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 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 $p$.

Related facts

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. 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.