For each $n\geqslant 1$, there are uncountably many dense subgroups of $\mathbb{F}_p^I$ of index $p^n$. 
Let $p$ be a prime and let $I$ be an infinite set. We equip $F^I_p$ with the product topology (where $F_p$ is equipped with the discrete topology). Show that for each $n \geqslant 1$ there are uncountably many dense subgroups of $F^I_p$ of index $p^n$.

I have no idea about this problem. Any hints are appreciated. Thanks in advance.

Here's an answer you can like or not like depending on your background. Let $c$ be the continuum cardinal.

Proposition: for every infinite set $I$ and nontrivial finite group $G$, the product $G^I$ has at least $2^c$ normal subgroups of index $|G|$ (such that the quotient is isomorphic to $G$).

Proof: for every ultrafilter $\eta$ on $I$ the kernel of the map $G^I\to G$, $f\mapsto\lim_\eta f$ is a subgroup $K_\eta$ with $G^I/K_\eta\simeq G$, and $\eta\mapsto K_\eta$ is injective. Moreover since $I$ is infinite there are $\ge 2^c$ ultrafilters on $I$ (this is much more than just "uncountable"!).
This applies to your case since $\mathbf{F}_p^I$ is isomorphic to $(\mathbf{F}_p^n)^I$. But moreover in this case, the $K_\eta$ are ideals with quotient isomorphic to $\mathbf{F}_p^n$ as ring.

Edit: here's a linear algebra argument:
For $I$ infinite, $\mathbf{F}_p^I\sim\mathbf{F}_{p^n}^I$ as groups and now view the latter as vector space over $\mathbf{F}_{p^n}$, of cardinal $2^{|I|}$ and hence of dimension $2^{|I|}$ over $\mathbf{F}_{p^n}$. A linear form is determined by its restriction to a basis, and this restriction is arbitrary: so the set of linear forms is in bijection with $\mathbf{F}_{p^n}^{2^{|I|}}$, which has cardinal $2^{2^{|I|}}$. Except the zero map, the kernel of a linear form is a hyperplane, and since only proportional linear forms have the same kernel, we deduce that the set of hyperplanes has cardinal $2^{2^{|I|}}$. This gives as many subgroups of index $p^n$.
