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