On the proof of some theorem on locally nilpotent subgroups in division rings

TL;DR

This paper provides a new proof for a theorem stating that every locally nilpotent subnormal subgroup in a division ring is central, correcting a previous proof with an identified mistake.

Contribution

The authors present an alternative proof for a key theorem about locally nilpotent subnormal subgroups in division rings, ensuring mathematical correctness.

Findings

The theorem that such subgroups are central is validated.

The new proof rectifies previous errors in the literature.

The result reinforces understanding of subgroup structure in division rings.

Abstract

In Hai-Thin (2009), there is a theorem, stating that every locally nilpotent subnormal subgroup in a division ring $D$ is central (see Hai-Thin (2009, Th. 2.2)). Unfortunately, there is some mistake in the proof of this theorem. In this note we give the another proof of this theorem.

Equations

$$G=G_n◃G_{n-1}◃\dots ◃G_1◃G_0=D^{\ast }.$$

$$x_1:=[x+1,y],x_{i+1}:=[x_i,y],\forall i\ge 1.$$

$$f(x)=\sum _{i=0}^r{a}_i{x}^i=0\text{mbox}and{y}^{-1}f(x)y=\sum _{i=0}^r{a}_i{b}^i{x}^i=0.$$

$$H_0=H,H_1=[H_0,H_0]\text{mbox}and{H}_i=[H_{i-1},H]\text{mbox}foralli\ge 2.$$

$$H_s=⟨[a,b]|a\in H_{s-1},b\in H⟩\ne 1.$$

$$G=G_n◃G_{n-1}◃\dots ◃G_1◃G_0=D^{\ast }.$$

$$N=N_n◃N_{n-1}◃\dots ◃N_1◃N_0=D_1^{\ast }.$$

Full text

Key words:* locally nilpotent, central.*

Mathematics Subject Classification 2000*: 16K20*

In Hai-Thin (2009, Th. 2.2) there is the following theorem:

Theorem 1.* (see Hai-Thin (2009, Th. 2.2)) Let $D$ be a division ring with the center $Z(D)$. Then, every locally nilpotent subnormal subgroup of $D^{*}$ is central, i.e. it is contained in $Z(D)$. *

*Unfortunately, in the proof of this theorem there is some mistake. However, the theorem remains always true. Here we give the another proof of this theorem. *

Lemma 2.* *Let $D$ be a non-commutative division ring and suppose that $G$ is a subnormal locally nilpotent subgroup of $D^{*}$. If $x,y\in G$ with $[x,y]:=x^{-1}y^{-1}xy\neq 1$, then $[x,y]\not\in Z(D)$.

Proof.* Since $G$ is subnormal in $D^{*}$, there exists the following series of subgroups:*

[TABLE]

Suppose that $x,y\in G$ with $[x,y]\neq 1$ and $x,y\in Z(D)$. Set

[TABLE]

We shall prove by induction that $x_{k}\in G_{k},\forall k\in\{1,\ldots,n\}$. In fact, since $y\in G\leq G_{k},\forall k$, it follows that $x_{1}=(1+x)^{-1}y^{-1}(1+x)y\in G_{1}$. Now, suppose that $x_{k}\in G_{k}$, then, $x_{k+1}=[x_{k},y]=(x_{k}^{-1}y^{-1}x_{k})y\in G_{k+1}$. Thus, $x_{k}\in G_{k},\forall k\in\{1,\ldots,n\}$. In particular, $x_{n}\in G_{n}=G.$ By supposition, $G$ is locally nilpotent, so the subgroup $\langle x_{n},y\rangle$ generated by elements $x_{n}$ and $y$, is nilpotent. Hence, there exists some integer $t$ such that $x_{t}=1$. Let $P$ be a prime subfield of $D$. Then, by Scott(1987, 14.3.2, p. 432), there exists some polynomial $g\in P(b)[X]$ such that $g(x)=0$ and, moreover, $g$ is not dependent either on $x$ or $y$, but it is dependent only on $b$. Clearly, for any $m\in\mathbb{Z}$ such that $mx\neq 0$, we have $[mx,y]=[x,y]=b$. Therefore, $g(mx)=0,\forall m\in\mathbb{Z}$ such that $mx\neq 0$. Evidently, this forces $Char(D)=p>0$. Suppose that $f(X)=\displaystyle\sum_{i=0}^{r}a_{i}X^{i}\in P(b)[X]$ with $a_{r}=1$, is the minimal polynomial of the element $x$ over $P(b)$. Then

[TABLE]

*It follows that $a_{1}(b-1)+\ldots+a_{r}(b^{r}-1)x^{r-1}=0$. Therefore, $b^{r}-1=0$ or $b^{r}=1$. Hence $P(b)$ is a finite field. Thus, $x$ is algebraic over a finite field $P(b)$. By symmetry, we can conclude that $y$ is algebraic over $P(b)$ too. Set $K:=\{\sum a_{ij}x^{i}y^{j}|~{}a_{ij}\in P(b)\}$. Since $[x,y]=b\in Z(D),K$ is closed under the multiplicative operation in $D$, so, clearly it is a subring of $D$. Moreover, since $x$ and $y$ both are algebraic over a finite field $P(b)$, it is easy to see that $K$ is finite. Therefore, $K$ is a finite division ring and by Wedderburn’s Theorem $K$ is a field. In particular, $x$ and $y$ commute with each other and that is a contradiction. *

Proof of Theorem 1.* Clearly we can suppose that $D$ is non-commutative. In the first, we shall prove that $G$ is abelian. Thus, suppose that $G$ is non-abelian. Then, there exist elements $x,y\in G$ such that $[x,y]\neq 1$. Let $H=\langle x,y\rangle$ be the subgroup of $G$ generated by $x$ and $y$. Set*

[TABLE]

Since $H$ is non-abelian nilpotent subgroup, there exists the integer $s\geq 1$ such that $H_{s}\neq 1$ and $H_{s+1}=1$. By definition we have

[TABLE]

Therefore, there exist $a_{0}\in H_{s-1},b_{0}\in H$ such that $c:=[a_{0},b_{0}]\neq 1$. Since $c\in H_{s}$ and $H_{s+1}=1,c$ commutes with each element from $H$. In particular, $c$ commutes with both $x$ and $y$. Now, suppose that

[TABLE]

By setting $D_{1}=C_{D}(c),N_{i}=G_{i}\cap D_{1}$ for all $i\in\{1,\ldots,n\}$, we obtain

[TABLE]

*So $N$ is a subnormal locally nilpotent subgroup of $D_{1}^{*}$. Since $x$ and $y$ are both commute with $c,H=\langle x,y\rangle\leq D_{1}^{*}$. It follows that $H\leq D_{1}^{*}\cap G_{n}=N_{n}=N$. Hence $a_{0},b_{0}\in N$. Since $c=[a_{0},b_{0}]\neq 1$, by Lemma 2, $c\not\in Z(D_{1})$. In the other hand, since $D_{1}=C_{D}(c),c\in Z(D_{1})$ and we have a contradiction. Thus, $G$ is abelian. Now, in view of Scott (1987, 14.4.4, p.440) $G$ is contained in $Z(D)$. *

Acknowledgement.* The authors thank Mojtaba Ramezannassab for some his remarks concerning the proof of our theorem in Hai-Thin (2009, Th.2.2).*