Exponent of finitely generated group Let $G$ be a finitely generated group with generators $X=\{g_1,g_2,\dots,g_m\}$ and $n$ be a positive integer such that the $n$-th power of every element in $X$ is the identity.
Is it true that $\mathrm{exp}(G) | n$? Clearly for abelian groups it is true. I think that it can develop to other specific groups.
 A: No - there exist finitely generated torsion groups with unbounded exponent.
The Burnside problem asks if there exists a finitely-generated torsion group. Golod and Shafarevich proved that there does exist such a group, but the group they construct has elements of unbounded order. Novikov and Adian proved that there exists such a group of bounded order, while Ol'shanskii proved that there exists groups where every proper, non-trivial subgroup  has order $p$ for some fixed prime $p>>1$ (such groups are called Tarski Monsters).
The links are all from wikipedia, as the original papers are all in Russian. Ol'shanskii wrote a book containing his proof, which you can try and read (although it is pretty close to unreadable...). It is called "Geometry of Defining Relations in Groups".
EDIT: I should say that we don't have to reach as far as infinite groups. For example, the group of symmetries of an icosahedron can be generated by and element $a$ of order two and an element $b$ of order three. However, $ab$ has order five. Of course, then one can take $a, b, ab$ as a generating set, or more generally the whole group as a generating set, to side-step this issue. One cannot do this with the infinite torsion groups I mentioned.
A: An easy counterexample is given by any symmetric group of degree $\geq 3$.
Each symmetric group is generated by the set of its transpositions, which gives $n = 2$. However starting with $S_3$, the exponent of the symmetric group is clearly not a divisor of $2$.
