Unipotent groups have central series with quotients embedded in G_a

By [F1] fix a closed embedding $j:G\hookrightarrow U_n$. Put $G_i=G\cap U_n^{(i)}$ for $0\le i\le m$, where $U_n^{(i)}$ is the central series of [F2]; the $G_i$ are closed subgroup schemes of $G$ with $G_0=G$ and $G_m=1$.

The series is central scheme-theoretically. For every commutative $k$-algebra $R$, the commutator of $g\in G(R)$ and $h\in G_i(R)$ lies in $U_n^{(i+1)}(R)$ by [F2], and it lies in $G(R)$. Hence it lies in $G_{i+1}(R)$, the fibre-product intersection. These factorizations are natural in $R$, so the commutator morphism $G\times G_i\to G$ factors through $G_{i+1}$. The same argument gives conjugation stability of each $G_i$, since $gh{g}^{-1}=[g,h]h\in G_i(R)$ for every $R$. Thus the terms are closed normal subgroup schemes and $[G,G_i]\subseteq G_{i+1}$ by the minimality definition in [F3].

Restrict to $G_i$ the homomorphism $U_n^{(i)}\to U_n^{(i)}\mathrm{/}{U}_n^{(i+1)}\cong {\mathbf{G}}_a$. Its scheme kernel is exactly $G\cap U_n^{(i+1)}=G_{i+1}$. By [F4] its represented quotient $G_i\mathrm{/}{G}_{i+1}$ is the closed image in ${\mathbf{G}}_a$. This proves the successive-quotient assertion for nonreduced groups as well.

By induction ${\gamma }_{i+1}G\subseteq G_i$: the case $i=0$ is equality, and if ${\gamma }_{i+1}G\subseteq G_i$, step 2.1 makes the commutator morphism $G\times {\gamma }_{i+1}G\to G$ factor through $G_{i+1}$, so its smallest closed subgroup image ${\gamma }_{i+2}G$ lies there by [F3]. Hence ${\gamma }_{m+1}G=1$ and $G$ is nilpotent. Likewise $D^{0}G=G_0$, and if $D^{i}G\subseteq G_i$, the commutator morphism on $D^{i}G\times D^{i}G$ factors through $G_{i+1}$ by step 2.1; the definition of the derived subgroup [F5] gives $D^{i+1}G\subseteq G_{i+1}$. Thus $D^{m}G=1$, proving solvability. All factorizations were checked on every base algebra, so no smoothness or reducedness is needed.