I found the following exercise in lectures notes from group cohomology. Let $G$ be a finite group and $P$ be a Sylow $p$-subgroup. Let $N_G(P)$ be the normalizer of $P$ in $G$. If $P$ is abelian, there is an isomorphism $$ H^*(G; \mathbb{F}_p) \cong H^*(N_G(P); \mathbb{F}_P)$$ induced by the restriction map.
We know that the composite $tr_{N}^G \circ res_{N}^G : H^*(G;\mathbb{F_p}) \rightarrow H^*(G;\mathbb{F}_p)$ is the multiplication by $[G:N_G(P)] \equiv 1$ mod $p$ (this is a consequence of the Sylow theorems).
Using that $P$ is Abelian, by a classical result of Burnside, $N_G(P)$ controls $p$-fusion in $G$ (namely, any two conjugated elements of $P$ in $G$ are conjugated in $N_G(P)$), but I do not see how to use this fact to conclude that the restriction map is an isomorphism.