I'm currently reading Carter's Introduction to algebraic groups and Lie algebras and the following example is confusing me:
"let $ G = GL_n(k)$. Then $G$ is a connected reductive group. The diagonal subgroup $T=D_n(k)$ is the maximal torus in G. The subgroup $B = T_n(k)$ of upper-triangular matrices is a Borel subgroup of G and the subgroup $ U = U_n(k)$ of upper unitriangular matrices is the unipotent radical of $B$..."
As far as I understand the definition of a 'reductive group' is one for which $R_u(G) = 1$, where $R_u(G)$ is the maximal closed connected normal subgroup of $G$ all of whose elements are unipotent.
If this is correct then how can $U$ be the unipotent radical of $B$ which is also a subgroup of $G$, while $G$ has unipotent radical $1$?
i.e. how is it that $R_u(B) = U$ when $U \leq G$ and $R_u(G) = 1$?, is it literally just because $U$ is normal in $B$, but not in $G$? Or have I misunderstood?