In this wikipedia article about $SU(1,1)$, it is stated that
This group [$SU(1,1)$] is isomorphic to $SO(2,1)$ and $SL(2,\mathbb ℝ)^{[17]}$
I'm confused by the relation of $SO(2,1)$ and $SL(2,\mathbb R)$. I know that they are locally isomorphic as Lie groups (as their Lie algebras are isomorphic). Topologically, they are different - $SL(2,\mathbb R)$ is connected (by row-echelon-reduction), where $SO(2,1)$ is not connected (there is the component fixing the upper hyperboloid and the component interchanging the hyperboloids). Therefore, they are not isomorphic as Lie groups.
So, are $SO(2,1)$ and $SL(2,\mathbb R)$ isomorphic as abstract groups? Why (not)?
The source $^{[17]}$ (Gilmore's Lie Groups, Lie Algebras and some of their Applications, p.201-205) from wikipedia seems only to show an isomorphism between the Lie algebras.
This question does not show that there is no isomorphism, but gives a 2-1-map, which is not surjective.
They are not isomorphic as abstract groups: $SL(2, {\mathbb R})$ has nontrivial center ($\pm I$), while $SO(2,1)$ has trivial center.
Edit. Here is how to see that $SO(2,1)$ has trivial center.
Since 3 is an odd number, every element $g\in SO(2,1)$ has at least one real eigenvalue, $\lambda$. Let $E_\lambda$ denote the corresponding eigenspace in ${\mathbb R}^3$. Since $3$ is an odd number, $E_\lambda$ is a proper subspace of ${\mathbb R}^3$ unless $g=I$. The centralizer of $g$ in $GL(3,{\mathbb R})$ has to preserve $E_\lambda$. This is a pleasant linear algebra exercise which works in all dimensions:
If $g_1, g_2\in GL(n,{\mathbb R})$ commute and $g_1 v= \lambda v$, $v\in E_\lambda$, then $g_2 g_1 v= \lambda g_2 v= g_1 g_2(v)$, since $g_1, g_2$ commute. Hence $$ g_1(g_2 v)= \lambda (g_2 v), $$ i.e. $g_2v\in E_\lambda$.
Thus, if $g$ were central in $SO(2,1)$, then $SO(2,1)$ would have an invariant line or plane $E_\lambda$. Using the invariance of the Lorentzian inner product, we see that $SO(2,1)$ then would have an invariant line in ${\mathbb R}^3$ (if $E_\lambda$ is a plane, take its Lorentzian orthogonal complement). The orbit of every nonzero vector $v$ under $SO(2,1)$ is either the null-cone (if the vector is null) or the set of positive vectors of the given "length" $\langle v,v\rangle$ (if the vector is positive) or the set of negative vectors of the given "length" (if the vector is negative). In any case, it is not contained in a line, thus, $SO(2,1)$ has trivial center.
The faulty wikipedia page was corrected.