torus in $SU(2)$ yields a torus in $SO(3)$

Question

in John Stillwell's book "Naive Lie Theory" there is an exercise to explain why a torus in $SU(2)$, (sub group that is isomorphic to $S^1 \times S^1$) yields a torus in $SO(3)$ (in order to prove that there is no torus in $SU(2)$).
I'm assuming that you have to use the fact that \begin{equation*}SU(2)/\{\pm1 \}\cong SO(3) \end{equation*} but I can't see how to use it.

Answer 1

Let $T$ be a torus in $SU(2)$. If $-1 \notin T$ then the image of $T$ in $SU(2)/ \{\pm1\}$ under projection is isomorphic to $T$, because else the projection would not be injective, meaning the there is a $\sigma \in SU(2)$ such that $\pm \sigma \in T$ with implies $-\sigma\sigma^{-1} = -1 \in T$, a contradiction.
Now if $-1\in T$. Since $S^1/\{\pm1\} \cong S^1$ by the map taking each element to its square and $-1 \in T$ is identified with some element of the form $(\pm1, ..., \pm1)$ in $T^n$, the n-dimensional torus since all elements of the Ker of the squaring map take this form. Therefore by looking under the identification at each coordinate we have $T/\{\pm1\} \cong T \cong T^n$.
I hope this helps