Exercise 18, § 5

A group G is called locally finite if every subgroup of G that admits a finite generating family is finite.

a) Let G be a group and H a normal subgroup of G. Prove that if H and $G/H$ are locally finite, the same holds for G.

b) Deduce from a) that a solvable group whose elements are all of finite order is locally finite.

c) Let K be a commutative field, V a finite-dimensional vector space over K, and G a subgroup of Aut(V) that admits a finite generating family consisting of elements of finite order. Prove that the order of the elements of G is bounded.

(Reduce to the case when K is a purely transcendental extension of a prime field P by considering the subfield K generated by the coefficients of a finite generating family of G. Then, prove that for every element $s$ of $\mathbf{G}\mathbf{L}_n(K)$ of finite order, we have $\varphi (s)\leqslant n$ if $P =\mathbf{Q}$ and $s\leqslant p^n-1$ if $P =\mathbf{F}_p.$) d) Prove that every subgroup H of Aut(V) whose elements are all of finite order is locally finite.[^1]

(Reduce to the case when K is algebraically closed, and use induction on dim(V). Let G be a subgroup of H that admits a finite generating family. If V is a simple K[G]-module, apply Exercise 17; if V contains a nontrivial subspace W that is stable under G, consider the image of G in Aut(V) $\times$ Aut(V$/W$), and apply a) and b)).

[^1]: For every odd integer $n\geqslant 665$, there exists a group in which the order of every element divides $n$ but that is not locally finite; cf. S. I. Adian, The Burnside problem and identities in groups, Ergebnisse der Math. 95, Springer-Verlag (1979). An analogous result holds when $n\geqslant 2^{48}$ and $2^9$ divides $n$ (cf. S. V. Ivanov, On the Burnside problem for groups of even exponent, Proceedings of the International Congress of Mathematicians, Vol. II (Berlin, 1998). Doc. Math. 1998, Extra Vol. II, 67–75).