Nilpotent group and exponent If $G$ is a (possibly infinite) nilpotent group generated by elements of bounded order, does it follow that $G$ has finite exponent?
 A: The answer seems to yes, but the only proof I can think of uses some deep results. According to Theorem 2.4 of Primož Moravec, Schur multipliers and power endomorphisms of groups. J. Algebra 308 (2007), no. 1, 12–25 (there is a version of this that can be freely downloaded), if a locally finite group $G$ has exponent $n$, then the exponent of the Schur Multiplier $M(G)$ is bounded by a function $f(n)$ of $n$. The proof of this result uses the positive solution to the Restricted Burnside Problem.
Given this, the result in the question is straightforward to prove by induction on the nilpotency class of $G$. It is true for abelian groups, and for nonabelian $G$, if $N$ is the last nontrivial term in the lower central series of $G$, then $G/N$ has exponent bounded by some number $e$ by inductive hypothesis and, since $N$ is a quotient of $M(G/N)$, $N$ has exponent at most $f(e)$, and hence $G$ has exponent at most $ef(e)$. (The fact that $G$ is locally finite also follows by induction.)
Note that it is a standard result that the torsion elements of a nilpotent group form a subgroup, from which it follows that $G$ is a torsion group. In particular, if $G$ is finitely generated nilpotent and generated by torsion elements, then it is finite.
