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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.

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    $\begingroup$ R.Thompson's group $F$? $\endgroup$
    – markvs
    Commented Feb 4, 2021 at 13:52
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    $\begingroup$ Here's Brown's computation of the cup product structure for $F$: pi.math.cornell.edu/~kbrown/papers/homology.pdf (I don't know enough about cohomology to actually answer your question, but this paper should hopefully clarify whether $F$ works as an example). $\endgroup$
    – Matt Zaremsky
    Commented Feb 4, 2021 at 14:11
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    $\begingroup$ Awesome, thanks! $\endgroup$
    – Jens Reinhold
    Commented Feb 4, 2021 at 14:57
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    $\begingroup$ It seems @MattZaremsky's link answers the question in Theorem~5.1 since it gives a subring which is a divided powers polynomial ring which seems to give non-nilpotent elements in degree 1 $\endgroup$
    – Benjamin Steinberg
    Commented Feb 4, 2021 at 18:05
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    $\begingroup$ This probably doesn't shed much light on the actual question, but does the fact that $H^\ast(BC_p ; \mathbb F_p)$ contains a polynomial subring imply that every finite group's cohomology has nonnilpotent elements in positive degree when the coefficients are of characteristic not coprime to the group order? $\endgroup$
    – Tim Campion
    Commented Feb 5, 2021 at 2:55

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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) 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.)

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    $\begingroup$ Thanks for correcting me. I had somehow missed that the divided powers polynomial ring was living in even degrees when I wrote my comment $\endgroup$
    – Benjamin Steinberg
    Commented Feb 5, 2021 at 3:19
  • $\begingroup$ @Benjamin Steinberg: Note that cohomology rings of groups are graded commutative. So elements of odd degree are either 2-torsion or square to zero. $\endgroup$
    – tj_
    Commented Feb 7, 2021 at 11:49

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