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Claim: let $R$ be a principal ideal domain and $Q$ be its quotient field. Then every finite subgroup of $GL_n(Q)$ is conjugated to a subgroup of $GL_n(R)$.

So I want to solve this exercise but I really have no idea where to start (how to use the fact that $R$ is PID for example). I kinda see that the request for the subgroup to be finite is probably relevant to get a common denominator (gcd?) but it's just a qualitative observation. I would really appreciate some input (I'm not really asking for a full solution). Is any representation theory knowledge needed?

Martin Brandenburg
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F13
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    Don't know the answer offhand, by my intuition would be to look at the LCM of denominators in all entries. Perhaps start with some simple examples, like $R=\mathbf Z$ or a DVR like $\mathbf Z_{(p)}$. – tomasz Sep 14 '24 at 13:46
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    Hint: start with $n=1$. – Martin Brandenburg Sep 14 '24 at 13:51
  • @MartinBrandenburg wait maybe I'm leading towards a weird direction but: if $n=1$ then for every $q \in Q$ then $q^{-1}Gq=G$ so the conjugacy classes are trivial. This is really weird, because this would require that every finite subgroup of $Q$ is actually a subgroup of $R$. I'll try to prove this but I am not really convinced about my own reasoning. – F13 Sep 14 '24 at 14:35
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    $G$ is acting on a vector space $V$ with basis $B$. The $R$-submodule $M$ spanned by ${gb :g \in G, b \in B}$ is $G$-invariant. $M$ is torsion-free and hence (since $R$ is a PID) free. – Derek Holt Sep 14 '24 at 16:00
  • @F13 Your approach for $n=1$ looks fine so far. – Martin Brandenburg Sep 14 '24 at 18:25
  • @MartinBrandenburg so I managed to prove that a finite PID is a field so in particular $R=Q$. This is obviously false for infinite ones (example: $\mathbb{Z}$), and in this case I'm finding it hard to come up with a simple answer. – F13 Sep 16 '24 at 14:22
  • Wait I may got it. Suppose $R$ is infinite, then also $Q$ is infinite because $R \subseteq Q$. Then consider $G \leq Q*$ and suppose that $G$ is finite and $q \in G \setminus R$. Then in particular $ \leq G$ so $q$ has finite order call that $n$. I have $q^n = 1 = q^{-n}$. Call $p = q^{-1}$ then $p^n = 1$ so $p \cdot p^{n-1} = 1$ so $q = p^{n-1} \in R$: contraddiction. – F13 Sep 16 '24 at 14:35

1 Answers1

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The subgroup $G \subseteq \mathrm{GL}_n(Q)$ can be seen as an action of $G$ on the vector space $Q^n$ over $Q$. Choose the standard basis $e_1,\dotsc,e_n$ of $Q^n$. We may also regard $Q^n$ as an $R$-module, and it is torsionfree. Consider the $R$-submodule $M \subseteq Q^n$ generated by the $g e_i$ for $g \in G$, $1 \leq i \leq n$. It is a finitely generated torsionfree $R$-module, hence it is free (by the classification of f.g. modules over PIDs). Notice $M \otimes_R Q = Q^n$ (since it contains all $e_i$). This shows that the rank of $M$ over $R$ is $n$. Let's say $b_1,\dotsc,b_n$ is a basis of $M$ over $R$. Now $G$ also acts on $M$ by $R$-linear automorphisms. This means we get a homomorphism $G \to \mathrm{Aut}_R(M) \cong \mathrm{GL}_n(R)$, let's say the image is $H \subseteq \mathrm{GL}_n(R)$. Define the $Q$-linear isomorphism $\sigma : Q^n \to Q^n$ by mapping $b_i \mapsto e_i$, or equivalently, $\sigma \in \mathrm{GL}_n(Q)$. Then $G = \sigma H \sigma^{-1}$.

Martin Brandenburg
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  • Hi thank you very much for you answer and your time. This exercise is supposed to be solvable without involving modules and related theory. Anyways I'm curious about those things so I may start giving a look at these topics by myself – F13 Sep 15 '24 at 13:07