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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?

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    Are you assuming that $\tilde u$ is an isomorphism for all rings $k$, or just for one specific ring $k$? – 2017-01-28
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    say a commutative ring with 1 – 2017-01-28
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    Oke, but it is an isomorphism for a single specific commutative ring with $1$, or for all commutative rings with $1$? – 2017-01-28
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    for every comm. ring with 1 – 2017-01-28
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    Oke, see my answer. Turns out having an isomorphism for a single ring is enough – 2017-01-28

1 Answers 1

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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.

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    Thank you very much. I have to admit, that i have some diff. understanding the last part after the kernel. I am just an undergrade. I realy thought, that it must have something to do with the Hom(-,B) Functor...!?! – 2017-01-28
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    @E.J.K. Do you know which ring I mean when I write $R[G]$? And which $\alpha$ I mean when I write $\alpha:G\to Aut_R(R[G])$? – 2017-01-28
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    maybe you could specify that one please, that would be great – 2017-01-28
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    @E.J.K. added some more details. don't hesitate to ask for clarification – 2017-01-28
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    would it be possible if you could post the surjectivity of the above lemma? that would be awesome. – 2017-03-02
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    Looking at the sequence $$\alpha:G_1\stackrel{u}{\to}G_2\stackrel{\tilde u^{-1}(\alpha)}{\to}Aut(M)$$ this just involves showing that given $g\in G_2$, there is some representation of $G_2$ such that $g$ acts non-trivially. This is the same argument as before. Let me know if you need more details. – 2017-03-06