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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.

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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.

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    $\begingroup$ 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. $\endgroup$ Jan 13 '19 at 18:54

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