LOCALLY NILPOTENT SUBGROUPS OF $\mathrm {GL}_n(D)$

Abstract

Let A be an F-central simple algebra of degree $m^2=\prod _{i=1}^k p_i^{2\alpha _i}$ and G be a subgroup of the unit group of A such that $F[G]=A$. We prove that if G is a central product of two of its subgroups M and N, then $F[M]\otimes _F F[N]\cong F[G]$. Also, if G is locally nilpotent, then G is a central product of subgroups $H_i$, where $[F[H_i]:F]=p_i^{2\alpha _i}$, $A=F[G]\cong F[H_1]\otimes _F \cdots \otimes _F F[H_k]$ and $H_i/Z(G)$ is the Sylow $p_i$-subgroup of $G/Z(G)$ for each i with $1\leq i\leq k$. Additionally, there is an element of order $p_i$ in F for each i with $1\leq i\leq k$.

1 Introduction

The multiplicative group of a noncommutative division ring has been investigated in various papers by Amitsur [3], Herstein [13, 14], Hua [15, 16], Huzurbazar [17] and Scott [23, 24]. Given a noncommutative division ring D with centre $Z(D) = F$ , the structure of the skew linear group $\mathrm {GL}_n(D)$ for $n \geq 1$ is generally unknown. A good account of the most important results concerning skew linear groups can be found in [25], as well as in [26] particularly for linear groups. For instance, it is shown in [12] that there is a close connection between the question of the existence of maximal subgroups in the multiplicative group of a finite-dimensional division algebra and Albert’s conjecture concerning the cyclicity of division algebras of prime degree. In this direction, in [20], it is shown that when D is a central division F-algebra of prime degree p, then D is cyclic if and only if $D^*$ contains a nonabelian soluble subgroup. Furthermore, a theorem of Albert (see [6, page 87]) asserts that D is cyclic if $D^*/F^*$ contains an element of order p.

The structure of locally nilpotent subgroups of $\mathrm {GL}_n(D)$ is studied in many papers. The basic structure of locally nilpotent skew linear groups over a locally finite-dimensional division algebra was studied by Zaleeskii [30]. One important problem raised by Zaleeskii remains open, namely, is every locally nilpotent subgroup of $\mathrm {GL}_n(D)$ hypercentral. In [10], Garascuk proved a theorem that shows this question has a positive answer in the case where $[D:F]< \infty $ . A treatment of such results which is both more elaborate and more refined may be found in [4, 25–29]. For example, it is shown in [29] that when H is a locally nilpotent normal subgroup of the absolutely irreducible skew linear group G, then H is centre-by-locally-finite and $G/C_G(H)$ is periodic. In special cases, the structure of maximal subgroups of $\mathrm {GL}_n(D)$ has been investigated (see [1, 2, 5, 7, 9]). For instance, it is shown in [1] that when D is a finite-dimensional division ring with infinite centre F and M is a locally nilpotent maximal subgroup of $\mathrm {GL}_n(D)$ , then M is an abelian group. Also, by [25, Theorem 3.3.8], when D is an F-central locally finite-dimensional division algebra, every locally nilpotent subgroup of $\mathrm {GL}_n(D)$ is soluble.

Another important property of locally nilpotent subgroups arises in crossed product constructions. Let R be a ring, S a subring of R and G a group of units of R normalising S such that $R=S[G]$ . Suppose that $N=S\cap G$ is a normal subgroup of G and $R=\oplus _{t \in T} tS$ , where T is some transversal of N to G. Set $H=G/N$ . We summarise this construction by saying that $(R, S, G, H)$ is a crossed product. Sometimes, we say that R is a crossed product of S by H. Let $\mathcal {O}$ be the class of all groups H such that every crossed product of a division ring by H is an Ore domain. In [25, Remark 1.4.4], it is shown that the group ring $EG$ is an Ore domain for any division ring E and any torsion-free locally nilpotent group G. In addition, any hyper torsion-free locally nilpotent group is in $\mathcal {O}$ .

Let D be an F-central division algebra and G a subgroup of $\mathrm {GL}_n(D)$ . The F-algebra of G, that is, the F-subalgebra generated by elements of G over F in $M_n(D)$ is denoted by $F[G]$ . Further, G is absolutely irreducible if $F[G]=M_n(D)$ . When $M_n(D)$ is a crossed product over a maximal subfield K, from [6, page 92], $K/F$ is Galois and we can write $M_n(D)=\oplus _{\sigma \in \mathrm {Gal}(K/F)}Ke_{\sigma }$ , where $e_{\sigma }\in \mathrm {GL}_n(D)$ and for each $x\in K$ and $\sigma \in \mathrm {Gal}(K/F)$ , there exists $\sigma (x)\in K$ such that $e_{\sigma }x=\sigma (x)e_{\sigma }$ . Several recent papers investigate the group theoretical properties which give useful tools to realise maximal Galois subfields of central simple algebras in terms of absolutely irreducible subgroups (see [1, 8, 9, 11, 18–20]).

We say a group G is a central product of two of its subgroups M and N if $G=MN$ and $M\subseteq C_G(N)$ . In fact, a central product of two groups is a quotient group of $M\times N$ . If F is a field and $FG$ denotes the group algebra of G, then it is well known that $FM\otimes _F FN\cong F(M\times N)$ . We prove a similar result for skew linear groups. Let A be an F-central simple algebra of degree $n^2=\prod _{i=1}^k p_i^{2\alpha _i}$ and G be a subgroup of the unit group of A such that $F[G]=A$ . We prove that if G is a central product of two of its subgroups M and N, then $F[M]\otimes _F F[N]\cong F[G]$ . Also, if G is locally nilpotent, then G is a central product of subgroups $H_i$ , where $[F[H_i]:F]=p_i^{2\alpha _i}$ , $A=F[G]\cong F[H_1]\otimes _F \cdots \otimes _F F[H_k]$ and $H_i/Z(G)$ is the Sylow $p_i$ -subgroup of $G/Z(G)$ for $1\leq i\leq k$ . Additionally, there is an element of order $p_i$ in F for $1\leq i\leq k$ .

2 Notation and conventions

We recall here some of the notation that we will need throughout this article. Given a subset S and a subring K of a ring R, the subring generated by K and S is denoted by $K[S].$ The unit group of R is written as $R^*$ . For a group G and subset $S\subset G$ , we denote by $Z(G)$ and $C_G(S)$ the centre and the centraliser of S in G and the same notation is applied for R. We use $N_G(S)$ for the normaliser of S in G and $G'$ for the derived subgroup of G. A group G is a central product of its subgroups $H_1,\ldots ,H_k$ if $G=H_1\cdots H_k$ and $H_i\subseteq C_G(H_j)$ for each $i\neq j$ .

Let F be a field, and A and B be two unital F-algebras. Let H be a subgroup of $A^*$ and G be a subgroup of $B^*$ . We define $H\otimes _F G$ by

$$ \begin{align*}H\otimes_F G=\{a\otimes b \mid a\in H, b\in G\}.\end{align*} $$

Note that $(a\otimes b)^{-1}=a^{-1}\otimes b^{-1}$ , so it is easily checked that $H\otimes _F G$ is a subgroup of $(A\otimes B)^*$ . Also, $F[H]\otimes _F F[G]=F[H\otimes _F G]$ in $A\otimes _F B$ .

Given a division ring D with centre F and a subgroup G of $\mathrm {GL}_n(D)$ , the space of column n-vectors $V=D^n$ over D is a $G\text {--}D$ bimodule; G is called irreducible, completely reducible or reducible according to whether V is irreducible, completely reducible or reducible as a $G\text {--}D$ bimodule.

An irreducible group G is said to be imprimitive if for some integer $m\geq 2,$ there exist subspaces $V_1,\ldots ,V_m$ of V such that $V=\oplus _{i=1}^m V_i$ and for any $g\in G$ , the mapping $V_i\rightarrow gV_i$ is a permutation of the set $\{V_1,\ldots ,V_m\}$ ; otherwise, G is called primitive.

The following important results on central simple algebras will be used later.

Theorem 2.1 (Double centraliser theorem; [6, page 43]).

Let $B\subseteq A$ be simple rings such that $K:=Z(A)=Z(B)$ . Then, $A\cong B\otimes _K C_A(B)$ whenever $[B:K]$ is finite.

Theorem 2.2 (Centraliser theorem; [6, page 42]).

Let B be a simple subring of a simple ring A, $ K:=Z(A)\subseteq Z(B)$ and $n:=[B:K]$ be finite. Then:

1. (1) $C_A(B)\otimes _K M_n(K) \cong A \otimes _K B^{\mathrm {op}}$ ;

2. (2) $C_A(B)$ is a simple ring;

3. (3) $Z(C_A(B))=Z(B)$ ;

4. (4) $C_A(C_A(B))=B$ ;

5. (5) if $L:=Z(B)$ and $r:=[L:K]$ , then $A \otimes _K L \cong M_r(B) \otimes _L C_A(B) $ ;

6. (6) A is a free left (right) $C_A(B)$ -module of unique rank n;

7. (7) if, in addition to the above assumptions, $m:=[A:K]$ is also finite, then A is a free left (right) B-module of unique rank $m/n=[C_A(B):K]$ .

Theorem 2.3 [6, page 30].

Let $A,B$ be K-algebras, $K:=Z(A)\subseteq Z(B)$ a field and either $[A:K]$ or $[B:K]$ finite. Then, $A\otimes _K B$ is a simple ring if and only if A and B are simple rings.

3 Central products of skew linear groups and tensor products of central simple algebras

In this section, we prove a theorem which relates a central decomposition of an absolutely irreducible group G to the tensor product decomposition of $F[G]$ .

It is well known that every finite dimensional division algebra is isomorphic to a tensor product of division algebras of prime power degree [6, page 68]. Since each central simple algebra is isomorphic to some $M_n(D)$ , we easily obtain the following result.

Lemma 3.1. Let A be an F-central simple algebra of degree $m^2=\prod _{i=1}^k p_i^{2\alpha _i}$ . Then, $A\cong A_1\otimes _F\cdots \otimes _F A_k$ , where $A_i$ is a unique (up to isomorphism) F-central simple algebra of degree $p_i^{2\alpha _i}$ .

Additionally, we have the following easy lemma.

Lemma 3.2. Let $A,B$ be two F-central simple algebras, and $M\leq A^*$ and $N\leq B^*$ . Then, M and N are absolutely irreducible if and only if $M\otimes _F N$ is an absolutely irreducible subgroup of $A\otimes _F B$ .

Lemma 3.3. Let F be a field, $A,B$ be two unital F-algebras and $a\in A,b\in B$ . Then, $a\otimes b=1\otimes 1$ if and only if $a,b \in F$ and $ab=1$ .

Proof. First, if $a,b \in F$ and $ab=1$ , then $a\otimes b=ab\otimes 1=1\otimes 1$ .

Conversely, assume $a\otimes b=1\otimes 1$ . It is clear that $a\neq 0$ and $b\neq 0$ . First, assume that $a,b \notin F^*$ . Then, $\{1,a\}$ is an F-linearly independent set in A and $\{1,b\}$ is an F-linearly independent set in B. By [6, Theorem 4.3], $\{a\otimes b,1\otimes 1\}$ is an F-linearly independent set in $A\otimes _F B$ . Therefore, $a\otimes b \neq 1\otimes 1$ . Next, assume that $a \notin F^*$ and $ b\in F^*$ . Then, $ab\notin F^*$ and $\{1,ab\}$ is an F-linearly independent set in B. Thus, $\{1\otimes ab,1\otimes 1\}$ is an F-linearly independent set in $A\otimes _F B$ and $a\otimes b=1\otimes ab\neq 1\otimes 1$ . When $b \notin F^*$ and $ a\in F^*$ , the proof is similar. We conclude that if $a\otimes b=1\otimes 1$ , then $a,b \in F^*$ . Now, we have $1\otimes 1=a\otimes b=ab\otimes 1=ab(1\otimes 1)$ . Consequently, $ab=1$ , as we desired.

The following result shows that any absolutely irreducible skew linear group can be viewed as an absolutely irreducible linear group.

Proposition 3.4. Let F be a field and D be a finite dimensional F-central division algebra such that $[D:F]=n^2$ . Let K be a maximal subfield of D and G be an absolutely irreducible subgroup of $\mathrm {GL}_m(D)$ . Then, $M_m(D)\otimes K\cong M_{mn}(K)$ and $G\otimes _F 1$ is an absolutely irreducible subgroup of $U(M_m(D)\otimes _F K)\cong \mathrm {GL}_{nm}(K)$ isomorphic to G.

Proof. By [21, Propositions 13.5 and 13.3], there exists a maximal subfield K of D such that $[D:K]=[K:F]=n$ and $D\otimes _F K\cong M_n(K)$ . Therefore, $M_m(D)\otimes _F K\cong M_m(F)\otimes _F(D\otimes _F K)\cong (M_m(F)\otimes _F M_n(F))\otimes _F K \cong M_{mn}(K)$ . Now, by Lemma 3.3, the map $\phi : G\rightarrow G\otimes _F 1$ given by $\phi (g)=g\otimes 1$ is an isomorphism. However, G is an absolutely irreducible subgroup of $\mathrm {GL}_m(D)$ , so $F[G]=M_m(D)$ . Also, $M_m(D)\otimes _F K=F[G]\otimes _F K=K[G\otimes _F K^*]\subseteq K[G\otimes _F 1]\subseteq M_m(D)\otimes _F K$ . Consequently, $K[G\otimes _F 1]=M_m(D)\otimes _F K$ . This means $G\otimes _F 1$ is an absolutely irreducible subgroup of $\mathrm {GL}_m(D)\otimes _F K^*$ isomorphic to G. In addition, G is isomorphic to an absolutely irreducible subgroup of $\mathrm {GL}_{nm}(K)$ .

Corollary 3.5. Let F be a field and D be a finite dimensional F-central division algebra. Assume that G is a subgroup of $\mathrm {GL}_m(D)$ such that $F[G]$ is a simple ring. Then, there exists an absolutely irreducible linear group H isomorphic to G.

Theorem 3.6 [25, page 7].

Let F be a field, D a locally finite-dimensional division F-algebra and G a subgroup of $\mathrm {GL}_n(D)$ . Set $R=F[G]\subseteq M_n(D)$ .

1. (1) If G is completely reducible, then R is semisimple Artinian.

2. (2) If G is irreducible, then R is simple Artinian.

Using Theorem 3.6, we obtain the following result.

Corollary 3.7. Let F be a field and D be a finite dimensional F-central division algebra. If G is an irreducible subgroup of $\mathrm {GL}_m(D)$ , then there exists an absolutely irreducible linear group H isomorphic to G.

When F is a field, a subgroup G of $\mathrm {GL}_n(F)$ is said to be absolutely irreducible if it is an irreducible subgroup of $\mathrm {GL}_n(K)$ for any extension K of F. Hence, we obtain the following result.

Corollary 3.8. Let F be a field and D be a finite dimensional F-central division algebra. If G is an irreducible subgroup of $\mathrm {GL}_m(D)$ such that either G is irreducible or $F[G]$ is a simple ring, then there exists an algebraically closed field $\Omega $ and an irreducible $\Omega $ -linear group H isomorphic to G.

Theorem 3.9 [25, page 8].

Let F be a field, D a division F-algebra and G a subgroup of $\mathrm {GL}_n(D)$ . Set $R=F[G]\subseteq M_n(D)$ .

1. (1) If R is semiprime (for example, if R is semisimple Artinian), then G is isomorphic to a completely reducible subgroup of $\mathrm {GL}_n(D)$ .

2. (2) If R is simple Artinian, then for some $m\leq n$ , the group G is isomorphic to an irreducible subgroup of $\mathrm {GL}_m(D)$ .

Using Theorem 3.9, we obtain the following result.

Corollary 3.10. Let F be a field and D be a finite dimensional F-central division algebra such that $[D:F]=n^2$ . Let $A=M_m(D)\subseteq M_{n^2m}(F)=B$ be an F-central simple algebra. If G is a subgroup of $\mathrm {GL}_m(D)$ such that either G is irreducible or $F[G]$ is a simple ring, then for some $s\leq mn^2$ , the group G is isomorphic to an irreducible subgroup of $\mathrm {GL}_{s}(F)$ .

Theorem 3.11 [26, page 111].

Let V be a finite dimensional linear space over a division ring D and G an irreducible subgroup of $\mathrm {GL}(V)$ which can be represented in the form $G=HF$ , where H and F are elementwise permutable normal subgroups of G. Then, the irreducible components of $H(F)$ are pairwise equivalent.

Proposition 3.12. Let F be a field and D be a finite dimensional F-central division algebra. Assume that G is an absolutely irreducible subgroup of $\mathrm {GL}_n(D)$ . If $G=MN$ is a central product decomposition of G, then $F[M]\otimes _F F[N]\cong F[G]$ and under this isomorphism, $M\otimes _F N\cong G$ . Additionally, $F[M]$ and $F[N]$ are F-central division algebras.

Proof. By [25, Theorem 1.2.1], G is irreducible. Using [25, Theorem 1.1.7] and Theorem 3.11, we conclude that M is a homogeneous completely irreducible subgroup. So Theorem 3.11 implies $D^n\cong V^m$ , where V is an irreducible $M-D$ bimodule. Hence, $F[N]\subseteq A=C_{M_n(D)}(M)=\mathrm {End}_{M-D}(D^n)\cong M_m(E)$ , where $E=\mathrm {End}_{M-D}(V)$ is a division ring by Schur’s lemma. Note that $F[N]\otimes F[M]\leq A\otimes _F C_{M_n(D)}(A)$ . Hence, by the centraliser theorem, $[F[M]:F]FN]:F]\leq [A:F][C_{M_n(D)}(A):F]=n^2[D:F]$ . Furthermore, $F[M],F[N]\subseteq F[G]$ implies that there is a surjective homomorphism f from $F[N]\otimes _F F[M]$ onto $F[G]{\kern-1pt}={\kern-1pt}M_n(D)$ such that $f(a\otimes b){\kern-1pt}={\kern-1pt}ab$ for each $a{\kern-1pt}\in{\kern-1pt} M,b{\kern-1pt}\in{\kern-1pt} N$ . So $F[M]\otimes _F F[N]\cong F[G]$ by dimension counting. It is clear that $\overline {f}$ , the restriction of f to $M\otimes _F N$ , is a surjective homomorphism on G. If $\overline {f}(a\otimes b)=ab=1$ , then $a=b^{-1}\in M\cap N\subseteq Z(G)\subseteq F$ . Hence, $a\otimes b=b^{-1}\otimes b=1\otimes b^{-1}b=1\otimes 1$ . So, $\ker (\overline {f})$ is trivial and $\overline {f}$ is an isomorphism from $M\otimes _F N$ to G. Consequently, $F[M]$ and $F[N]$ are F-central division algebras by Theorem 2.3.

The following example shows that the above result is not true in semisimple rings.

Example 3.13. Let $A=F\times F$ , $G=\{(1,1),(1,-1),(-1,1),(-1,-1)\}$ , $M=\{(1,1), (1,-1)\},N=\{(1,1),(-1,1)\}$ . Then, G is a central product of M and N. However, $ [F[M]\otimes _F F[N]:F]=4$ . So $F[M]\otimes _F F[N]\ncong F[G]=A$ .

Next we introduce some notation from [26]. Let V be a finite dimensional linear space over a division ring D and G a completely irreducible subgroup of $\mathrm {GL}(V)$ . Let $D^n=V=L_1\oplus \cdots \oplus L_r $ and suppose that $L_i$ is a G-invariant G-irreducible subspace of V for $1\leq i\leq r$ . We determine the irreducible components of G, that is, the irreducible representations $d_i$ of the form

$$ \begin{align*}d_i:G\rightarrow \text{GL}(L_i), \quad g\rightarrow g\mid L_i, \quad i=1, \ldots, r.\end{align*} $$

By [26, Lemma 13.1], the irreducible components $d_i$ and $d_j$ of G are equivalent if and only if there exists a module isomorphism $\Psi : L_i\rightarrow L_j$ such that for any $y \in G$ ,

$$ \begin{align*}d_j(y)=\Psi d_1(y)\Psi^{-1}.\end{align*} $$

In addition, these representations are equivalent if and only if the modules $L_i$ and $L_j$ have respective bases $B_1 $ and $B_2$ such that for any $y\in G$ , the matrix of the endomorphism $d_i(y)$ in $B_1$ is the same as that of $d_j(y)$ in $B_2$ . This observation gives the following result.

Lemma 3.14. Let G be a completely irreducible subgroup of $\mathrm {GL}_n(D)$ such that the irreducible components of G are pairwise equivalent. Let r be the degree of an irreducible component of G and $n=rs$ . Then, there is an isomorphism f with $f:M_n(D)\longrightarrow M_r(D)\otimes _F M_s(F)$ and an irreducible subgroup H of $\mathrm {GL}_r(D)$ such that $f(G)=H\otimes \{1\}$ .

4 Locally nilpotent subgroups of $\textrm {GL}_n(D)$

In this section, we prove that every absolutely irreducible locally nilpotent subgroup of $\mathrm {GL}_n(D)$ is a central product of some of its subgroups which gives a decomposition of $M_n(D)$ as a tensor product of central simple algebras of prime power degree. First, we recall the following general results which play a key role in proving our main theorems.

Theorem 4.1 [26, page 216].

Let F be an arbitrary field and G be an absolutely irreducible locally nilpotent subgroup of $\mathrm {GL}_n(F)$ . Then, $G/Z(G) $ is periodic and $\pi (G/Z(G))=\pi (n)$ .

Theorem 4.2 [29].

Let H be a locally nilpotent normal subgroup of the absolutely irreducible skew linear group G. Then, H is centre-by-locally finite and $G/C_G(H)$ is periodic.

Theorem 4.3 [22, page 342].

Let G be a locally nilpotent group. Then, the elements of finite order in G form a fully invariant subgroup T (the torsion subgroup of G) such that $G/T$ is torsion and T is a direct product of p-groups.

Theorem 4.4 [5].

Let N be a normal subgroup in a primitive subgroup M of $\mathrm {GL}_n(D)$ . Then:

1. (1) $F[N]$ is a prime ring;

2. (2) $C_{M_n(D)}(N)$ is a simple Artinian ring;

3. (3) if $C_{M_n(D)}(N)$ is a division ring, then N is irreducible.

Theorem 4.5 [18].

Let D be a finite dimensional F-central division algebra. Then, $M_m(D)$ is a crossed product over a maximal subfield if and only if there exists an absolutely irreducible subgroup G of $M_m(D)$ and a normal abelian subgroup A of G such that $C_G(A)=A$ and $F[A]$ contains no zero divisor.

Theorem 4.6. Let $A=M_n(D)$ be an F-central simple algebra of degree $m^2=\prod _{i=1}^k p_i^{2\alpha _i}$ and G be an absolutely irreducible locally nilpotent subgroup $A^*$ . Then:

1. (1) $G/Z(G) $ is locally finite and $\pi (G/Z(G))=\pi (m)$ ;

2. (2) $G/Z(G) $ is a p-group for some prime p if and only if m is a pth power.

Proof. (1) By Theorem 4.2, G is centre-by-locally finite. Let K be a maximal subfield of D. By Proposition 3.4, G is isomorphic to an absolutely irreducible subgroup of $\mathrm {GL}_{m}(K)$ . Now, Theorem 4.1 asserts that $\pi (G/Z(G))=\pi (m)$ .

(2) This statement is clear from item (1).

Corollary 4.7. Let $A=M_n(D)$ be an F-central simple algebra of degree $m^2=\prod _{i=1}^k p_i^{2\alpha _i}$ and G be an absolutely irreducible locally nilpotent subgroup of $A^*$ . Then:

1. (1) $G/Z(G) $ is locally finite and $\pi (G/Z(G))=\pi (m^2/[C_{M_n(D)}:F])\subseteq \pi (m)$ ;

2. (2) if $G/Z(G) $ is a p-group for some prime p, then $[F[G]:F]$ is a pth power;

3. (3) if m is a pth power for some prime p, then $G/Z(G) $ is a p-group.

Proof. By Theorem 3.6, $F[G]$ is a simple ring. From the centraliser theorem, $[F[G]:F][C_{M_n(D)}:F]=m^2$ . The reminder of the proof is similar to the proof of Theorem 4.6.

Now we are ready to prove the main theorem of this article.

Theorem 4.8. Let $A=M_n(D)$ be an F-central simple algebra of degree $m^2=\prod _{i=1}^k p_i^{2\alpha _i}$ and G be an absolutely irreducible locally nilpotent subgroup $A^*$ . Then:

1. (1) $G/Z(G)$ is the internal direct product of ${H_1}/{Z(G)},\ldots , {H_k}/{Z(G)}$ , where $H_i/Z(G)$ is the Sylow $p_i$ -subgroup of $G/Z(G)$ ;

2. (2) G is the central product of $H_1,\ldots ,H_k$ ;

3. (3) $A=F[G]\cong F[H_1]\otimes _F \cdots \otimes _F F[H_k]$ and $G\cong H_1\otimes _F \cdots \otimes _F H_k $ under this isomorphism and, for each i, $A_i=F[H_i]$ is an F-central simple algebra and $[F[H_i]:F]={p_i}^{2\alpha _i}$ .

Proof. (1) The statement follows from Theorems 4.3 and 4.6.

(2) Let $i \neq j$ and take $a \in H_i,b \in H_j$ . Then, $ab=\lambda ba$ with $\lambda \in Z(G) \subseteq F^*$ . Now, $a^{{p_i}^\gamma } \in F^*$ and $b^{{p_j}^\delta }\in F^*$ , so $\lambda ^{{p_i}^\gamma }=\lambda ^{{p_j}^\delta }=1$ , which gives $\lambda =1$ and $ab=ba$ . So, $H_i\subseteq C_G(H_j)$ and G is the central product of $H_1,\ldots ,H_k$ .

(3) This statement follows from Proposition 3.12 and induction on k.

Corollary 4.9. Keep the notation and assumptions of Theorem 4.8. If $n=1$ and $F[H_i]=D_i$ , then $D\cong D_1\otimes _F \cdots \otimes _F D_k$ , where $i(D_i)={p_i}^{\alpha _i}$ .

Using [19, Theorem 2.4], we have the following proposition.

Proposition 4.10. Keep the notation and assumptions of Theorem 4.8. Then, $F[G]=M_n(D)$ is a crossed product over a maximal subfield K if and only if for each i, $F[H_i]$ is a crossed product over a maximal subfield $K_i$ . In addition, under these circumstances, $K\cong K_1 \otimes _F\cdots \otimes _F K_k$ and $\mathrm {Gal}(K/F)\cong \mathrm {Gal}(K_1/F)\times \cdots \times \mathrm {Gal}(K_k/F). $

Theorem 4.11. Let D be an F-central finite dimensional division algebra. Assume that G be a primitive absolutely irreducible locally nilpotent subgroup of $\mathrm {GL}_n(D)$ . Then, $M_n(D)$ is a crossed product over a maximal subfield K. With the notation and assumptions of Theorem 4.8:

1. (1) there exists an abelian normal subgroup S of G such that $G/S$ and $\mathrm {Gal}(K/F)$ are finite nilpotent groups and $\mathrm {Gal}(K/F)\cong N_{\mathrm {GL}_n(D)}(K^*)/K^*\cong G/S$ ;

2. (2) for each i, there exists an abelian subgroup $A_i$ of $H_i$ such that $F[H_i]$ is a crossed product over a maximal subfield $K_i$ and, in addition, $H_i/A_i$ and $Gal(K_i/F)$ are finite nilpotent groups and $\mathrm {Gal}(K_i/F)\cong N_{F[H_i]^*}(K_i^*)/K_i^*\cong H_i/A_i$ ;

3. (3) $S\cong A_1\otimes _F \cdots \otimes _F A_k $ , $K\cong K_1\otimes _F \cdots \otimes _F K_k $ and $S=A_1 \cdots A_k $ .

Proof. By [25, Theorem 3.3.8], G is soluble. Now, using [26, Theorem 6, page 135], G contains a maximal abelian normal subgroup, say S, such that $|G/S|< \infty $ . By Theorem 4.4, $K=F[S]$ is a field and by a result in [10], G is hypercentral. Hence, by an exercise from [22, page 354], we conclude that every maximal abelian normal subgroup of G is self-centralising. Now, using Theorem 4.5, we conclude that $M_n(D)$ is a crossed product over a maximal subfield K. By a result of [6, page 92], $K/F$ is Galois and we can write $M_n(D)=\oplus _{\sigma \in \mathrm {Gal}(K/F)}Ke_{\sigma }$ , where $e_{\sigma }\in \mathrm {GL}_n(D)$ and for each $x\in K$ and $\sigma \in \mathrm {Gal}(K/F)$ , there exists $\sigma (x)\in K$ such that $e_{\sigma }x=\sigma (x)e_{\sigma }$ . So, $e_{\sigma }\in N_{\mathrm {GL}_n(D)}(K^*)$ . Now, using the Skolem–Noether theorem [6, page 39] and the fact that $C_{M_n(D)}(K)=K$ , we obtain $\mathrm {Gal}(K/F)\cong N_{\mathrm {GL}_n(D)}(K^*)/K^*$ . However, consider the homomorphism $\sigma : G\rightarrow \mathrm {Gal}(K/F)$ given by $\sigma (x)=f_x$ , where $f_x(k)=xkx^{-1}$ for $k\in K$ . Clearly, $\ker (\sigma )=C_G(K)$ . Since $S\subseteq C_G(K)\subseteq C_G(S)=S$ , we have $C_G(K)=S$ . Choose an element $a \in \mathrm {Fix}(\mathrm {Im}\, \sigma )$ . For any $x\in G$ , we have $f_x(a)=a$ and hence $xa=ax$ . This shows that $\mathrm {Fix}(\mathrm {Im}\, \sigma )\subseteq C_K(G)\subseteq C_{M_n(D)}(G)=F$ . Hence, $F=\mathrm {Fix}(\mathrm {Im}\, \sigma )$ and $\sigma $ is surjective. Therefore, $\mathrm {Gal}(K/F)\cong G/S$ , as we claimed.

The proof is completed by using Theorem 4.8 and Proposition 4.10.

We can immediately deduce the following theorem.

Theorem 4.12. Let D be an F-central finite dimensional division algebra such that $[D:F]=i(D)^2=\prod _{i=1}^k{p_i}^{2\alpha _i}$ . If $D^*$ contains an absolutely irreducible locally nilpotent subgroup G, then D is a crossed product over a maximal subfield K. With the notation and assumptions of Theorems 4.8 and 4.11, $D\cong D_1\otimes _F \cdots \otimes _F D_k$ , where $F[H_i]=D_i$ and $D_i$ is a crossed product over a maximal subfield $K_i$ .

Proposition 4.13. Let $A=M_n(D)$ be an F-central simple algebra of degree $m^2=\prod _{i=1}^k p_i^{2\alpha _i}$ and G be an absolutely irreducible locally nilpotent subgroup $A^*$ . Then, there is an element of order $p_i$ in F for $1\leq i\leq k$ .

Proof. Keep the notation and assumptions of Theorem 4.8, so that $[F[H_i]:F]={p_i}^{2\alpha _i}$ . Since $F[H_i]$ is a central simple algebra, $F[H_i]\cong M_{{p_i}^{\beta _i}}(D_i)$ , where $D_i$ is an F-central division algebra of degree a power of $p_i$ . Assume that $K_i$ is a maximal subfield of $D_i$ . By [26, Theorem 27.6] and Proposition 3.4, $K_i$ contains an element b, say, of order $p_i$ . Now, $[F(b):F]\leq p_i-1$ and $[F(b):F]\mid [K_i:F]$ . However, $[K_i:F]$ is a power of $p_i$ , which implies $[F[b]:F]=1$ , that is, $b\in F$ .

Proposition 4.14. Let D be an F-central finite dimensional division algebra and suppose that for $p \in \pi (n)$ , there is an element of order p in F, when $n>1$ . Then, $\mathrm {GL}_n(D)$ contains a finite irreducible nonabelian nilpotent subgroup G such that $F[G]=M_n(F)\subseteq M_n(D)$ .

Proof. By [26, Theorem 27.6], there exists a finite nilpotent subgroup G of $\mathrm {GL}_n(F)$ such that $F[G]=M_n(F)\subseteq M_n(D)$ . We show that G is an irreducible subgroup of $\mathrm {GL}_n(D)$ . In contrast, assume that G is reducible in $\mathrm {GL}_n(D)$ . By [25, Theorem 1.1.1], there exists a matrix $P \in \mathrm {GL}_n(D)$ such that

$$ \begin{align*}P(F[G])P^{-1}\subseteq\left[\begin{matrix} M_r(D) & B \\ 0_{(n-s)\times r} & M_{n-s}(D)\\ \end{matrix}\right].\end{align*} $$

This means that we can define a homomorphism from $M_n(F)$ to $M_r(D)$ . However, $M_n(F)$ is a simple ring. Hence, this map is an injection. This contradicts [25, Theorem 1.1.9], which asserts that the matrix ring $M_r(D)$ contains at most r nonzero pairwise orthogonal idempotents.

Example 4.15. The multiplicative group of the real quaternion division algebra contains the quaternion group which is an absolutely irreducible $2$ -group. By [8, Corollary 3.5], if D is a noncommutative finite dimensional F-central division algebra and $D^*$ contains an absolutely irreducible finite p-subgroup for some prime p, then D is a nilpotent crossed product with $[D : F] = 2^m$ for some $m \in \mathbb {N}$ .