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MATHEMATICS
M. Sh. KHUZURBAZAR
THE MULTIPLICATIVE GROUP OF A DIVISION RING
(Presented by Academician P. S. Aleksandrov on 26 XII 1959)
Let \(K\) be an associative noncommutative division ring, and \(Z\) the center of \(K\). The set of all nonzero elements of the division ring \(K\) forms a group under multiplication, which we shall denote by \(K^*\). If \(Z^*\) is the set of all nonzero elements of \(Z\), then \(Z^*\) is the center of the group \(K^*\). The group \(K^*\) has so far been little studied. Hua \((^{1,2})\) proved that \(K^*\) is not a (finite) solvable group and that the factor group \(K^*/Z^*\) has no center. Scott \((^3)\) generalized this, proving that the factor group \(K^*/Z^*\) has no nontrivial abelian normal divisors.
In the present article these results are generalized. The following main theorem holds:
Theorem 1. The group \(K^*\) is not locally nilpotent. Moreover, every locally nilpotent normal divisor of the group \(K^*\) is contained in the center \(Z^*\). Consequently, the factor group \(K^*/Z^*\) has no nontrivial locally nilpotent normal divisors.
In other words, using the terminology and results from the works of B. I. Plotkin \((^{4,5})\), \(Z^*\) is the upper radical of the group \(K^*\). The group \(K^*\) is not a radical group. In particular, \(K^*\) is not an \(RN^*\)-solvable group, i.e., \(K^*\) does not possess an ascending solvable normal series. Indeed, every \(RN^*\)-solvable normal divisor of the group \(K^*\) is contained in the center \(Z^*\).
Thus, one may say that the division ring \(K\) is strongly noncommutative.
The proof of the theorem is based on the following four lemmas. Denote
\([a,b]=aba^{-1}b^{-1}\) for nonzero elements \(a\) and \(b\) of the division ring \(K\).
Lemma 1. Let \(x\) and \(y\) be elements of the division ring \(K\) such that
\[
[y,x]=yxy^{-1}x^{-1}=\lambda\ne 1,
\]
where \(\lambda\) commutes with each of the elements \(x\) and \(y\). Let \(\Lambda\) be the ring of polynomials in \(\lambda\) with integral coefficients (the integers being taken from the prime field \(P\) of the division ring \(K\)). Then, if \(x\) is algebraic over \(\Lambda\), \(\lambda\) is a root of unity. If, in addition, \(y\) is also algebraic over \(\Lambda\), then the characteristic of the division ring \(K\) is zero.
Proof. Let \(r\) be the least degree of an equation over \(\Lambda\) satisfied by the element \(x\):
\[
f(x)=\alpha_0+\alpha_1x+\cdots+\alpha_rx^r=0,\qquad \alpha_i\in\Lambda.
\]
Using the relation \(yx=\lambda xy\), we obtain
\[
yf(x)y^{-1}=\alpha_0+\alpha_1\lambda x+\cdots+\alpha_r\lambda^r x^r=0.
\]
Therefore,
\[
\alpha_1(\lambda-1)+\alpha_2(\lambda^2-1)x+\cdots+\alpha_r(\lambda^r-1)x^{r-1}=0.
\]
If \(\lambda^r\ne 1\), then \(x\) satisfies an equation of degree less than \(r\), which is impossible. Suppose now that the element \(y\) is also algebraic over \(\Lambda\) and that the characteristic of the skew field \(K\) is finite. Since \(\lambda\) is a root of unity, \(\Lambda\) is a finite field. Further, the set \(T\) of all finite sums of the form \(\sum a_{ij}x^i y^j,\ a_{ij}\in\Lambda\), is a finite skew field and, consequently, a field. Since \(x\) and \(y\in T\), we have \(yx=xy\), which contradicts the original assumption \(\lambda\ne 1\).
Lemma 2. Let \(x,y,\lambda,\Lambda\) be defined as in Lemma 1. Define
\[ x_1=[y,1+x],\qquad x_{i+1}=[y,x_i],\qquad i=1,2,3,\ldots \]
If \(x_n=1\) for some \(n\), then \(x\) is algebraic over \(\Lambda\). Further, let \(m\) be an integer from the prime field \(P\) of the skew field \(K\). Define analogously for the element \(mx\)
\[ (mx)_1=[y,1+mx],\qquad (mx)_{i+1}=[y,(mx)_i],\qquad i=1,2,3,\ldots \]
If for every integer \(m\) from \(P\) there exists an index \(N_m\) such that \((mx)_{N_m}=1\), then the characteristic of the skew field \(K\) is finite.
Proof. We first note that, by virtue of the relation \(yx=\lambda xy\) and the commutativity of the element \(\lambda\) with \(x\) and \(y\), we obtain \(y f(x)=f(\lambda x)y\) for an arbitrary rational function \(f(x)\) with integer coefficients. If \(f(x)\ne 0\), then \([y,f(x)]=f(\lambda x)(f(x))^{-1}\). Hence
\[ x_1=[y,1+x]=(1+\lambda x)(1+x)^{-1},\qquad x_2=[y,x_1]=(1+\lambda^2x)(1+\lambda x)^{-2}(1+x); \]
in general,
\[ x_n=[y,x_{n-1}]=\prod_{i=0}^{n}(1+\lambda^i x)^{\left[\begin{smallmatrix} n\\ i\end{smallmatrix}\right]\cdot(-1)^{\,n-i}}, \]
where \(\left[\begin{smallmatrix} n\\ i\end{smallmatrix}\right]=\dfrac{n!}{i!(n-i)!}\).
If now \(x_n=1\), then the equality
\[ (\lambda-1)^n+a_1x+a_2x^2+\cdots+a_r x^r=0,\qquad a_i\in\Lambda,\qquad r=2^{\,n-1}-2, \]
holds, i.e. \(x\) is algebraic over \(\Lambda\).
Let for every integer \(m\), \((mx)_{N_m}=1\). Then \(mx\) satisfies a polynomial over \(\Lambda\), namely
\[ L_m(t)=(\lambda-1)^{N_m}+a_1^{(m)}t+a_2^{(m)}t^2+\cdots+a_{r_m}^{(m)}t^{r_m}, \]
where
\[ a_i^{(m)}\in\Lambda,\qquad r_m=2^{N_m-1}-2. \]
Let \(q\) be the least degree of a polynomial over \(\Lambda\) satisfied by the element \(x\); \(G(x)=b_0+b_1x+b_2x^2+\cdots+b_qx^q=0,\ b_i\in\Lambda\). Then \(q\) is the least degree of a polynomial over \(\Lambda\) satisfied by the element \(mx\); as such a polynomial one may take
\[ G_m(t)=b_0m^q+b_1m^{q-1}t+b_2m^{q-2}t^2+\cdots+b_qt^q. \]
Applying the division theorem (see \((^6)\), p. 30) to the polynomials \(L_m(t)\) and \(G_m(t)\), we find a polynomial \(Q_m(t)\) with coefficients in \(\Lambda\) such that
\[ b_q^{k_m}L_m(t)=Q_m(t)G_m(t),\qquad \text{where } k_m=r_m-q+1. \]
Comparing constant terms, we obtain \(b_q^{k_m}(\lambda-1)^{N_m}=c_0^{(m)}b_0m^q\), where \(c_0^{(m)}\) is the constant term in \(Q_m(t)\). Consequently, \(b_q^{k_m}(\lambda-1)^{N_m}\) is divisible by \(m\).
Suppose now that the characteristic of the field \(K\) is zero. Let \(\lambda\) be a primitive \(l\)-th root of unity; \(1=s_1,s_2,\ldots,s_h\) are natural numbers less than \(l\) and relatively prime to \(l\). For each polynomial \(f(\lambda)\) in \(\lambda\) with integral coefficients, define the norm
\[ \operatorname{Norm} f(\lambda)=\prod_{j=1}^{h} f(\lambda^{s_j}). \]
It is clear that the norm has the multiplicative property and that its values are integers.
Since \((\operatorname{Norm} b_q)\cdot(\operatorname{Norm}(\lambda-1))\) has only a finite number of prime divisors, there is a prime number \(m\) which does not divide \((\operatorname{Norm} b_q)\cdot(\operatorname{Norm}(\lambda-1))\). Then \(m\) does not divide \(b_q^{k_m}(\lambda-1)^{N_m}\). This is a contradiction.
The following result is obvious, but since it is essential, we formulate it as a lemma.
Lemma 3. In every nonabelian nilpotent group one can find elements \(x,y\) such that \([y,x]=yxy^{-1}x^{-1}=\lambda\ne 1\), where \(\lambda\) commutes with the elements \(x\) and \(y\).
Lemma 4. Every abelian normal divisor of the group \(K^*\) is contained in the center \(Z^*\).
This follows from the Cartan—Brauer—Hua theorem (see (7), p. 186).
Proof of Theorem 1. Let \(H\) be a locally nilpotent normal divisor and \(H\nsubseteq Z^*\). Then, by Lemmas 3 and 4, in \(H\) there are elements \(a,b\) such that \([a,b]=aba^{-1}b^{-1}=\lambda\ne 1\), where \(\lambda\) commutes with the elements \(a\) and \(b\). Let \(m\) be an arbitrary integer from the prime field \(P\) of the field \(K\). Define \((mb)_1=[a,1+mb]\), \((mb)_{i+1}=[a,(mb)_i]\), \(i=1,2,3,\ldots\). It is clear that \((mb)_1\in H\), and, by the local nilpotence of the group \(H\), the elements \(a\) and \((mb)_1\) generate a nilpotent subgroup.
Consequently, \((mb)_{N_m}=1\) for some \(N_m\). Now it follows from Lemma 2 that the element \(b\) is algebraic over \(\Lambda\) and the characteristic of the field \(K\) is finite. Similarly it is proved that the element \(a\) is also algebraic over \(\Lambda\). But then Lemma 1 implies that the characteristic of the field \(K\) is zero. This is a contradiction.
Remark. In the proof of the theorem, the local nilpotence of the group \(H\) is not quite essential, since only the fact is used that any two elements of the group \(H\) generate a nilpotent subgroup. Groups in which any two elements generate a nilpotent subgroup were called weakly nilpotent by B. I. Plotkin (5). However, it is still unknown whether there exist weakly nilpotent but not locally nilpotent groups.
The center \(Z^*\) of the group \(K^*\) has yet another interesting property, namely, it is a primary normal divisor in the sense of K. K. Shchukin (8). Recall that a normal divisor \(N\) of a group \(G\), different from the group itself, is called a primary normal divisor if from the relation \([A,B]\subseteq N\), where \([A,B]\) is the mutual commutator of the invariant subgroups \(A\) and \(B\) in \(G\), it follows that at least one of the subgroups \(A,B\) is contained in \(N\). A group \(G\) is called primary if its identity subgroup is a primary normal divisor in \(G\). In (8) it is proved that \(N\) is a primary normal divisor of the group \(G\) if and only if the factor group \(G/N\) is primary.
Theorem 2. \(Z^*\) is a primary normal divisor of the group \(K^*\) and, consequently, \(K^*/Z^*\) is a primary group.
First note that, by the Cartan—Brauer—Hua theorem (7), the following lemma holds:
Lemma 5. If \(N\) is a normal divisor of the group \(K^*\) and \(N\nsubseteq Z^*\), then the centralizer of the subgroup \(N\) in \(K^*\) coincides with \(Z^*\).
Proof of Theorem 2. Let \(A\) and \(B\) be normal subgroups of the group \(K^*\), with \(B \nsubseteq Z^*\). Then from \([A,B]\subseteq Z^*\) it follows at once that \([[A,A],B]=1\), i.e., \([A,A]\) is contained in the centralizer of the subgroup \(B\) in \(K^*\). By Lemma 5, \([A,A]\subseteq Z^*\). In view of Theorem 1, \(A\subseteq Z^*\).
I express my deep gratitude to Prof. A. G. Kurosh for his supervision of this work.
Moscow State University
named after M. V. Lomonosov
Received
25 XII 1959
References
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\(^{3}\) W. R. Scott, Proc. Am. Math. Soc., 8, No. 2 (1957).
\(^{4}\) B. I. Plotkin, Tr. Moskovsk. matem. obshch., 6, 299 (1957).
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\(^{8}\) K. K. Shchukin, Uch. zap. Kishinevsk. gos. univ., 39, 209 (1959).