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).

  • $\begingroup$ $U$ isn't unipotent. For $G=SL_2$, $U=G$. $\endgroup$ – Jack Schmidt Feb 26 '14 at 18:46
  • $\begingroup$ For instance $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \in U_1$ and $\begin{bmatrix} 1& 0 \\1 & 1 \end{bmatrix} \in U_{-1}$ then their product $\begin{bmatrix}2 & 1\\1&1\end{bmatrix}$ has minimal polynomial $x^2-3x+1 \neq (x-1)^2$, so is not unipotent. $\endgroup$ – Jack Schmidt Feb 26 '14 at 18:48
  • $\begingroup$ @JackSchmidt Oh yes, rookie mistake : $U$ is only generated by unipotent elements… Tanks a lot. $\endgroup$ – Pece Feb 26 '14 at 18:51
  • $\begingroup$ No problem. :-) $\endgroup$ – Jack Schmidt Feb 26 '14 at 18:51
  • $\begingroup$ @JackSchmidt One thing : for $G=SL_2$, if $U = G$, then the maximal torus is trivial ; but isn't $\{ \begin{bmatrix} \lambda & 0 \\ 0 & \lambda^{-1} \end{bmatrix} \}$ a maximal torus of $SL_2$ ? $\endgroup$ – Pece Feb 26 '14 at 18:54

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.

  • $\begingroup$ By the way, do we have any criterion for $G_u$ (the set of unipotent elements of $G$) to be a subgroup of $G$ (not necessarily reductive here) ? I know it is the case if $G$ is connected and solvable by Lie-Kolchin's theorem (in fact $G_u$ is even normal then). $\endgroup$ – Pece Feb 26 '14 at 19:21
  • $\begingroup$ I don't know of any, but I tend to work in finite groups and there the condition is called having a normal Sylow $p$-subgroup, so is pretty rare. $\endgroup$ – Jack Schmidt Feb 26 '14 at 19:31

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