Maps from a profinite group to $\operatorname{GL}_n(\mathbb R)$. Suppose $G$ is a profinite group (a topological group which is compact, Hausdorff and totally disconnected). Let $r:G \to \operatorname{GL}_n(\mathbb R)$ be a group homomorphism.
How to prove: $r$ is continuous $\Leftrightarrow$ $\ker(r)$ is open in $G$?
I want to prove $\operatorname{Im}(r)$ is discrete when $r$ is continuous, but I don't know how to achieve it.
 A: The simplest argument to this is probably something called a "small subgroup argument". Anyway, here's a different approach:
Let $G$ be a profinite group and let $H$ be a (real) Lie group and let $f:G \to H$ be a continuous homomorphism. I claim that $\mathrm{Im}(f)$ is finite. We can endow $K:=\mathrm{Im}(f)$ with the subspace topology. $K$ is compact and hence closed in $H$. Thus by a theorem of Cartan in Lie theory, $K$ is a Lie subgroup of $H$.
On the other hand, the canonical map $G/\mathrm{ker}(f) \to K$ is a continuous bijection from a compact space to a Hausdorff space and hence a homeomorphism, thus an isomorphism of topological groups. $\mathrm{ker}(f)$ is closed and thus $G/\mathrm{ker}(f)$ is profinite. Thus $K$ is profinite.
Finally, every manifold is locally connected and every profinite group is totally disconnected. But a space that is both totally disconnected and locally connected must be discrete. Thus $K$ is discrete, and by compactness, finite. This implies that $\mathrm{ker}(f)$ is of finite index and thus (as it is also closed) open.
A: This is a useful argument called the no-small subgroups argument.
Lemma: any sufficiently small open neighborhood of the identity in $GL_n(\mathbb{R})$ contains no subgroups.
The lemma is clear since if $M$ is any matrix whose determininant is slightly greater than $1$, then $M^n$ has unbounded determinant.
The result follows: say $r$ is a continuous homomorphism. Now $G$, being profinite, has a neighborhood basis $\{ G_n \}$ of open subgroups of $1$. Suppose that none of the $G_n$ satisfies $r(G_n) = \{1\}$. Pick a neighborhood $U$ of the identity in $GL_n(\mathbb{R})$ not containing any of the $r(G_n)$. Then $r^{-1}(U)$ is open but doesn't contain any $G_n$, contradiction.
