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Let be $G \subset GL_n(\mathbb{C})$ such that there is some $r \in \mathbb{N}^{*}$ and $g_1, \ldots, g_r \in G$ so that for all $g \in G$, $g$ is conjugated to some $g_i, i \in [[1, r]]$ in $G$.

Is $G$ finite? I feel like that yes, I tried to poke the stabilizer, but I have no reason to think that it'd be finite.

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    $\begingroup$ Is the conjugation by elements of $G$ or by elements of $\mathrm{GL}_n$? $\endgroup$ – Mathmo123 Jan 27 '19 at 11:21
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    $\begingroup$ If conjugacy by elements of $GL_n$, consider the group of all $\begin{pmatrix}1&z\\0&1\end{pmatrix}$, which is isomoirphic to $\Bbb C$. $\endgroup$ – Hagen von Eitzen Jan 27 '19 at 11:23
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    $\begingroup$ @Raito that's still a bit unclear. Certainly the $g_i$ are in $G$. By assumption, if $g\in G$, then $g = hg_i h^{-1}$ for some $h$. My question is: is $h\in G$, or is $h\in \mathrm{GL}_n$? $\endgroup$ – Mathmo123 Jan 27 '19 at 11:25
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    $\begingroup$ @Raito you said that $g$ and the $g_i$ are in $G$, you also said they are conjugated, i.e., that $g=hg_ih^{-1}$ for some $h$, but Mathmo's question is whether $h\in G$ $\endgroup$ – Hagen von Eitzen Jan 27 '19 at 11:25
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    $\begingroup$ @HagenvonEitzen Right, sorry, the conjugacy is in G. $\endgroup$ – Raito Jan 27 '19 at 11:26
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$G$ must be finite. Here's a proof working for $G$ arbitrary (Derek's proof works only when $G$ is finitely generated).

Passing to a finite index subgroup, we can suppose that $G$ has connected Zariski closure.

The assumption implies that $G$ has finitely many traces, and using connectedness of the Zariski closure, it has a single trace. So $\mathrm{Tr}(g(g'-g''))=0$ for all $g,g',g''\in G$.

If we assume in addition that $G$ is irreducible and hence linearly generates the space of matrices, since the Trace yields a nondegenerate bilinear form, we deduce that $g'-g''=0$ for all $g',g''\in G$, i.e., $G=\{1\}$.

In general, this applies to the projection on all irreducible diagonal blocks, which are thus one-dimensional.

In other words, this proves that if $G$ is a subgroup of $\mathrm{GL}(n,\mathbf{C})$ with finitely many traces, then $G$ has a finite index subgroup $H$ (namely, the intersection with the identity component of its Zariski closure) that is conjugated to a group of upper triangular unipotent matrices.

If $G$ has finitely many conjugacy classes, so does $H$. Then $H$ has a normal series in which all subquotients are torsion-free abelian groups (it inherits this from the group of unipotent upper triangular matrices). This implies that if $H\neq\{1\}$ then it has an infinite abelian quotient (namely a nontrivial torsion-free abelian quotient). This implies that $H$ has an infinite number of conjugacy classes.

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Yes, $G$ must be finite.

There are only finitely many finite groups (up to isomorphism of course) with a bounded number of conjugacy classes (see here for a proof).

But it is well known that linear groups are residually finite, and so if there were an infinite example, then it would have arbitrarily large finite quotients, each with only $r$ conjugacy classes.

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  • $\begingroup$ "Linear groups are residually finite": this is only true for finitely generated groups. $\endgroup$ – YCor Jan 31 '19 at 6:07

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