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

Statement

Adjoint group version

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., $[x,y]:=xy-yx$.

Algebra group version

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., $[x,y]:=xy-yx$.