# Isomorphism of representation induced by a morphism of groups

Let $$G_1$$,$$G_2$$ be groups. The question is to prove that if $$u: G_1 \rightarrow G_2$$ induces an isomorphism $$\overline{u}: \operatorname{Hom}(G_2,Gl(M)) \rightarrow \operatorname{Hom}(G_1,Gl(M))$$ for all $$M$$ module over $$\mathbb{Z}$$, then $$u$$ is an isomorphism.

I proved that is true for $$G_1$$, $$G_2$$ abelian, but I'm in trouble proving it is true in general.

Thanks.

First, we prove that $$u$$ is injective. Let $$M$$ be the free abelian group with basis indexed by $$G_1$$. Then there is an injective morphism $$G_1\to Gl(M)$$ given by the action of $$G_1$$ on $$M$$ by permutation of its basis elements. But if $$u$$ is not injective, then no element in the image of $$\bar u$$ is injective. Thus $$u$$ has to be injective, otherwise $$\bar u$$ would not be surjective.
Next, we prove that $$u$$ is surjective. Assume that it is not, and consider the set $$G_2/u(G_1)$$ of cosets of $$u(G_1)$$ in $$G_2$$. Let $$M$$ be the free abelian group with basis indexed by $$G_2/u(G_1)$$. The group $$G_2$$ acts on $$G_2/u(G_1)$$ by translation of cosets; therefore, there is a corresponding non-trivial morphism $$\phi:G_2\to Gl(M)$$ given by the action of $$G_2$$ on $$M$$ by permutation of its basis elements. Note that $$\bar u (\phi)$$ is the trivial morphism, since the elements of $$u(G_1)$$ act trivially on $$M$$. Thus, $$\bar u$$ is not injective, a contradiction.
Therefore $$u$$ is bijective, so it is an isomorphism of groups.
• Thanks you. Anyway I bypassed the abelian case with the fact that $Hom(G, Gl(M))$ is isomorphic to $Hom(\mathbb{Z}[G], End(M))$ and so i can prove an isomorphism between $\mathbb{Z}[G_1]$ and $\mathbb{Z}[G_2]$ , but it is induced by $u$ and so $u$ is isomorphism. Jan 13 '19 at 18:54