How do morphism of Groups be the same as the Group-Representation 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
$\tilde{u} : Rep(G_2,Mod(k)) \overset{\sim}{\longrightarrow} Rep(G_1, Mod(k))$,
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 Iso.
I am thinking this (seemingly easy) Problem for some time now, but i have no Idea how this works. Maybe someone can help me please?
 A: An element of $\phi\in Rep(G,Mod_R)$ is just a morphism $\tilde\phi:G\to Aut_R(M)$. 
We show that $u:G_1\to G_2$ must be injective. We first rewrite this into an equivalent statement. Let 
$$\alpha:G_1\to Aut(M)$$
 be a representation. The fact that $\tilde u$ is an isomorphism means that $\alpha$ factors as 
$$\alpha:G_1\stackrel{u}{\to}G_2\stackrel{\tilde u^{-1}(\alpha)}{\to}Aut(M)$$
Now let $g\in \ker u$. Then we see that for all $\alpha$ we have $\alpha(g)=\tilde u^{-1}(\alpha)(u(g))=\tilde u^{-1}(\alpha)(0)=id_M$, and hence for all module $M$ and for all $\alpha:G_1\to Aut(M)$ we have $\alpha(g)=id_M$. But then consider
$$\alpha:G\to Aut_R(R[G])$$
Where $R[G]=\oplus_{g\in G}R$, the free module over $R$ with basis $G$, and $\alpha:g\mapsto [1\cdot h\mapsto 1\cdot gh]$ (we specify the automorphism of $R[G]$ by giving the images of the basis). Then clearly $\alpha(g)=id_{R[G]}$ if and only if $g=1_G$, and hence by what we said above we conclude that we must have $\ker u=\{1_G\}$.
Surjectivity is similar.
