A. A proof for Linear groups

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
*g*_{1},..., *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(*r*^{n}) = 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^{ . }*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 *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 *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 *Q* we have roots of unity of
degree at most *n*^{ . }*d* over *Q*. There are only finitely many
such roots of unity. Thus the order of *G* is bounded.

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