} $$

Hence, $\sigma |_{E}$ is an automorphism of E.

Lemma 2.4 enables us to prove the next lemma.

Lemma 2.5. Let K be a field and $\sigma $ a nonidentity automorphism of K with fixed subfield $k=\{a\in K: \sigma (a)=a\}$ . Assume that there exists $a\in K\setminus k$ such that $E=k(a,\sigma (a), \sigma ^2(a),\ldots )$ is a finitely generated extension of k. If N is a subnormal subgroup of $K(x,\sigma )^*$ containing x, then N contains a free subgroup.

Proof. By Lemma 2.4, the restriction $\sigma |_{E}$ is an automorphism of E of finite order. Since $\sigma |_{E}(a)=\sigma (a)\neq a$ , the automorphism $\sigma |_{E}$ is a nonidentity. Thus, $R=E(x,\sigma |_{E})$ is a noncommutative centrally finite division subring of $K(x,\sigma )$ . Because N is subnormal in $K(x,\sigma )^*$ , also $H=N\cap R^*$ is subnormal in $R^*$ . However, H is noncentral since $x\in H$ . According to Lemma 2.2, H contains a nonabelian free subgroup. Hence, N contains a free subgroup.

Theorem 2.6. Let K be a field and $\sigma $ a nonidentity automorphism of K. Assume that N is a subnormal subgroup of $K(x,\sigma )^*$ . If $x\in N$ , then N contains a free subgroup.

Proof. Let $D=K(x,\sigma )$ , where K is a field and $\sigma $ is a nonidentity automorphism of K with fixed subfield $k=\{a\in K:\sigma (a)=a\}$ . Select $a\in K\setminus k$ . Put $E=k(a,\sigma (a), \sigma ^2(a),\ldots )$ , $R=E(x,\sigma |_{E})$ and $H=N\cap E^*$ . Then H is a subnormal subgroup of $R^*$ containing x. It follows from Lemma 2.5 that if $E/k$ is a finitely generated extension, then H contains a free subgroup. Otherwise, by Lemma 2.3, H also contains a free subgroup. Hence, in both cases, N contains a free subgroup. The proof is complete.

Proof of the Main Theorem.

Assume that N is a subnormal subgroup of $D^*$ containing a nonabelian solvable subgroup G. Suppose that the derived length of G is $r\geq 1$ with the derived series

$$ \begin{align*}\langle 1\rangle=G^{(r)} \trianglelefteq G^{(r-1)} \trianglelefteq \ldots\trianglelefteq G^{(1)} \trianglelefteq G^{(0)}=G.\end{align*} $$

Clearly, $G^{(r-2)}$ is a nonabelian solvable subgroup. Replace G by $G^{(r-2)}$ . Then $G'$ is a nontrivial abelian subgroup of G. By Zorn’s lemma, the set

$$ \begin{align*}\mathcal{U}=\{A\leq G: A \text{ is abelian and } G'\subseteq A\}\end{align*} $$

has a maximal element M. Due to the maximality, $M\not \subseteq \mathrm {Z}(G)$ . Additionally, since $G'\subseteq M$ , M is normal in G. Select $g\in G$ and $u\in M$ such that $ug\neq gu$ . The normality of M in G leads to $u_n:=g^{n}ug^{-n}\in M$ for every $n\in \mathbb {Z}$ , and so all these elements commute. Put $F=\mathrm {Z}(D)$ . Then the division subring $K=F(u_n\mid n\in \mathbb {Z})$ is a field and the map $\sigma :K\to K$ , $z\mapsto gzg^{-1}$ is a nonidentity automorphism of K.

Let $R=K[g]$ be the subring of D generated by $K\cup \{g\}$ and $K[x,\sigma ]$ the ring of skew polynomials in an indeterminate x. We observe that the map $\phi :K[x,\sigma ] \to R$ , $\sum \nolimits _{i}a_ix^i\mapsto \sum \nolimits _{i}a_ig^i$ is a surjective F-algebra homomorphism. Put $I=\mathrm {ker}\phi $ . We consider two cases.

Case 1: $I=(0)$ . Then $R\cong K[x,\sigma ]$ , and thus the division ring of fractions $\overline {R}$ of R is isomorphic to $K(x,\sigma )$ . The image of $N\cap \overline {R}^*$ is a subnormal subgroup of $K(x,\sigma )^*$ and contains x. By Theorem 2.6, this subgroup contains a free subgroup. Consequently, $N\cap \overline {R}^*$ and also N contain a free subgroup.

Case 2: $I\neq (0)$ . Then $R\cong K[x,\sigma ]/I$ . This means $K[x,\sigma ]/I$ is a domain and I is a prime ideal. Additionally, $K[x,\sigma ]$ is a principal left ideal domain. This implies that $I=(p)$ for some irreducible polynomial p in $K[x,\sigma ]$ , that is, I is a maximal ideal of $K[x,\sigma ]$ , and so $K[x,\sigma ]/I$ is a division ring. However, by the divide algorithm for $K[x,\sigma ]$ (see [12, The Euclidean algorithm]), $K[x,\sigma ]/I$ is a finite dimensional K-vector space. Therefore, R is a centrally finite division subring. By Lemma 2.1, $N\cap R^*$ contains a free subgroup. Hence, N also contains a free subgroup.

Finally, the Main Theorem gives the affirmation of Conjecture 1.3 as follows.

Corollary 2.7. Let D be a division ring. Then every locally solvable subnormal subgroup of $D^*$ is central.

Proof. Let D be a division ring and N be a locally solvable subnormal subgroup of $D^*$ . Assume that N is noncentral. Then N contains a finitely generated subgroup that is nonabelian and solvable. It follows from the Main Theorem that N contains a free subgroup. This is a contradiction to the local solvability of N. Hence, N is central.