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The General Burnside Problem

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In 1902 William Burnside posed the following question concerning finitely generated groups:

Bounded Burnside Problem. If G is a finitely generated group and there is an integer n such that g^n = 1 for every g\in G, then must G be finite?

The problem also has the following variant:

General Burnside Problem. If G is a finitely generated group and every element of G has finite order, then must G be finite?

The answer to both questions was expected to be “yes”’; the solution to the General Burnside Problem was therefore anticipated to be somewhat harder than that of the Bounded Burnside Problem. However, the answer to both problems turned out to be “no”. A counterexample to the General Burnside problem was given by a beautiful and elegant construction of Golod and Shafarevich in 1964; and a counterexample to the Bounded Burnside Problem was given by Novikov and Adjan in 1968.

The Burnside Problem has the following ring-theoretic analogue:

Kurosh’s Problem. If A is a finitely generated algebra over a field F and every element of A is nilpotent, then must A be nilpotent?

(A element a \in A is nilpotent if a^n=0 for some n; and A is itself nilpotent if there is an m such that a_1a_2\cdots a_m = 0 for all a_1,a_2,\ldots,a_m \in A.) The Golod-Shafarevich construction also provides a counter-example to Kurosh’s Problem. We sketch the construction below.

The Golod-Shafarevich Theorem

Let F be a field and let T = F\langle x_1,x_2,\ldots x_d \rangle be the free non-commutative algebra over F generated by the variables x_1,x_2,\ldots,x_d. Let T_n denote the subspace of T consisting of all linear combinations of monomials of degree n. The elements of T_n are the homogeneous elements of degree n. Let I be a two-sided ideal of T generated by a set A of homogeneous elements, each of degree at least 2. Suppose that A has at most r_i elements of degree i for i \geq 2. The following (which we do not prove here) is the main computational result of the Golod-Shafarevich construction:

Theorem. The quotient algebra T/I is infinite dimensional over F if the coefficients in the power series expansion of (1 - dz + \sum_{i=2}^\infty r_i z^i)^{-1} are nonnegative.

Using this theorem, one constructs a counterexample to Kurosh’s Problem as follows.

Counterexample to Kurosh’s Problem

Let T = F\langle x_1,x_2,x_3 \rangle be the free algebra over a countable field F. Let T' be the ideal of T consisting of all elements of T without constant term. Enumerate the elements of T' as t_1, t_2, \ldots. Choose an integer m_1 \geq 2 and write t_1^{m_1} = t_{1,2} + t_{1,3} + \cdots + t_{1,k_1}, where each t_{1,j} \in T_j. Choose another positive integer m_2 sufficiently large so that t_2^{m_2} = t_{2,k_1+1} + t_{2,k_1+2} + \cdots + t_{2,k_2} for some k_2 > k_1. Continue in this way for sufficiently large powers of t_3, t_4, \cdots. Now let I be the ideal generated by the t_{i,j} defined in the process above. Consider the quotient T'/I. The construction of I guarantees that each element of T'/I is nilpotent; but the theorem above ensures that T'/I is infinite dimensional over F (and hence not nilpotent). Thus T'/I is a counterexample to Kurosh’s Problem. From this construction, we can in turn build a counterexample to the General Burnside Problem.

Counterexample to the General Burnside Problem

Let us suppose now that p is a prime number and F is the field with p elements. Let T and I be as defined above. Let a_1, a_2, a_3 be the elements x_1 + I, x_2 + I, x_3 + I of the quotient T/I respectively. Let G be the multiplicative semigroup in T/I generated by the elements 1+a_1, 1+a_2, and 1+ a_3. An element of G has the form 1+a for some a \in T'/I. The element a is nilpotent by the construction of T'/I, and so for sufficiently large n, we have a^{p^n} = 0. Since we are in characteristic p we have (1+a)^{p^n} = 1 + a^{p^n} = 1. It follows that 1+a has an inverse, whence G is a group. Moreover every element 1+a of G has finite order (indeed order a power of p). Thus G satisfies the conditions of the General Burnside Problem. It remains to show that G is infinite. If G were finite, then the linear combinations of its elements would form a finite dimensional algebra B over F. Moreover, since both 1 and 1+a_i are in G, the linear combination (1+a_i) - 1 = a_i is in B. Thus, 1, a_1, a_2, a_3 are all in B. But 1, a_1, a_2, a_3 suffice to generate T/I, which was previously shown to be infinite dimensional. The algebra B is therefore also infinite dimensional, a contradiction. Thus G is infinite, as required.

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Written by uncudh

September 4, 2009 at 12:20 am