Exponent of associated Lie ring divides exponent of group

Statement
Suppose $$G$$ is a group of finite exponent and $$L(G)$$ is the associated Lie ring of $$G$$. Then, the additive group of $$L(G)$$ is also a group of finite exponent and the exponent of this additive group divides the exponent of $$G$$.

In particular, if $$G$$ has exponent equal to a positive integer $$m > 1$$, this forces that $$L(G)$$ has the structure of a $$\mathbb{Z}/m\mathbb{Z}$$-Lie algebra.

Note that the exponents of the two groups need not be equal in general.

Related facts

 * Order of associated Lie ring equals order of group for nilpotent group
 * Nilpotency class of associated Lie ring equals nilpotency class of quotient of group by nilpotent residual