Let $R$ be a DVR with fraction field $K$ and a perfect residue field $k$. Consider a commutative connected group scheme $G$ over $R$, say smooth and of finite type over $R$. So we may use Chevalley's Theorem on algebraic groups and obtain a unique exact sequence $ 0 \to H_K \to G_K \to A_K \to 0, $ where $H_K$ is a connected affine group scheme and $A_K$ is an abelian variety. $H_K$ contains a maximal torus $T_K$, more exactly $H_K \cong T_K \times U_K$, where $U_K$ is unipotent and both are unique. This is the argumentation in [2] on p. 4. The same argumentation can be applied to the special fiber $G_k$ of $G$. Denote by $U_k$ the unique unipotent subgroup of $G_k$ obtained by the above method.
Now the authors claim, that $U_K = 0$ iff $U_k = 0$.
Question: Why is that true?