Is $H^4(PSL(2,\mathbb{Z}),\mathbb{Z})$ known? I ask this in response to the recent calculation of the same cohomology group for $\mathrm{Co}\_0$ and $\mathrm{Co}\_1$.



Fourth cohomology of the modular group
From Jeremy Rickard's comments, the group cohomology (with coefficients a module with trivial action) of a free product of (discrete) groups is sent to direct sum (eg Proposition 1.3.16.3 in C. Löh, *[Group Cohomology & Bounded Cohomology](https://wwwmath.uni-muenster.de/u/clara.loeh/topologie3_ws0910/prelim.pdf)* (pdf)), so 
$$
H^n(PSL(2,\mathbb{Z}),\mathbb{Z}) \simeq H^n(C\_2\ast C\_3,\mathbb{Z})\simeq H^n(C\_2,\mathbb{Z})\oplus H^n(C\_3,\mathbb{Z}),
$$ 
and positive, even-degree cohomology of cyclic groups is $H^{2k}(C\_n,\mathbb{Z}) = \mathbb{Z}/n$ ($k\gt 0$) and $0$ in odd degree, hence $H^4(PSL(2,\mathbb{Z}),\mathbb{Z}) = \mathbb{Z}/2 \oplus \mathbb{Z}/3$.


