V. A. KREKNIN

ON THE SOLVABILITY OF LIE ALGEBRAS WITH A REGULAR AUTOMORPHISM OF FINITE PERIOD

(Presented by Academician I. M. Vinogradov on 14 XII 1962)

1. An automorphism \(\Phi\) of a Lie algebra (ring) \(\mathfrak L\) is called regular if \(\Phi(x) \ne x\) for all \(x \ne 0,\ x \in \mathfrak L\). In the present note the following assertion is proved.

Theorem. A Lie algebra \(\mathfrak L\) of arbitrary dimension over a field of characteristic \(p \geqslant 0\), admitting a regular automorphism of finite period \(n\) \((\Phi^n = 1)\), is solvable, and the length of its derived series does not exceed \(2^{\,n-1}\).

From this theorem and Theorem B \((^3)\) the following result is obtained.

Corollary. A Lie ring \(\mathfrak L\) with a regular automorphism of prime period \(q\) is nilpotent, and its nilpotency class does not exceed

\[ \frac{(q-1)^{2^{q-1}} - 1}{q - 2}. \]

The nilpotency of a Lie ring with a regular automorphism of period \(q\) was proved earlier by Higman \((^1)\). In the same work it was proved that the nilpotency class of such a Lie ring does not exceed some number \(k(q)\), depending only on \(q\). From the results of Thompson \((^2)\) and Higman (Theorem 3 \((^1)\)) it follows that the number \(k(q)\) is also an upper bound for the nilpotency class of a finite group with a regular automorphism of prime period \(q\). The values of \(k(q)\) for \(q = 2, 3, 5\) are respectively equal to \(1, 2, 6\). For \(q \geqslant 7\) it is known only that \(k(q) \geqslant (q^2 - 1)/4\). The corresponding example was given by Higman \((^1)\). The corollary stated above gives a certain upper bound for the nilpotency class of a Lie ring (and consequently also of a finite group) with a regular automorphism of period \(q\), although, undoubtedly, this bound is considerably larger than the number \(k(q)\). To obtain the exact value of \(k(q)\) for all prime \(q\), more refined methods are apparently necessary.

1. The situation changes substantially if the Lie algebra has a regular automorphism of arbitrary period \(n\).

First, such an algebra may be solvable but nonnilpotent. We give an example for any composite \(n\). Let \(n = q \cdot m\), where \(q\) is some prime divisor of the number \(n\), \(m \ne 1\). Consider the Lie algebra \(\mathfrak L\) generated by the elements \(x, y_{1+iq},\ i = 0, 1, \ldots, m - 1\), with the following commutation law: \([y_{1+iq}, y_{1+jq}] = 0\) for all \(i, j\); \([x, y_{1+iq}] = y_{1+(i+1)q}\), where \((i+1)\) is taken modulo \(m\). The ground field \(\mathfrak f\) will be assumed algebraically closed of characteristic \(p,\ (p,n)=1\). Let \(\Phi\) be the linear transformation of the vector space \(\mathfrak L\):

\[ \Phi(x)=\xi^q x,\quad \Phi(y_{1+iq})=\xi^{1+iq}y_{1+iq},\quad i=0,1,\ldots,m-1; \]

\(\xi\) is a primitive \(n\)-th root of unity. It is easily verified that \(\Phi\) is a regular automorphism of the algebra \(\mathfrak L\) with period \(n\), and that the algebra \(\mathfrak L\) is solvable but nonnilpotent.

Second, a nilpotent Lie algebra with a regular automorphism of arbitrary period \(n\) may have an arbitrarily large nilpotency class. Let the ground field \(\mathfrak f\) and the number \(n\) be the same as in the preceding example. Suppose further that the algebra \(\mathfrak L\) is generated by the elements \(x, y^s_{1+iq},\ i = 0, 1, \ldots, m - 1;\ s = 1, 2, \ldots, t\), where \(t\) is an arbitrary integer. Form...

commuting algebras \(\mathfrak L\) commute with one another according to the following rules: \([y^s_{1+iq}, y^r_{1+jq}] = 0\) for all indices \(i,j,r,s\); \([x, y^s_{1+iq}] = y^{s+1}_{1+(i+1)q}\), \((i+1)\) is taken modulo \(m\), and \([x, y^t_{1+iq}] = 0,\ i=0,1,\ldots,m-1\). The algebra \(\mathfrak L\) is nilpotent, and its nilpotency class is equal to \(t\). It is not difficult to see that the following linear transformation \(\Phi\) of the vector space \(\mathfrak L\) will be a regular automorphism of the algebra \(\mathfrak L\) with period \(n\): \(\Phi(x)=\xi^q x\), \(\Phi(y^s_{1+iq})=\xi^{1+iq}y^s_{1+iq}\); \(i=0,1,\ldots,m-1\); \(s=1,2,\ldots,t\); \(\xi\) is a primitive \(n\)-th root of 1. Since \(t\) can be chosen arbitrarily large, the nilpotency class of the algebra \(\mathfrak L\) can also be arbitrarily large.

In the present note we consider only the case where the regular automorphism has finite period. The paper \((^3)\) gives a certain approach to the study of finite-dimensional Lie algebras over a field of characteristic \(p \geqslant 0\) with an automorphism satisfying an arbitrary equation \(f(x)=0\), none of whose roots is equal to 1.

1. We shall prove the theorem formulated in item 1. Let there be an algebra \(\mathfrak L\) with a regular automorphism \(\Phi\) of period \(n\). Without loss of generality one may assume that the ground field \(\mathfrak f\) is algebraically closed. If \(p\) is the characteristic of the field \(\mathfrak f\), then \(\Phi^p\) is a regular automorphism. Suppose this is not so. Then the set of elements \(x \in \mathfrak L\) such that \(\Phi^p(x)=x\) is a nonzero \(\Phi\)-invariant subalgebra \(\mathfrak L_1\) of the algebra \(\mathfrak L\). Consequently, the algebra \(\mathfrak L_1\) over the field of characteristic \(p\) admits a regular automorphism \(\Phi\) of period \(p\), which is impossible \((^1)\). Thus, if the algebra \(\mathfrak L\) admits a regular automorphism of finite period, then it admits a regular automorphism whose period is not divisible by the characteristic of the ground field. We shall therefore assume from the outset that \((n,p)=1\). The latter means that all roots of the minimal equation for \(\Phi\) are distinct.

The algebra \(\mathfrak L\), considered as a vector space over the field \(\mathfrak f\), is represented in the form
\[ \mathfrak L=\sum_{i=1}^{n-1}\mathfrak L_i, \]
where \(\mathfrak L_i\) is the root subspace corresponding to the root \(\xi^i\) of the minimal equation for \(\Phi\); here \(\xi\) is a primitive root of degree \(n\) of 1. The subspace \(\mathfrak L_0=\{0\}\) by virtue of the regularity of \(\Phi\). Since \(\Phi\) is an automorphism, \([\mathfrak L_i\mathfrak L_j]\subseteq \mathfrak L_{i+j}\), \((i+j)\) is taken modulo \(n\). We agree to regard residues \(i\) modulo \(n\) as positive, \(1 \leqslant i \leqslant n-1\). The set of residues is ordered in the natural way:
\[ 1<2<\cdots<n-1. \]

Let \(\mathfrak L=\mathfrak L^{(0)} \supseteq \mathfrak L^{(1)} \supseteq \cdots \supseteq \mathfrak L^{(k)} \supseteq \cdots\) be the derived series of the algebra \(\mathfrak L\). Since all \(\mathfrak L^{(s)}\), \(s=1,2,\ldots,k,\ldots\), are \(\Phi\)-invariant, they are also representable in the form
\[ \mathfrak L^{(s)}=\sum_{i=1}^{n-1}\mathfrak L^{(s)}_i, \]
where \(\mathfrak L^{(s)}_i\) is the root subspace of the vector space \(\mathfrak L^{(s)}\) belonging to the root \(\xi^i\). Obviously,
\[ \mathfrak L^{(s)}_i=\mathfrak L_i\cap \mathfrak L^{(s)} \]
and \(\mathfrak L^{(s)}_i \subseteq \mathfrak L^{(t)}_i\) for \(t \leqslant s\). Elements of \(\mathfrak L^{(s)}_i\) will be denoted by \(x_i^{(s)}\) (for simplicity one symbol is used, although in the expressions below different elements of \(\mathfrak L^{(s)}_i\) may occur). We shall further denote by \(L_k\) the subalgebra generated by the subspaces \(\mathfrak L_i\), \(i \geqslant k\). Since \(\mathfrak L^{(1)}=[\mathfrak L,\mathfrak L]\), the subspace \(\mathfrak L^{(1)}_1\) consists of elements
\[ x^{(1)}_1=\sum_{i=2}^{\left[\frac{n+1}{2}\right]}[x_i,x_{n+1-i}],\quad x_i\in\mathfrak L_i \]
and their linear combinations. This means that \(x^{(1)}_1 \in L_2\), i.e. \(\mathfrak L^{(1)}_1\subseteq L_2\). Suppose, by induction, that the subspaces
\[ \mathfrak L^{(2^k-1)}_i\subseteq L_k,\quad i\leqslant k-1. \]
We shall then show that
\[ \mathfrak L^{(2^k-1)}_i\subseteq L_{k+1},\quad i\leqslant k. \]
Dei-

indeed, as above, the subspace \(\Omega_k^{(2^{k}-1)}\) consists of elements

\[ x_k^{(2^{k}-1)} = \sum_{j=1}^{\left[\frac{k}{2}\right]} \left[ x_j^{(2^{k-1}-1)},\, x_{k-j}^{(2^{k-1}-1)} \right] + \sum_{j=k+1}^{\left[\frac{n+k}{2}\right]} \left[ x_j^{(2^{k-1}-1)},\, x_{n+k-j}^{(2^{k-1}-1)} \right] \tag{*} \]

and their linear combinations. Here and below in the proof the following elementary fact is used: if \(i_0 \equiv i_1+i_2 \pmod n\), where \(0<i_j<n\), \(j=0,1,2\), then either both summands on the right-hand side of the congruence are less than \(i_0\), or both are greater than \(i_0\). Indeed, if \(i_0>i_1\), then \(i_2=i_0-i_1<i_0\); if, however, \(i_0<i_1\), then \(i_2=n+i_0-i_1=(n-i_1)+i_0>i_0\). From this remark it follows that the second sum on the right-hand side of \((*)\) is contained in \(L_{k+1}\).

By the induction hypothesis, \(\Omega_i^{(2^{k-1}-1)} \subset L_k\) for \(i\leq k-1\). Consequently, an arbitrary element

\[ x_{k-j}^{(2^{k-1}-1)} = \sum [x_{i_1},x_{i_2},\ldots,x_{i_m}], \quad \text{where } 1\leq j\leq \left[\frac{k}{2}\right],\ x_{i_s}\in \Omega_{i_s},\ i_s\geq k, \]

where under the summation sign there stands a left-normed product. Substitute the value of \(x_{k-j}^{(2^{k-1}-1)}\) in \((*)\) and consider an arbitrary term of the resulting sum:

\[ \left[ x_j^{(2^{k-1}-1)}[x_{i_1},\ldots,x_{i_m}] \right] = \sum \alpha_\pi \left[ x_j^{(2^{k-1}-1)}x_{\pi i_1},\ldots,x_{\pi i_m} \right], \]

where \(\pi\) is some permutation of the symbols \(i_1,i_2,\ldots,i_m\), and under the summation sign there stand left-normed products. Two cases are possible: \(\pi i_m=k\) and \(\pi i_m>k\). If \(\pi i_m=k\), then \(j+\sum_{s=1}^{m-1}\pi i_s=0\), and, consequently,

\[ [x_j^{(2^{k-1}-1)},x_{\pi i_1},\ldots,x_{\pi i_m}]=0. \]

If, however, \(\pi i_m>k\), then \(j+\sum_{s=1}^{m-1}\pi i_s>k\). Then

\[ [x^{(2^{k-1}-1)},x_{\pi i_1},\ldots,x_{\pi i_{m-1}}] = y_t\in \Omega_t,\quad t=j+\sum_{s=1}^{m-1}\pi i_s>k; \]

hence

\[ [x^{(2^{k-1}-1)}x_{\pi i_1},\ldots,x_{\pi i_m}] = [y_t,x_{\pi i_m}] \in L_{k+1}. \]

Thus, the first sum on the right-hand side of \((*)\) belongs to \(L_{k+1}\), i.e. \(\Omega_k^{(2^k-1)}\subset L_{k+1}\). The preceding arguments are applicable to any algebra with a regular automorphism of period \(n\), in particular to the algebra \(\mathfrak N=\Omega^{(2^k-1)}\). By the induction hypothesis, \(\mathfrak N_i^{(2^{k-1}-1)}\subset N_k\), \(i\leq k-1\). The subalgebra \(N_k\) in \(\mathfrak N\) corresponds to the subalgebra \(L_k\) in \(\Omega\). But, by what has been proved, \(\Omega_k^{(2^k-1)}=\mathfrak N_k\subset L_{k+1}\), and since \(N_{k+1}\subset L_{k+1}\), it follows that \(\Omega_i^{(2^k-1)}=\mathfrak N_i^{(2^{k-1}-1)}\subset L_{k+1}\), \(i\leq k-1\). Since \(\Omega_k^{(2^k-1)}\subset\Omega_k^{(2^{k-1})}\), we have \(\Omega_k^{(2^n-1)}\subset L_{k+1}\), and the induction hypothesis is justified. For \(k=n-2\), \(\Omega_i^{(2^{n-2}-1)}\subset L_{n-1}\), \(i\leq n-2\), i.e. \(\Omega^{(2^{n-2}-1)}\subset L_{n-1}\). From this it is not difficult to obtain that \(\Omega_{n-1}^{(2^{n-2})}=\{0\}\) and \(\Omega^{(2^n-1)}=\{0\}\), as was required to be proved.

In the case when the period of the regular automorphism is equal to a prime number \(q\), Higman’s construction \((^{1})\), p. 327, makes it possible to apply the technique of the proof of the preceding theorem and Theorem B \((^{3})\) to an arbitrary Lie ring admitting the indicated automorphism. Hence, as was already said in item 1, our corollary is obtained.

I take this opportunity to express my gratitude to A. I. Kostrikin for valuable advice and remarks.

Mathematical Institute named after V. A. Steklov
Academy of Sciences of the USSR

Received
13 XII 1962

REFERENCES

\(^{1}\) G. Higman, J. Lond. Math. Soc., 32 (3), 321 (1957).
\(^{2}\) J. G. Thompson, Proc. Nat. Acad. Sci., 45, 578 (1959).
\(^{3}\) V. A. Kreknin, A. I. Kostrikin, DAN, 149, No. 2 (1963).