# If $G \subset GL_n(\mathbb{C})$ has finitely many conjugacy classes for all element, is it finite?

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.

• Is the conjugation by elements of $G$ or by elements of $\mathrm{GL}_n$? – Mathmo123 Jan 27 '19 at 11:21
• If conjugacy by elements of $GL_n$, consider the group of all $\begin{pmatrix}1&z\\0&1\end{pmatrix}$, which is isomoirphic to $\Bbb C$. – Hagen von Eitzen Jan 27 '19 at 11:23
• @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 $h\in \mathrm{GL}_n$? – Mathmo123 Jan 27 '19 at 11:25
• @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 $h\in G$ – Hagen von Eitzen Jan 27 '19 at 11:25
• @HagenvonEitzen Right, sorry, the conjugacy is in G. – Raito Jan 27 '19 at 11:26

$$G$$ 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.

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.

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