The image of an Artin representation is finite First let us define the necessary terms for the problem:

Definition

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*Let $G$ be a profinite group and let $k$ be a topological field. An $n$-dimensional representation of $G$ over $k$ is a continuous homomorphism of groups $\rho: G \to \operatorname{GL}_n(k)$.


*An $n$-dimensional representation of $G$ over $k$ is called an Artin representation if $k \subset \mathbb{C}$, i.e. $k$ is a topological subfield of $\mathbb{C}$.

Problem: Now I would like to show that the image of an Artin representation is finite.
My own progress so far:

*

*Since $\require{enclose} \enclose{horizontalstrike}{G}$ is a profinite group we know that $\require{enclose} \enclose{horizontalstrike}{G}$ is isomorphic to an inverse limit of a projective system of finitely many groups $\require{enclose} \enclose{horizontalstrike}{\{G_1,\dots,G_m\}}$. In other words,

$$ \require{enclose} \enclose{horizontalstrike}{ G = \varprojlim_{i} G_i = \left\{ (\gamma_1, \dots, \gamma_m) \in \prod_{i=1}^m G_i \: \middle| \: f^j_i(\gamma_j)=\gamma_i \: \forall j \geq i \right\}.}$$
That means that we can insert $m$-tuples $\require{enclose} \enclose{horizontalstrike}{ (\gamma_1,\dots,\gamma_m) }$ in our presentation $\rho$ which gives us elements $\require{enclose} \enclose{horizontalstrike}{ \rho(\gamma_1,\dots,\gamma_m) \in \operatorname{GL}_n(k) }$.
Edit: I made a huge mistake! In a projective limit, the groups are finite, not the system of groups! Now I have no idea at all again.
But now I do not know how to continue so solve this problem anymore.
My specific questions (besides "How can I show this statement?"):

*

*Why exactly do I need that $k$ is a topological subfield in $\mathbb{C}$?

*How do I handle the group homomorphisms $f^j_i : G_j \to G_i$ in our projective system? Do they play an important role for our problem?

Any help is really appreciated.
 A: This comes down to the "no small subgroup argument."
Lemma: Let $H = \operatorname{GL}_n(k)$, where $k$ is a topological subfield of $\mathbb C$.  There exists an open neighborhood $U$ of the identity matrix $I$ which contains no subgroups of $H$ besides the trivial subgroup.
A consequence of the lemma is that if $\rho: G \rightarrow \operatorname{GL}_n(k)$ is a continuous homomorphism, then the kernel of $\rho$ is an open subgroup of $G$.  Since $G$ is profinite, it is compact, so the kernel of $\rho$ is of finite index in $G$.  This immediately implies that the image of $\rho$ is finite.
To see that the kernel of $\rho$ is open, take an open neighborhood $U$ as in the lemma, and consider the preimage $\rho^{-1}(U)$, which is an open neighborhood of $1_G$.  It's a general fact about profinite groups that inside any open neighborhood of the identity, you can find an open subgroup.  So let $N$ be an open subgroup of $G$ contained in $\rho^{-1}(U)$.  Then $\rho(N)$ is a subgroup of $\operatorname{GL}_n(k)$ which is contained in $U$.  By the hypothesis on $U$, $\rho(N)$ must be the trivial group, i.e. $N$ must be contained in $\operatorname{Ker}\rho$.  Thus $\operatorname{Ker} \rho$ contains an open subgroup, so it is itself open.
Proof of the lemma: It suffices to prove the lemma in the case $k = \mathbb C$, since if $U$ is an open neighborhood of $I$ which works for $\operatorname{GL}_n(\mathbb C)$, then $U \cap \operatorname{GL}_n(k)$ is one which works for $\operatorname{GL}_n(k)$.
The case $n =1$ (where you are working with $\operatorname{GL}_1(\mathbb C) = \mathbb C^{\ast}$) is obvious.
For arbitrary $n$, I have seen several proofs of this.  My favorite one comes from differential geometry.  The same link provides alternative proofs.  
