$G$ is a connected linear algebraic group which is not solvable, and $T$ is a maximal torus of $G$, with $\textrm{Dim } T = 1$. $B$ is a Borel subgroup of $G$ containing $T$, and $U$ is the group of unipotent elements of $B$. The radical $R(G)$ is the unique maximal closed, connected, normal, solvable subgroup of $G$. Equivalently, $R(G)$ is the identity component of the intersection of all Borel subgroups of $G$. I have checked all the details of this lemma, except I am stuck on the last sentence. Why can't $R(G)$ contain a torus?
This lemma comes from the book Linear Algebraic Groups by T.A. Springer.
If $R(G)$ contained a torus, such a torus would automatically be maximal. So it would contain some conjugate $yTy^{-1}T$ of $T$. But $R(G)$ is normal in $G$, so this means that it would contain all maximal tori. I still don't see what the contradiction is.