I have frequently seen results like: There are 4 isomorphism types of finite subgroups of $SL_2(\mathbb{Z})$, namely $\mathbb{Z}_2,\mathbb{Z}_3,\mathbb{Z}_4,\mathbb{Z}_6$.

I wonder what is known of we replace $\mathbb{Z}$ by the ring of integers $\mathcal{O}_K$ of an algebraic number field $K$. I've found these groups discussed in several contexts, but I could not find my question discussed, although I expect of course it has been....any hint or reference would be greatly appreciated. Thanx!

More concretely I wonder whether the finite groups $SL_2(3)$ and $SL_2(5)$ do appear in this way as a finite subgroup, maybe for $SL_2(\mathcal{O}_K)$ for $K=\mathbb{Q}(e^{2\pi i /3}),\mathbb{Q}(e^{2\pi i /4}), \mathbb{Q}(e^{2\pi i /5}),\mathbb{Q}(e^{2\pi i /6})$.

  • $\begingroup$ Hopefully such results should be stated in another way: if you want to describe finite subgroups up to conjugacy, you don't expect to end up with a finite list of isomorphism type of subgroups: in general isomorphic subgroups are not conjugate, even if it's the case in $\mathrm{SL}_2(\mathbf{Z})$. $\endgroup$ – YCor Mar 17 '17 at 19:19
  • $\begingroup$ Of course you are right, thank you. I have edited the question accordingly. $\endgroup$ – Simon Lentner Mar 17 '17 at 20:11
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    $\begingroup$ $SL_2(\mathcal{O}_K)$ embeds into $SL_2(\mathbb{C})$; the finite subgroups of $SL_2(\mathbb{C})$ are well-known since Felix Klein. $\endgroup$ – abx Mar 17 '17 at 20:19
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    $\begingroup$ Also, if you are thinking about $GL_n(\mathcal{O}_K)$ for $n$ larger than $2$, then Jordan's theorem en.wikipedia.org/wiki/Jordan%E2%80%93Schur_theorem is relevant: For every $n$ there is an $N$ such that, if $G$ is a finite subgroup of $GL_n$, then $G$ is an extension $1 \to A \to G \to H \to 1$ with $A$ abelian and $\#(H) < N$. Requiring $G$ to lie in $GL_n(K)$ as well should put strong restrictions on $A$. $\endgroup$ – David E Speyer Mar 18 '17 at 1:13
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    $\begingroup$ abx's comment does not close the question of describing finite subgroups up to isomorphism: it remains to determine which ones of the finite subgroups of $\mathrm{SL}_2(\mathbf{C})$ appear. But I still believe that the question up to conjugacy is more natural. $\endgroup$ – YCor Mar 19 '17 at 2:36

I'll comment on the last question, with the group $G = {\rm SL}(2,5).$ All $2$-dimensional irreducible complex representations of $G$ are real valued, so all have Schur index dividing $2$ by a standard theorem ( Speiser?). Since $G$ has a quaternion subgroup of order $8$, the Schur index of such a representation is necessarily $2$. Since the ring of integers of each of the fields mentioned in the question is a PID ( by Minkowski for example, though that is overkill) , it suffices to check whether $G$ is isomorphic to a subgroup of ${\rm SL}(2,K)$ for each of the listed fields. However any field over which such a representation can be realised must certainly contain the field generated by its character. This field necessarily contains $\sqrt{5}$ as it contains $\omega + \overline{\omega}$ for $\omega = e^{\frac{2 \pi i}{5}}.$ Hence the only possible choice for the listed $K$ we need to concern ourselves with is $K = \mathbb{Q}[e^{\frac{2 \pi i}{5}}],$ since the other listed fields do not contain $\sqrt{5}.$

Now for this choice of $\omega$, we note that ${\rm SL}(2,K)$ contains a double cover $E$ of a dihedral group of order $10$, which is isomorphic to a Sylow $5$-normalizer of $G,$ namely $E = \langle \left(\begin{array}{clcr}0 & \omega\\-\overline{\omega} & 0\end{array}\right), \left(\begin{array}{clcr}\omega & 0\\0 & \overline{\omega}\end{array}\right) \rangle.$ Identifying $E$ with a Sylow $5$-normalizer of $G,$ and inducing the character $\mu$ of this representation, we find that ${\rm Ind}_{E}^{G}(\mu)$ has irreducible constituents of degree $2,4$ and $6.$ By the theory of the Schur index and the discussion above we know that the representation affording the $2$-dimensional constituent $\chi$ may be afforded over $K$ ( and hence over the ring of integers of $K$, since that ring of integers is a PID). For the Schur index $m_{K}(\chi)$ divides $2$, but we have exhibited a $K$-representation of $G$ which contains the character $\chi$ with multiplicity one. (Note that $K$ is not the field generated by the character values of $\chi,$ but a degree $2$-extension of that field, so this is consistent with earlier statements).

  • $\begingroup$ Though the more general approaches above were also very instructive to me...in particular abx comment on SL2(C) which I embarrassingly haven't seen. There of course I find all the groups I look for realiezed within different number fields. $\endgroup$ – Simon Lentner Mar 22 '17 at 10:31

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