UDC 519.41

MATHEMATICS

A. L. SHMEL'KIN

ON SOLVABLE VARIETIES OF GROUPS

(Presented by Academician A. I. Mal'cev, March 13, 1967)

A variety of groups is called solvable if all its groups are solvable. Such is, in particular, every product \(\mathfrak N_k\mathfrak A\) of the variety of all nilpotent groups of class \(\leqslant k\) and the variety \(\mathfrak A\) of all abelian groups (recall that the product \(\mathfrak U\mathfrak V\) of varieties \(\mathfrak U\) and \(\mathfrak V\) consists of all possible extensions of groups from \(\mathfrak U\) by means of groups from \(\mathfrak V\)). We shall call two-step solvable groups, and varieties consisting of metabelian groups, metabelian; all metabelian groups form the variety \(\mathfrak S_2=\mathfrak A^2\).

Our main result is the following.

Theorem 1. Let \(\mathfrak B\) be a variety contained in \(\mathfrak N_k\mathfrak A\) for some \(k\geqslant 1\), but not containing the variety \(\mathfrak S_2\) of all metabelian groups. Then every group \(G\in\mathfrak B\) has a series of normal subgroups

\[ G \supseteq U \supseteq T \supseteq 1, \]

where \(G/U\) and \(T\) have finite exponents, and \(U/T\) is nilpotent. In particular, in the variety \(\mathfrak B\) every torsion-free group with a finite number of generators has a nilpotent normal subgroup of finite index.

Lemma 1. Let \(G\) be a free group of rank 2 of some variety \(\mathfrak B\subseteq\mathfrak S_2\), \(\mathfrak B\ne\mathfrak S_2\). If the group \(G\) is torsion-free, then in it, for some \(n>0\), the relation

\[ \bigl[u,\underbrace{x,x,\ldots,x}_{n},\underbrace{x^2,\ldots,x^2}_{n-1},\ldots,\underbrace{x^n}_{1}\bigr]=1 \]

holds for every \(u\in G'\) and every \(x\) which is a free generator modulo \(G'\) (in other words, for every \(x\) from which, modulo \(G'\), no root of any degree \(>1\) can be extracted).

Proof. If \(x,y\) are free generators of the group \(G\), then \(G'\) is a cyclic \(G/G'\)-module with generator \([x,y]\). Since \(G\) is not a free metabelian group (see (1)), \(G'\) is not a free module. Denote by \(O\) the annihilator of the module \(G'\) in the integral group ring \(Z(G/G')\). An element lies in \(O\) if and only if it annihilates some generator of \(G/G'\), for example \([x,y]\). The ideal \(O\) has the following properties: 1) if \(mf(x,y)\in O\), where \(f(x,y)\in Z(G/G')\) and \(m\) is an integer, then \(f(x,y)\in O\)—this follows from the fact that the group \(G'\) is torsion-free; 2) if \(x'=x^\alpha y^\beta,\ y'=x^\gamma y^\delta\),

\[ \begin{vmatrix} \alpha & \beta\\ \gamma & \delta \end{vmatrix} =\pm1 \]

and \(f(x,y)\in O\), then \(f(x',y')\in O\). Indeed, in this case there exist free generators \(\bar x,\bar y\) of the group \(G\) such that \(\bar x\equiv x',\ \bar y\equiv y'\pmod {G'}\); therefore \(f(x',y')\) annihilates \([\bar x,\bar y]\), which is a generator of the module \(G'\).

We first show that \(O\) contains some polynomial \(f(y)\). Obviously, \(O\) contains some polynomial \(f(x,y)\). Arrange it according to powers of \(x\):

\[ f(x,y)=f_0(y)+f_1(y)x+\cdots+f_n(y)x^n. \]

Let \(f(x,y)\) be chosen so that \(n\) is minimal. Then, making the substitution \(x'=xy,\ y'=y\), we obtain

\[ f(x',y')=f_0(y)+f_1(y)yx+\cdots+f_n(y)y^n x^n\in O. \]

Then \(x^{-1}(f(x',y')-f(x,y))\in O\) and has smaller degree with respect to \(x\). Hence \(n=0\), and \(O\) contains some polynomial \(f(y)\ne0\). Let
\(f(y)=a_0+a_1y+\cdots+a_ny^n\). Making \(n\) substitutions \(x'=x,\ y'=yx^i,\ i=1,\ldots,n\), and substituting in \(f(y)\), we obtain

\[ \begin{gathered} a_0+a_1y+\cdots+a_ny^n=f(y)\in O,\\ a_0+a_1yx+\cdots+a_ny^nx^n=f(xy)\in O,\\ \cdot\quad\cdot\quad\cdot\quad\cdot\quad\cdot\quad\cdot\quad\cdot\quad\cdot\\ a_0+a_1yx^n+\cdots+a_ny^nx^{n^2}=f(x^ny)\in O . \end{gathered} \tag{1} \]

If this system of equalities is regarded as a linear system relating
\(a_0,a_1y,\ldots,a_ny^n\), then its determinant is the Vandermonde determinant
\(W(1,x,\ldots,x^n)\). Hence we obtain that, for any \(0\le j\le n\),

\[ a_jy^jW(1,x,\ldots,x^n)\in O . \]

By the invertibility of \(y\) and property 1 of the ideal \(O\), we have

\[ W(1,x,\ldots,x^n)\in O . \]

Multiplying \(W(1,x,\ldots,x^n)\) by a suitable negative power of \(x\), we obtain

\[ (x-1)^n(x^2-1)^{n-1}\cdots(x^n-1)\in O, \tag{2} \]

and from property 2 of the ideal \(O\) it follows that here, as \(x\), one may take any element from which roots are not extracted in \(G/G'\). The inclusion obtained, translated into the language of the group \(G\), is precisely the assertion of the lemma.

Lemma 2. Let \(A\) be a certain module over a free abelian group \(B\) of finite rank, and suppose that \(A\), under addition, is a free abelian group of finite rank. If

\[ a(x-1)^n(x^2-1)^{n-1}\cdots(x^n-1)=0 \]

for every \(a\in A\) and every free generator \(x\) of the group \(B\), then for some \(s,t\)

\[ A(B^s-1)^t=0. \]

Proof. We proceed by induction on the rank of the group \(A\). If \(A\) is cyclic, then \(A(B^2-1)=0\), since every automorphism of the group \(A\) has order 2. Suppose that our assertion has been proved in the case where \(A\) has rank \(<r\), and let \(A\) have rank \(r\). Suppose that \(A\) has a submodule \(A_1\ne0\) of rank \(<r\). We may assume that the group \(A/A_1\) has no elements of finite order, for these can be adjoined to \(A_1\). By the induction hypothesis,
\((A/A_1)(B^{s_1}-1)^{t_1}=0,\ A_1(B^{s_2}-1)^{t_2}=0\).
Then for the module \(A\) it suffices to take \(s=s_1s_2,\ t=t_1+t_2\). Suppose now that \(A\) has no submodules of smaller rank. If \(x\) is some free generator of the group \(B\), then every submodule
\(A(x-1)^{\alpha_1}(x^2-1)^{\alpha_2}\cdots(x^n-1)^{\alpha_n}\),
\(\alpha_i\le n+1-i\), is either zero or contains some submodule of the form \(mA\). By the condition of the lemma, for some \(i>0\),
\(mA(x^i-1)=0\), whence \(A(x^i-1)=0\).

Since the group \(B\) has finite rank, \(A(x^i-1)=0\) will hold for every \(x\) from a fixed system of generators of the group \(B\), for some \(i>0\). But then from the equality
\((xy)^i-1=(x^i-1)(y^i-1)+(x^i-1)+(y^i-1)\) it follows that
\(A(B^i-1)=0\). The lemma is proved.

Lemma 3. The group \(G\) of Lemma 1 has a nilpotent normal subgroup of finite index.

Proof. The group \(G'\), isomorphic as a module to \(Z(G/G')/O\), has a finite number of generators.

For example, for the additive group \(Z(G/G')/O\), generators are
\(x^iy^j,\ 0\le i,j<r\), where \(r\) is the degree of the polynomial in (2). This follows from the fact that,

that the polynomial (2) and the analogous one written for \(y\) have leading and trailing coefficients \(\pm 1\). Hence Lemma 2 is applicable modulo \(G'\). Thus,

\[ [G',\underbrace{G^s,G^s,\ldots,G^s}_{t\ \text{times}}]=1 \tag{3} \]

whence \(G^s\) is nilpotent of class \(\leq t+1\). Since the group \(G\) is solvable, \(G/G^s\) is finite. The lemma is proved.

Lemma 4. Let the abelian torsion-free group \(A\) be a module over an abelian group \(B\), and suppose that for every \(x\in B\), \(A(x^s-1)^t=0\). Then

\[ A(B^s-1)^{2t}=0. \]

The proof is analogous to one argument of K. Gruenberg (2) and is carried out by induction on \(t\). For \(t=1\) the assertion is obvious. Suppose it has already been proved for \(t\); we prove it for \(t+1\). Clearly, in the endomorphism ring of the group \(A\), for \(x,y\in B\) we have

\[ 0=(x^s y^s-1)^{t+1}=\bigl((x^s-1)(y^s-1)+(x^s-1)+(y^s-1)\bigr)^{t+1}= \]

\[ =\sum_{i+j\geq t+1} a_{ij}(x^s-1)^i(y^s-1)^j. \]

Multiplying this equality by \((x^s-1)^{t-1}\) and taking into account the condition of the lemma for \(t+1\), applied both to \(x\) and to \(y\), we obtain

\[ a_{1t}(x^s-1)^t(y^s-1)^t=0, \]

or, since \(A\) is torsion-free,

\[ (x^s-1)^t(y^s-1)^t=0. \tag{4} \]

Now consider the aggregate \(A_1\) of linear combinations of elements of the form \(a(x^s-1)^t\), for all possible \(a\in A,\ x\in B\). If in \(A/A_1\) there are elements of finite order, adjoin them to \(A_1\); then we obtain, obviously, a submodule \(A_2\) of the module \(A\), and in \(A/A_2\) and, by virtue of (4), in \(A_2\) the condition of the lemma for \(t\) is fulfilled. The rest is obvious.

Lemma 5. Let the group \(G\) be the same as in Lemma 1, but of arbitrary rank. Then for some \(s>0\), \(G^s\) is nilpotent.

Indeed, from Lemma 3 it follows that in every group of the variety \(\mathfrak V\) the identity

\[ [x^s,\underbrace{y^s,\ldots,y^s}_{t}]=1 \]

holds for some \(t>0\), and then it suffices to use Lemma 4.

Lemma 6. Suppose that in a certain group \(S\) there are a nilpotent normal torsion-free divisor \(N\) and the factor group \(S/I(N')\) by the isolator of the commutator subgroup \(N'\) in \(N\). Then the group \(S\) is also nilpotent.

Apparently this lemma is known. Its proof is not difficult and we omit it.

Proof of Theorem 1. It is clear that it suffices to prove the theorem for a \(\mathfrak V\)-free group \(G\) of countable rank. One may also assume that the variety \(\mathfrak V\) has exponent zero. Then all elements of finite order lie in \(G'\) and, by virtue of the nilpotency of \(G'\), form a subgroup \(P\). Consider the group \(H=G/P\). By the complete characteristicity of \(P\) in \(G\), \(H\) is free in some variety \(\mathfrak B\subset \mathfrak N_c\mathfrak A\). The group \(H/I(H'')\) satisfies the condition of Lemma 5; therefore for some \(s>0\), \((H/I(H''))^s\) is nilpotent. Applying Lemma 6 to the case where \(N=H'\), and \(S\) is the complete preimage of \((H/I(H''))^s\) in \(H\), we obtain that \(H^s\) is nilpotent of class \(c\). Take in \(G\) \(U=G^s\), and as \(T\) the subgroup generated by all commutators \([g_1^s,g_2^s,\ldots,g_{c+1}^s]\), \(g_1,\ldots,g_{c+1}\in G\). Since \(H^s\) is nilpotent of class \(c\), all these commutators have finite orders, and these orders are divisors of the order of the element \([x_1^s,x_2^s,\ldots,x_{c+1}^s]\),

where \(x_1, x_2, \ldots, x_{c+1}, \ldots\) is a system of free generators of the group \(G\). Obviously, in a nilpotent group (in our case in \(G'\)) any set of elements of bounded orders generates a subgroup of positive exponent. Thus, \(T\) has positive exponent. In the group \(G/T\) the identity \([x_1^s, x_2^s, \ldots, x_{c+1}^s] = 1\) holds; hence (by induction on \(c\)) it is not hard to obtain that \((G/T)^s = G^s/T\) is nilpotent of class \(\leq c\). The theorem is proved.

Remark. Obviously, one may also take \(G^s \cdot G'\) as \(U\), so that one may assume that \(G/U\) is abelian of finite exponent.

Theorem 2. Let \(G\) be a polycyclic group such that, in the variety \(\operatorname{var}(G)\) generated by \(G\), the free group on two generators is polycyclic. Then \(G\) has a nilpotent normal subgroup of finite index. Obviously, the converse is true for any number of generators.

By a theorem of A. I. Mal'cev \((^3)\), for some \(s\), \(G^s\) has a nilpotent commutant. We shall assume that the group \(G\) itself has a nilpotent commutant. Moreover, one may assume that \(G\) is torsion-free (otherwise it would have a torsion-free normal subgroup of finite index). If \(\operatorname{var}(G)\) contains \(\mathfrak{C}_2\), then the hypothesis of the theorem is violated. Hence Theorem 1 is applicable.

Moscow State University
named after M. V. Lomonosov

Received
24 XII 1966

CITED LITERATURE

\(^1\) G. Baumslag, B. H. Neumann et al., Math. Zs., 86, No. 1, 93 (1964).
\(^2\) K. W. Gruenberg, Illinois J. Math., 5, No. 3, 436 (1961).
\(^3\) A. I. Mal'cev, Matem. sborn., 28, No. 3, 567, 1951.