$T_1 \times T_2$ is a maximal torus? I have been working on teaching myself matrix groups and I have come across a problem about maximal tori. If I have a torus, $T_1 \subset G_1 $ and it is the maximal torus and If I have a torus, $T_2 \subset G_s $ and it is the maximal torus. Does that mean $T_1 \times T_2$ is the maximal torus of $G_1 \times G_2$? I want to take advantage of the fact that the n-dimensional torus is the group given by $T^n := U(1) \times U(1) \times U(1)$ and so on. So I know that torus created by  $T_1 \times T_2$ will be in the same dimensions as  $G_1 \times G_2$. And thus there cannot be a higher dimensional torus. Is this the right?
Thanks
 A: The answer is yes. Suppose $\pi_{1}, \pi_{2}$ are the projections onto the two factors $G_{1}$, $G_{2}$ of $G_{1} \times G_{2}$. Let $T$ be a maximal torus of $G_{1} \times G_{2}$. Then, since homomorphic images of tori are tori, $\pi_{1}(T), \pi_{2}(T)$ must be tori in $G_{1}, G_{2}$ respectively. Hence, there must be maximal tori $T_{1}', T_{2}'$, of $G_{1}, G_{2}$ respectively, containing these two tori. 
Since maximal tori are all conjugate, we have $g_{1}, g_{2}$ such that 
$$g_{1}\pi_{1}(T)g_{1}^{-1} \subseteq g_{1}T_{1}'g_{1}^{-1} = T_{1}$$
and similarly, 
$$g_{2}\pi_{2}(T)g_{2}^{-1} \subseteq g_{2}T_{2}'g_{2}^{-1} = T_{2}.$$
Hence, conjugating $T$ by $(g_{1}, g_{2})$, and calling the new maximal torus, $T'$, we can see that $T' \subseteq T_{1} \times T_{2}$. Now, $T'$ is a maximal torus, $T_{1} \times T_{2}$ is a torus. Thus, by maximality, $T' = T_{1} \times T_{2}$. Hence, $T_{1} \times T_{2}$ is a maximal torus of $G_{1} \times G_{2}$. 
Note that you do have to be careful about what you mean by maximal tori: there is a difference in the use of the term in compact Lie groups, noncompact Lie groups and complex groups. In the complex groups $GL_{n}(\mathbb{C})$, for example, the maximal torus is the entire group of diagonal matrices, which is isomorphic to $\mathbb{C^{*}}^{n}$ not $U(1)^{n}.$
