Lazard Lie group has the same order statistics as the additive group of its Lazard Lie ring

Statement
Suppose $$G$$ is a Lazard Lie group and $$L$$ is the Lazard Lie ring for $$G$$, i.e., the Lie ring obtained by applying Lazard's theorem. Then, the additive group of $$L$$ has the same order statistics as $$G$$. In other words, $$G$$ and the additive group of $$L$$ are fact about::order statistics-equivalent finite groups.

Examples
Let $$p$$ be an odd prime. Consider groups of order $$p^3$$:


 * For the non-abelian group of exponent p, the additive group of the Lazard Lie ring is the elementary abelian group of order $$p^3$$. Thus, both these groups have the same order statistics.
 * For the non-abelian group of prime-squared exponent, the additive group of the Lazard Lie ring is the direct product of the cyclic group of order $$p^2$$ and the cyclic group of order $$p$$.

Converse

 * Same order statistics as abelian p-group not implies Lazard Lie group

Other related facts

 * Order statistics of a finite group determine whether it is nilpotent
 * Finite abelian groups with the same order statistics are isomorphic
 * Finite group having the same order statistics as a cyclic group is cyclic

Facts used

 * 1) uses::Logarithm map from Lazard Lie group to its Lazard Lie ring is a 1-isomorphism

Proof
This follows directly from fact (1), i.e., the fact that the bijective logarithm map from $$G$$ to $$L$$ restrict to isomorphisms on cyclic subgroups.