# 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?

• Comparing their centers should work, I think. Feb 17 at 23:39
• @Thorgott Oh! I thought both centers were isomorphic to $\mathbb Z/2$ but actually I'm not quite sure about $Z(SO(3))$ now, I'll try to work out the details tomorrow Feb 17 at 23:49

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 did find the centers of $SU(n)$ as an exercise a while ago but I was somehow convinced that $SO(3)$ had nontrivial center too! Thanks for the thorough answer Feb 18 at 8:58