$\def\GL{\operatorname{GL}}\def\ZZ{\mathbb{Z}}\def\FF{\mathbb{F}}\def\Id{\mathrm{Id}}\def\fu{\mathfrak{u}}$Let $p$ be prime, let $n<p$, let $U_n(\FF_p)$ be the group of $n \times n$ upper triangular matrices with $1$'s on the diagonal, and let $\fu_n(\FF_p)$ be the $\FF_p$-Lie algebra of $n \times n$ upper triangular matrices with $0$'s on the diagonal. Then we have mutually inverse bijections $$ \exp : \fu_n(\FF_p) \to U_n(\FF_p) \quad \text{and} \quad \log : U_n(\FF_p) \to \fu_n(\FF_p)$$ given by the usual power series. These power series are actually polynomials since the matrices involved are nilpotent, and we never divide by $p$ since $n<p$.
Question: Suppose that $G$ is a subgroup of $U_n(\FF_p)$. Is $\log(G)$ a Lie-subalgebra of $\fu_n(\FF_p)$?
Motivation: See here for a related question where the answer turned out to be "no". I was motivated by this vexing question, but I no longer claim my approach there is useful. I wanted to turn the question of embedding groups into a question of embedding Lie algebras, and thus turn it into the question of effective bounds in Ado's theorem. In particular, I believed that I had a family of $2$-step nilpotent Lie algebras whose smallest faithful representation had dimension $\Omega(n^2)$. But, according to this survey on effective versions of Ado, for nilpotence degree $\leq 3$, $n+1$ always suffices.
Remark: The hard part is showing that $\log(G)$ is closed under addition. If so, I'm pretty sure I can show that it's closed under Lie bracket. I'll upload the details of this in the unlikely case that someone can do addition but not Lie bracket.
