Are $SO(3)$ and $SU(2)$ abstractly isomorphic? The groups $SO(3)$ and $SU(2)$ are usually "the" classic example of nonisomorphic Lie groups whose corresponding Lie algebras are isomorphic. While I understand the isomorphism of their algebras, it was not obvious at all to me that we must have $SU(2) \not\cong SO(3)$, I think one can conclude that by looking at the topological structure, for example the fact that their fundamental groups are nonisomorphic should suffice (I think).
What if we forget about the topology and consider them just as abstract groups? I've been trying to find some obstruction but all the facts I know about them, for example $SO(3) \cong SU(2)/Z(SU(2))$, don't seem to be enough! Am I missing something obvious?
 A: The standard way to distinguish these two is that $\pi_1(\operatorname{SU}(2))=0$ while $\pi_1(\operatorname{SO}(3))=\Bbb{Z}/2$. Actually, more is true. You can realize $\operatorname{SU}(2)$ as the unit quaternions (hence diffeomorphic to $S^3$), and $\operatorname{SO}(3)=\operatorname{SU}(2)/Z(\operatorname{SU}(2))$ realizes $\operatorname{SU}(2)\to\operatorname{SO}(3)$ as a $2-$fold covering map. From what we know about the topology of $S^3$ we get indeed that $\operatorname{SU}(2)$ is simply connected and hence that $\pi_1(\operatorname{SO}(3))\cong \Bbb{Z}/2.$
Clearly, I misread part of the question. Sorry – we can analyze the centres of these groups to find that $Z(\operatorname{SO}(3))=\{I_3\}$, while $Z(\operatorname{SU}(2))=\{\pm I_2\}$. This is a good exercise to do for yourself. The fact about $\operatorname{SU}(2)$ follows from thinking about the identification as quaternions, while the fact for $\operatorname{SO}(3)$ can be understood by thinking about $\operatorname{SO}(3)$ acting on $S^2$. You want to show that the centres are of the form $\{\text{Scalar matrices}\}\cap G$.
I think there is also a nice proof of this using Schur's Lemma.
