A question about reductive groups's structure. There is a flaw somewhere in the following argument but I can't track it.
Take a reductive connected affine algebraic group $G$ : by definition, its unipotent radical $R_u(G)$ is trivial. One have the following (non trivial) result about reductive groups :

Let $T$ be a maximal torus of $G$. Denote by $(U_\alpha)_{\alpha \in \Phi}$ an enumeration of the minimal closed unipotent normalized by $T$ subgroups of $G$. Then any closed connected subgroup $H$ of $G$ containing $T$ is generated by $T$ and the $U_\alpha$ that $H$ contains.

(Actually, the theorem is far stronger than that, but this part suffices for my purpose.)
In particular, $G$ is a closed connected subgroup of $G$ containing $T$ ; as it contains all the $U_\alpha$, one must have
$$ G = \langle T, U_\alpha : \alpha \in \Phi \rangle. $$
Denote $U = \langle U_\alpha : \alpha \in \Phi \rangle$ : this the subgroup of all unipotent elements of $G$. In particular, $U$ is a unipotent subgroup of $G$ and is closed (embedding $G$ into some $\mathrm{GL}_n$, this is the subgroup of $G$ satisfying the equation $(g-1)^n=0$). It is also connected as all the $U_\alpha$ are. But it is also normal in $G$ : $T$ normalizes all the $U_\alpha$, so normalizes $U$ ; and $U$ certainly normlizes itself. So at the end, $U$ is a closed connected normal unipotent subgroup of $G$, and hence must lie into $R_u(G) = \{1\}$.
This show that every reductive group is a torus… which is certainly false.

Edit. Jack Schmidt solved it in the comments. Actually $U$ is not unipotent (and is not the set of all unipotent elements of $G$ as I said) : it is just generated by unipotent elements. The set $G_u$ of all unipotent elements of $G$ is not necessarily a group, so the product of unipotent elements is not necessarily unipotent (see the example in $SL_2$ of Jack Schmidt in the comments).
 A: The subgroup generated by two unipotent subgroups need not be itself unipotent, since the product of two unipotent elements need not be unipotent.
For example, let $G = \operatorname{SL}_2(K)$ for some field $K$, and choose $T=\left\{ \begin{bmatrix} \lambda & 0 \\ 0 &\lambda^{-1} \end{bmatrix} : \lambda \in K^\times \right\} \cong K^\times$ to be the maximal torus. Then there are only two non-identity closed unipotent subgroups of $G$ that are normalized by $T$: $$U_1 = \left\{ \begin{bmatrix} 1 & \lambda \\ 0 & 1 \end{bmatrix} : \lambda \in K \right\} \cong K_+, \qquad U_{-1} = \left\{ \begin{bmatrix} 1 & 0 \\ \lambda & 1 \end{bmatrix} : \lambda \in K \right\} \cong K_+$$
Then $$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0\\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$$ has minimal polynomial $$(x-2)(x-1)-1 = x^2-3x+1 \neq (x-1)^2,$$ so it is not unipotent, even though it is a product of unipotent elements.
In case of $\operatorname{SL}_2$ we actually get a simpler equality: $G=\langle U_1, U_{-1} \rangle$. The subgroup $T$ is redundant here.  However for $\operatorname{GL}_2$ it is needed, and also I believe for $\operatorname{PGL}_2$ over many fields.
