Do you have an example of an infinite group K and a finite subgroup G such that the inclusion induces an isomorphism in cohomology?

Unfortunately, the link to Johnson's paper doesn't work any more. I would very much appreciate, if you could renew it. Thanks in advance.

@Theo Johnson-Freyd: Yes.

In fact (up to p=3) it isn't injective! For, since all maximal subgroups (in the exponent p case) are elementary abelian of rank 2, the kernel of the product map is the essential cohomology. Now a theorem of Minh ("Essential cohomology and extraspecial p-groups") says that the essential cohomology is non trivial, except in case p=3.

In general, one only knows that the kernel of the product of restrictions to maximal elementary abelian subgroups has nilpotent kernel. Do you have a reference for the stated injectivity?

"such as a group algebra over a finite field": Should that mean that all group algebras over a finite field are local or should it mean that there are group algebras over finite fields that are local? Note that a group algebra kG over a field of char. p and G finite is local iff G is a p-group.

It seems that your proof depends on the axiom of choice (or, at least, some weaker form of it).

... Conversely, let $K$ be an algebraic extension of $\mathbb{F}_p$. Let $M$ be a maimal ideal of $R$ containing $p$. Then $R/M$ is an algebraic closure of $\mathbb{F}_p$. Hence we can assume $K \le R/M$. Let $\pi: R \to R/M$ and let $S = \pi^{-1}(K)$. Then $L=S/(M \cap S)$. qed

The subring can be taken inside the algebraic integers iff $K$ is an algebraic extension of $\mathbb{F}_p$. For, let $R$ be the ring of algebraic integers in $\mathbb{C}$. Since any subring $S$ of $R$ is an algebraic extension of $\mathbb{Z}$, the quotient of $S$ by a maximal ideal $m$ is an algebraic extension of $\mathbb{F}_p$ where $(p)=\mathbb{Z}\cap m$. ...

An algebraic closure of $\mathbb{F}_p$ is given by $\bar{\mathbb{F}}_p = R/M$ where $R$ is the ring of algebraic integers and $M$ any maximal ideal containing $p$.

@YCor: "Every field automorphism of R is continuous": Even more is true: The identity is the only field automorphism of $\mathbb{R}$.