A. A proof for Linear groups

Let G be a finitely generated subgroup of the group of invertible n×n matrices over complex numbers. We give a proof for the Ordinary Burnside problem first.

Let R be the collection of complex linear combinations of elements of G. The R is a finite dimensional vector space over the field of complex numbers spanned by the elements of G; thus there are elements g₁,..., g_r of G which form a basis of R.

Now suppose r is an element of R such that the traces Trace(g_i ^. r) = 0 all vanish. Then we obtain Trace(rⁿ) = 0 for all positive integers n by expressing r^{n - 1} as a linear combination of the g_i. But then these identities imply that r = 0. Thus an element g of G is uniquely determined once we know Trace(g_i ^. g) for all i (if Trace(g_i ^. g) = Trace(g_i ^. h) then apply the above argument to r = g - h).

Now we are given that each element of G satisfies g^N = e. Thus the trace of any element of G is a sum of n numbers of the form exp(2$\pi$$\sqrt{-1}$ ^. k/N) for k = 1,..., N. But there are only finitely such sums. Thus by the previous paragraph there are only finitely many elements in G. (Exercise: use the above argument to provide an explicit bound).

We now show how to reduce the General Burnside Problem to the Ordinary Burnside Problem in this case. Let K be the field generated (over the field $\Bbb$Q of rational numbers) by the matrix coefficients of the finite collection of generators of G. Let L be the subfield of K consisting of all algebraic numbers (elements satisfying a polynomial with rational coefficients). Since K is finitely generated L is a finite extension of $\Bbb$Q.

Now any element g of G has finite order. Hence the eigenvalues of g are roots of unity. Moreover, the characteristic polynomial of g has coefficients in the field K; since its roots are algebraic numbers the coefficients are in L. Thus the eigenvalues are roots of unity satisfying a polynomial of degree n over L; hence if d is the degree of the field extension L of $\Bbb$Q we have roots of unity of degree at most n ^. d over $\Bbb$Q. There are only finitely many such roots of unity. Thus the order of G is bounded.

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