Direct proof that closed subgroups of profinite groups are profinite Each of the references that I check prove first the relatively involved characterization of profinite groups as compact Hausdorff totally disconnected topological groups, and then appeal to the fact that the closed subsets of a compact Hausdorff space are precisely the compact subsets, in order to establish the claim in my title. Is there, however, a direct proof of this?
For example, if $H \leq G$ is closed and $G$ is the inverse limit of groups $G_i$ with morphisms $f_i : G \to G_i$, then can we directly prove that $H$ is the inverse limit of the $\text{im } f_i|_H$? By the universal property, it would've been enough to show that $H$ is saturated with respect to each of the $f_i$, but I know that's not true if $H$ isn't open...
 A: Yes there is:
Suppose $(G_i, f_{i,j}, I)$ is an inverse system of finite groups. Their inverse limit $G$ is a subset of the product $\prod_{i \in I}G_i$. Denote the projection maps $G \to G_i$ as $\pi_i$ for each $i \in I$.
Denote $H_i = \pi_i(H) \leq G_i$, a finite group. For each $j \geq i$ $f_{i,j} \circ \pi_j = \pi_i$ so that $f_{i,j}(H_j) = f_{i,j}\circ \pi_j (H) = \pi_i(H) \subset H_i$. It's now easy to see that $(H_i, f_{{i,j}_{|H_j}}, I)$ is an inverse system and $H$ along with the maps $\pi_{i_|H}$ is compatible with it.
Denote the inverse limit of the latter system as $C \subset \prod_{i \in I}H_i$ and note it is clear that $H \leq C$.
You need a result (its proof does not use the alternative characterisation of profinite groups) that says:

If $Y$ is a subset of $C$ such that $\pi_{i_|H}(Y) = H_i$ for each $i
\in I$, then $Y$ is dense in $C$.

(this can be found in Wilson's Profinite Groups proposition 1.1.6)
Clearly $H$ has this property, and because it is closed, it is $C$.
