Examples of groups which are virtually isomorphic but not commensurable Let $G_1, G_2$ be groups. We say $G_1$ and $G_2$ are commensurable if there exist finite index subgroups $H_1 \leq G_1$, $H_2 \leq G_2$ such that $H_1 \simeq H_2$.
We say $G_1$ and $G_2$ are virtually isomorphic if there exist finite index subgroups $H_1 \leq G_1$, $H_2 \leq G_2$, and finite normal subgroups $N_1\trianglelefteq H_1$, $N_2 \trianglelefteq H_2$ such that $H_1 / N_1 \simeq H_2 / N_2$.
It is easy to see that commensurability implies virtual isomorphism, we just take the normal subgroups to be trivial. How can one go about finding a counterexample to the converse?
 A: There are examples of groups $G$ and central extensions
$$
1\to A\to \hat{G}\to G\to 1
$$
such that $A$ is finite and every finite index subgroup of $\hat{G}$ contains $A$, so that $G$ is not commensurable to $\hat{G}$ but is virtually isomorphic to it. A good example to work out is $G=PSL(2, {\mathbb R})$ and $\hat{G}=SL(2, {\mathbb R})$. In fact, this $G$ is simple as an abstract group (see here) and, thus, contains no proper finite index subgroups whatsoever. Similarly, $\hat{G}$: Its only proper normal subgroup is its center.  
There are similar (but more complicated) examples with $G$ finitely generated. See for instance: 
P. Deligne, Extensions centrales non résiduellement finies de groupes arithmétiques, C. R. Acad. Sci. Paris, série A–B, 287 (1978), 203–208.
R.M. Hill, Non-residually finite extensions of arithmetic groups. Res. Number Theory 5 (2019).
J. Millson, Real vector bundles with discrete structure group,Topology, 18 (1979), 83–89.
M. S. Raghunathan, Torsion in cocompact lattices in coverings of Spin (2,n), Math. Annalen, 266 (1984), 403–419.
