In a solvable algebraic group all maximal tori are conjugate to each other

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I want to prove the following statement:

All maximal tori in a solvable algebraic group are conjugate to each other.

I use the following facts:

  1. Theorem. Let $G$ be an irreducible solvable algebraic group and $T$ a torus complementary to its unipotent radical $U$. Then any semisimple element of $G$ is conjugate to some element of $T$.
  2. Lemma. Any torus has elements which are not contained in any of its proper algebraic subgroups. I call ad-hoc these element "bagels".

My idea is as follows:

Let $t_1 \in T_1$ be a bagel of a maximal torus $T_1$. It is semisimple. Then by Theorem 1 it is conjugate to some element of another torus, say $T_2$. Then $g t_1 g^{-1} \in T_2$. Since $g t_1 g^{-1}$ is also a bagel (I say that it is true because conjugation is an automorphism), then $T_2$ is maximal (since otherwise, if $T_2$ is a subgroup of another torus $T_3$ (i.e. $T_2 \subset T_3$) then $g t_1 g^{-1}$ is not a bagel since it is contained in a proper subgroup of a torus). We then can apply the same argument on bagels of $T_2$ and every other maximal torus.

I would like to know rather my argument is valid and if there are any gaps I need to close.

Added in Edit:

One gap I have found: I need to reduce the case of solvable algebraic group $G$ with maximal torus $T_1$ to a case of irreducible algebraic $H$ with $T$ is complementary to $U(H)$ = unipotent radical of $H$, in order to apply Theorem 1. How do I fix it?

I read in some book that if $U$ is the unipotent radical of $G$ then $T$, its complementary torus, is a maximal torus and the group $G$ containing it is irreducible. We also have $G = U \rtimes T$, so I think this close the first gap I mentioned.

Added in Edit 2:

I think I managed to proved this elementwise. However, I need to prove a stronger result:

let $T$ be the maximal torus complementary to the unipotent radical $U$. Then for every other maximal torus, $T'$, there exists $g \in G$ such that $ g T' g^{-1} = T $ (that is, $g$ is global to all the elements of the torus).

Added in Edit 3:

The globality of the conjugating element follows from that a torus is a group of commuting semisimple matrices and therefore can be simultaniously diagonalized.