Does there exist a finitely presented (preferably $\text{FP}\_{\infty}$) group $\Gamma$ and an element $\alpha \in \text{H}^{\ast>0}(B\Gamma;\mathbf{Q})$ that is not nilpotent?


If non-discrete groups were allowed, the Euler class $e \in \text{H}^2(BS^1;\mathbf{Q})$ would do the trick, and there are corresponding classes in $\text{H}^2(BC\_p;\mathbf{F})$ for $\mathbf{F}$ a finite field of chracteristic $p$, and $C\_p$ cyclic of order $p$.


One approach I thought of is to apply the Kan–Thurston theorem. But the (uncountable) groups appearing in their construction cannot be easily replaced by finitely generated ones, unless the complex one starts with is of low dimension. See the second half of sub-section 2.2 in their [paper](https://www.sciencedirect.com/science/article/pii/0040938376900409?via%3Dihub).



A finitely presented group whose rational cohomology is not nilpotent
Let me compile the comments into an official answer: yes, such a group exists. As @dodd predicted in a comment, Thompson's group $F$ does the trick. Brown's computation of the cohomology ring ([http://pi.math.cornell.edu/~kbrown/papers/homology.pdf](http://pi.math.cornell.edu/%7Ekbrown/papers/homology.pdf)) reveals non-nilpotent elements, e.g., the element he denotes $u$, which lives in degree 2 and generates a divided polynomial ring. (Thanks to @BenjaminSteinberg for the comment making me realize things were a lot more straightforward than I initially thought.)


