This is a question from Stillwell's "Naive Lie Theory" (3.5.2) which I am self studying.
There is a less explicit form of the question was asked here, but I don't fully grasp the answer and would like to be able to answer it in the terms it was asked.
torus in $SU(2)$ yields a torus in $SO(3)$
To be explicit, here is the actual question:
Explain why a $\mathbb{T}^2$ in $SU(2)$ yields a $\mathbb{T}^2$ in $SO(3)$, so $\mathbb{T}^1$ is maximal in $SU(2)$ Hint: Map each element $g$ in the $\mathbb{T}^2$ in $SU(2)$ to the pair $\pm g$ in $SO(3)$ and look at the images of the $\mathbb{S}^1$ factors of $\mathbb{T}^2$.
I do at least know $SO(2)/({\pm I})\cong $SO(3)$
Each $T^1$ in SO$(3)$ consists of the rotations about a fixed axis. So each $T^1$ is conjugate to the set of matrices of the form $$R_t=\pmatrix{\cos t&\sin t&0\\-\sin t&\cos t&0\\0&0&1}.$$ One gets a $T^2$ by putting together two $T^1$s that commute. One can do this in SO$(4)$: $$\pmatrix{\cos t&\sin t&0&0\\-\sin t&\cos t&0&0\\ 0&0&\cos u&\sin u\\0&0&-\sin u&\cos u}$$ for varying $t$ and $u$ is a $T^2$ in SO$(4)$. But in SO$(3)$ one can't do that. One can assume that one $T^1$ consists of the $R_t$. But the only rotations that commute with all the $R_t$ are the $R_t$ so you cannot get a second $T^1$.
Moving to SU$(2)$, each $T^1$ there is the inverse image of a $T^1$ in SO$(3)$. As there are no commuting $T^1$s in SO$(3)$, neither are there any in SU$(2)$.