## Posts Tagged ‘**group theory**’

## The General Burnside Problem

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

**Bounded Burnside Problem**. If is a finitely generated group and there is an integer such that for every , then must be finite?

The problem also has the following variant:

**General Burnside Problem**. If is a finitely generated group and every element of has finite order, then must 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 is a finitely generated algebra over a field and every element of is nilpotent, then must be nilpotent?

(A element is nilpotent if for some ; and is itself nilpotent if there is an such that for all .) The Golod-Shafarevich construction also provides a counter-example to Kurosh’s Problem. We sketch the construction below.

**The Golod-Shafarevich Theorem
**

Let be a field and let be the free non-commutative algebra over generated by the variables . Let denote the subspace of consisting of all linear combinations of monomials of degree . The elements of are the homogeneous elements of degree . Let be a two-sided ideal of generated by a set of homogeneous elements, each of degree at least 2. Suppose that has at most elements of degree for . The following (which we do not prove here) is the main computational result of the Golod-Shafarevich construction:

**Theorem**. The quotient algebra is infinite dimensional over if the coefficients in the power series expansion of are nonnegative.

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

**Counterexample to Kurosh’s Problem**

Let be the free algebra over a countable field . Let be the ideal of consisting of all elements of without constant term. Enumerate the elements of as . Choose an integer and write , where each . Choose another positive integer sufficiently large so that for some . Continue in this way for sufficiently large powers of . Now let be the ideal generated by the defined in the process above. Consider the quotient . The construction of guarantees that each element of is nilpotent; but the theorem above ensures that is infinite dimensional over (and hence not nilpotent). Thus 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 is a prime number and is the field with elements. Let and be as defined above. Let be the elements of the quotient respectively. Let be the multiplicative semigroup in generated by the elements , , and . An element of has the form for some . The element is nilpotent by the construction of , and so for sufficiently large , we have . Since we are in characteristic we have . It follows that has an inverse, whence is a group. Moreover every element of has finite order (indeed order a power of ). Thus satisfies the conditions of the General Burnside Problem. It remains to show that is infinite. If were finite, then the linear combinations of its elements would form a finite dimensional algebra over . Moreover, since both 1 and are in , the linear combination is in . Thus, are all in . But suffice to generate , which was previously shown to be infinite dimensional. The algebra is therefore also infinite dimensional, a contradiction. Thus is infinite, as required.