$H^1(G,A)\rightarrow H^1(H,A)$ is onto

Question

Note that $H^1(G,A)\rightarrow H^1(H,A)^{G/H}$ where $A$ is $G$-module and $H$ is a subgroup in $G$

But I suspect that ${\rm Res}\ : \ H^1(G,A)\rightarrow H^1(H,A)$ may be onto

since $ f\in C^n (G,A)\mapsto f|_{H\times \cdots \times H} \in C^n(H,A)$ is onto.

Do I miss some ?

Answer 1

No, the map isn't onto in general. It's easy to construct counterexamples.

In fact, every non-abelian p-group (more general: nilpotent group) yields a counterexample. For, let $G$ be a non-abelian p-group. Since each non-trivial normal subgroup intersects the center non-trivially, we find a central subgroup $1 \neq C \subseteq [G,G] \subsetneqq G$.

$\text{Res}: H^1(G,\mathbb{F}_p) \to H^1(C,\mathbb{F}_p)=H^1(C,\mathbb{F}_p)^{G/C}$ is the restriction $Hom(G,\mathbb{F}_p) \to Hom(C,\mathbb{F}_p)$ which is zero because $f \in Hom(G,\mathbb{F}_p)$ vanishes on $[G,G]$.