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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?

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    $\begingroup$ Comparing their centers should work, I think. $\endgroup$
    – Thorgott
    Feb 17 at 23:39
  • $\begingroup$ @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 $\endgroup$
    – Ettore
    Feb 17 at 23:49
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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.

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    $\begingroup$ This distinguishes them topologically, but does it prove they are not isomorphic as (just) groups? $\endgroup$
    – Randall
    Feb 18 at 0:51
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    $\begingroup$ Whoops. I guess I missed part of the question. Anyway, the centres will do the trick. $\endgroup$ Feb 18 at 0:56
  • $\begingroup$ 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 $\endgroup$
    – Ettore
    Feb 18 at 8:58

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