Let $G$ be be a connected real algebraic reductive Lie group. Is it always possible to find finitely many maximal algebraic $\mathbf{R}$-tori $\{T\_i\}\_{i=1}^n$ such that the group generated by the $T\_i$'s is $G$.


P.S. The assumption for $G$ to be reductive is essential since $\mathbb{G}\_a$ does not contain any real torus. 


**added:** The algebraicity assumption was added in order to avoid one-parameter subgroups like $\gamma:\mathbf{R}^{\times}\rightarrow GL\_2(\mathbf{R})$:
$$
t\mapsto
\left(
\begin{array}{cc}
1 & \log|t| \newline
0 & 1
\end{array}
\right)
$$ 



Generating a reductive real Lie group with finitely many maximal real tori
At least for affine algebraic groups this is true. Indeed, let $G$ be a connected affine real algebraic group. Assume $G$ contains at least one diagonalizable element. Then diagonalizable elements are dense in $G$. Let $U$ be a neighborhood of 0 in the Lie algebra $g$ of $G$ such that $exp|U$ is a diffeomorphism. Choose a basis $e\_1,\ldots,e\_k\in U$ of $g$ that consists of diagonalizable elements.


Set $G\_k$ to be the set of the real points of the Zariski closure of $\{exp(te\_k)\mid t\in\mathbb{R}\}$. The map $\mathbb{R}^k\to G$ defined as $(t\_1,\ldots, t\_k)\mapsto exp(te\_1)\cdot\cdots\cdot exp(te\_k)$ is a diffeomorphism in a neighborhood of 0. So every element of $G$ that is sufficiently close to the unit can be written as a product of elements of $G\_i$'s. Since $G$ is connected, so can be any element of $G$. On the other hand, each $G\_i$ is a torus and as such is included in a maximal torus.


