TITLE: Examples of a group $G$ and an $F$-representation $V$ where $\cup:H^1(G,F)\otimes H^1(G,V)\to H^2(G,V)$ is injective
QUESTION [5 upvotes]: Let $G$ be a group and $F$ a field. I am particularly interested in the case where $G$ is a uniform lattice in a Lie group and $F=\mathbb{F}_2$, or in finite groups $G$ where
$\operatorname{char} F$ divides
$|G|$, but the discussion applies to general $G$ and $F$.
Let $\rho:G\to \mathrm{GL}(V)$ be a representation of $G$ on a finite-dimensional $F$-vector space $V$. Letting $G$ act trivially on $F$, we may form the cohomology ring $H^*(G,F)$, which acts via the cup product on the graded module $H^*(G,V)$. In particular, the cup product induces a linear map
$$
p: H^1(G,F)\otimes_F H^1(G,V)\to H^2(G,V).
$$
My question is whether there are  $G$ and $V$ for which:

$p$ is injective,
$\dim H^1(G,V) \geq 1$,
$\dim H^1(G,F)\geq 2$.

More generally, is there a recipe for constructing such examples? I am not aware of even a single example of this kind.
The injectivity of $p$ is the real issue. For example, if we take $V=F$, then the cup product $\cup :H^1(G,F)\times H^1(G,F)\to H^2(G,F)$ is well-known
to be anti-symmetric, which means that
$\dim \ker p\geq \frac{1}{2}\dim H^1(G,F)(\dim H^1(G,F)-1)$ if the characteristic of $F$ is $2$, or $\dim \ker p\geq \frac{1}{2}\dim H^1(G,F)(\dim H^1(G,F)+1)$ if $F$ is not of characteristic $2$, so $\ker p$ would not be trivial when $\dim H^1(G,F)\geq 2$.
(This special example is studied in a paper of Hillman.)

REPLY [3 votes]: $\newcommand{\bZ}{\mathbb{Z}}$Let $G$ be a finite group of order divisible by $p:=\mathrm{char}\, F$ such that $\dim_F \mathrm{Hom}(G,F)\geq 2$ (e.g. $G=\bZ/p\times\bZ/p$). Take $V$ to be the kernel of the augmentation map $e(\sum a_g\cdot g)=\sum a_g$ from $F[G]$ to $F$. Since  $F[G]$ is an injective $G$-module, the short exact sequence $0\to V\to F[G]\xrightarrow{e} F\to 0$ yields isomorphisms $H^i(G, F)\simeq H^{i+1}(G,V)$ for $i\geq 0$ (this is automatic for $i>0$ and for $i=0$ follows from the fact that $e:F[G]^G\to F$ is the zero map because the order of $G$ is divisible by $p$).
These isomorphisms are compatible with the cup-product in the sense that the following diagram commutes for every $i$ (this follows from the associativity of cup-product using the observation that the isomorphism $H^i(G,F)\to H^{i+1}(G,V)$ is obtained by cupping with a class in $H^1(G,V)$)
$\require{AMScd}$
\begin{CD}
H^1(G,F)\otimes H^{i+1}(G,V) @>\cup >> H^{i+2}(G,V)\\
@|   @|\\
H^1(G,F)\otimes H^{i}(G,F) @>>\cup > H^{i+1}(G,F)
\end{CD}
For $i=0$ the bottom map is obviously an isomorphism, hence the top map is an isomorphism, as desired.
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