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Exerc. 8.1.13 in D.J.S. Robinson, A Course in the Theory of Groups, 2d edition, p. 222, is as follows : Let F be a field, let $n$ be a natural number, let $G$ be a subgroup of $GL(n,F)$. Assume that the class number (number of conjugacy classes) of $G$ is finite. Then $G$ is finite.

I think I found a proof, but I wonder if it is the best possible. I use two lemmas :

Lemma 1. Let H be a group with finite class number $c$. Let $N$ a normal subgroup of finite index of $H$. Then the class number of $N$ is finite (and $\leq c \ [H:N]$).

Lemma 2. Let H be a soluble group wih finite class number $c$. Then $H$ is finite (and $\vert H \vert \leq c^{2^{s}-1}$, where $s$ is the derived length of $H$).

Now, I sketch the proof of the satement of the exercise. Without loss of generality, we may assume that $F$ is algebraically closed. The statement is equivalent to the same statement for a subgroup $G$ of $GL(V)$, where $V$ is a $F$-vector space of dimension $n$. We use induction on $n$. The case $n=0$ being trivial, we asume $n > 0$. Assume first that $G$ is irreducible. Since the class number of $G$ is finite, there is only a finite number of possible values for the traces of the elements of $G$. This, with the irreducibility of $G$, implies that $G$ is finite (Robinson, 8.1.9, p. 220). Thus the statement is true if $G$ is irreducible. If not, there is a subspace $U$ of $V$, invariant by $G$, with $0 < U < V$. Let $L_{1}$ denote the subgroup of $G$ formed by the elements of $G$ whose restriction to $U$ is identiy. Let $L_{2}$ denote the subgroup of $G$ formed by the element of $G$ acting on $V/U$ as the identity (with an evident definition). Put $L = L_{1} \cap L_{2}$. As in the proof of Robinson, 8.1.11, (i), p. 221, $L$ is a normal subgroup of $G$ and the induction hypothesis implies that $G/L$ is finite; moreover, $L$ is isomorphic to a subgroup of the group of unitriangular matrices in $GL(n,F)$, thus $L$ is nilpotent. By lemma 1, the class number of $L$ is finite. By lemma 2 (and the fact that $L$ is nilpotent and thus soluble), $L$ is finite, thus $G$ is finite.

I wonder if the lemmas are really needed, because Robinson doesn't speak of them, alhough they don't seem trivial to me.

Do you know a simpler proof ? Thanks in advance.

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Your line of reasoning is perfectly right, see also a spelled out proof in the book of T.Y. Lam, A First Course in Non-Commutative Rings, Theorem (9.5).

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