ELEMENTS OF ORDER TWO IN ENGEL LIE RINGS

A. B. Yadryshnikov

(Presented by the Department of Higher Mathematics)

The purpose of the present work is to simplify one point in the proof of the nilpotency of Engel Lie rings. We shall give here results concerning elements of order two. We shall say that an element \(a \in L\), where \(L\) is a Lie ring satisfying the \(n\)-th Engel condition

\[ [uv^n]=0 \]

for all \(u,\ v \in L\) of characteristic \(p>n\) (\(p\) a prime number) or \(p=0\), has order \(k\) if

\[ a^{k-1}\ne 0,\quad a^k=0; \]

obviously, for any element of order two the relations

\[ c^2=cuc=0 \]

hold for any

\[ u\in L. \]

A. I. Kostrikin [1] proved the following theorem: in an arbitrary Lie ring \(L\) with the \(n\)-th Engel condition and characteristic \(p>n\) (or \(p=0\)) there exists an element of order two.

The proof is carried out by the method of descent. It is proved that, if \(v^m=0\) \((4\le m\le n)\), then \([uv^{m-1}]^{m-1}=0\). Thus one can obtain elements of all orders from \(n\) down to 3 inclusive. For the passage from an element of order three to an element of order two, a series of identities is first proved, and then the required element is found by repeatedly decreasing the exponent in one of these identities.

In the proof proposed by us it is established that the sum of two elements of order two is an element of order three. Indeed, if \(a,b\in L\), \(a^2=b^2=0\), then

\[ (a+b)^2=ab+ba, \]

\[ (a+b)^3=(ab+ba)(a+b)=0. \]

Further,

\[ (a-b)^3=0. \]

Then, knowing that elements of order two exist, one can obtain them from elements of order three. Moreover, the left product of two elements of order two is also an element of order two.

Consequently, starting even from a pair of elements of order 2, one can obtain sufficiently many elements of order 2. But we can do without the assumption of the existence of at least two elements of order two. Namely, let \(v\) be an element of third order. Consider the element

\[ w=[[uv][uv^2]], \]

where \(u\) is an arbitrary element of the ring \(L\). Express the element \(w\) as a sum of products of powers of the elements \(u, v\):

\[ w=[[uv][uv^2]]=|uv|\,|uv^2|-|uv^2|\,|uv|= \]

\[ =-vu^2v^2-2uv^2uv+4vuvuv-2vuv^2u-v^2u^2v. \]

Now consider the square of the element \(w\):

\[ w^2=(-vu^2v^2-2uv^2uv+4vuvuv-2vuv^2u-v^2u^2v)^2; \]

the following table explains the course of the computations:

here the summands in the expression for \(w\) in the first degree are recorded. On the right their coefficients are indicated. Next are indicated all possible products of each of these elements by each:

\[ \begin{aligned} &vu^2v^2\cdot vu^2v^2 && 1,\\ &vu^2v^2\cdot uv^2uv && 2,\\ &vu^2v^2\cdot vuvuv && -4,\\ &vu^2v^2\cdot vuv^2u && 2,\\ &vu^2v^2\cdot v^2u^2v && 1,\\ &uv^2uv\cdot vu^2v^2 && 2,\\ &uv^2uv\cdot uv^2uv && 4,\\ &uv^2uv\cdot vuvuv && -8,\\ &uv^2uv\cdot vuv^2u && 4,\\ &uv^2uv\cdot v^2u^2v && 2,\\ &vuvuv\cdot vu^2v^2 && -4, && 1)\\ &vuvuv\cdot uv^2uv && -8, && 2)\\ &vuvuv\cdot vuvuv && 16, && 3)\\ &vuvuv\cdot vuv^2u && -8,\\ &vuvuv\cdot v^2u^2v && -4,\\ &vuv^2u\cdot vu^2v^2 && 2, && 4)\\ &vuv^2u\cdot uv^2uv && 4, && 5)\\ &vuv^2u\cdot vuvuv && -8, && 6)\\ &vuv^2u\cdot vuv^2u && 4, \end{aligned} \]

\[ \begin{aligned} &vuv^{2}u\cdot v^{2}u^{2}v &&2,\\ &v^{2}u^{2}v\cdot vu^{2}v^{2} &&1, &&7)\\ &v^{2}u^{2}v\cdot uv^{2}uv &&2, &&8)\\ &v^{2}u^{2}v\cdot vuvuv &&-4, &&9)\\ &v^{2}u^{2}v\cdot vuv^{2}u &&2,\\ &v^{2}u^{2}v\cdot v^{2}u^{2}v &&1. \end{aligned} \]

The summands not marked by a numeral with a parenthesis vanish by virtue of one of the relations:
\[ v^{3}=0, \]
\[ v^{2}uv=vuv^{2}, \]
\[ v^{2}uv^{2}=0, \]
which hold for any element \(v\) of order three and any element \(u\) of the ring \(L\). The sum of the terms marked by the numbers 2), 3), 6) gives zero, since these terms are made similar by a single or double application of the equality. The sum of the terms 1), 4), 5), 8), 9) is also equal to zero, because, applying, besides the identity \(v^{2}uv=vuv^{2}\), also the identity
\[ v^{2}u^{2}v^{2}uv=vuv^{2}uv^{2}, \]
we obtain that they are similar. There remains the sole term 7):
\[ v^{2}u^{2}v^{2}u^{2}v^{2}. \]
But it too is equal to zero by virtue of the identity
\[ v^{2}\left(u^{2}v^{2}\right)^{m}=0;\quad m=0,1,\ldots,n. \]

REFERENCES

1. A. I. Kostrikin. On the Burnside problem. Izvestiya AN SSSR, vol. 23, mathematical series, issue 1, 1959, pp. 3–34.