# Lazard Lie ring of adjoint group of a radical ring equals associated Lie ring of the radical ring under suitable nilpotency assumptions

Suppose $N$ is a radical ring such that there exists a positive integer $n$ so that the additive group of $N$ has powering threshold at least $n - 1$ and $x^n = 0$ for all $x \in N$. Suppose $N$ has adjoint group $G$. Suppose that $G$ is a Lazard Lie group. Then, the Lazard Lie ring of $G$ is isomorphic to the Lie ring associated to $N$, i.e., the Lie ring whose additive group is the same as that of $N$ and whose Lie bracket is the additive commutator in $N$, i.e., $[x,y] := xy - yx$.
Suppose $q$ is a power of a prime $p$ and $N$ is a nilpotent associative algebra over the finite field $\mathbb{F}_q$. Suppose, further, that $x^p = 0$ for all $x \in N$.
Let $G$ be the algebra group corresponding to $N$. Then, $G$ is a Lazard Lie group. Then, the Lazard Lie ring of $G$ is isomorphic to the Lie ring associated to $N$, i.e., the Lie ring whose additive group is the same as that of $N$ and whose Lie bracket is the additive commutator in $N$, i.e., $[x,y] := xy - yx$.