How do morphism of Groups be the same as the Group-Representation

Question

I am trying to prove the folling Lemma

Say I am given a Morphism of Groups

$u:G_1 \longrightarrow G_2$

that induces an Isomorphism

$\tilde{u} :Rep(G2,Mod(k))⟶Rep(G1,Mod(k))$,

where Rep(-,Mod(k)) is the category of Representations of Groups into the category of Modules over some Ring $k$.

Then $u$ is also an isomorphism.

I am thinking this (seemingly easy) Problem for some time now, but i have no Idea how this works. Maybe someone could post a proof for that. It probably has something to do with the $Hom(-,B)$ functor (for some fixed B) but iam not shure.

Answer 1

I'll assume that $k$ is not the zero ring.

If $u$ is not injective, then $\tilde{u}$ is not essentially surjective, since no faithful representation of $G_1$ is in the image.

Suppose $u$ is injective but not surjective (so $G_1$ is a proper subgroup of $G_2$ and you're asking whether restriction of representations is an isomorphism of categories). Let $kG_2$ be the regular $kG_2$-module. Then there are $k$-module endomorphisms of $kG_2$ that are $kG_1$-module endomorphisms but not $kG_2$-module endomorphisms, so $\tilde{u}$ is not a full functor. For example, the endomorphism that is defined on the obvious basis by $\varphi(g)=g$ for $g\in G_1$ and $\varphi(g)=0$ for $g\in G_2\setminus G_1$.