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.
Recall that in the product toppology, sets of the form $$\tag1U_{i,a}:=\{\,(x_\iota)_{\iota\in I}\colon x_i=a\,\}$$ are open, as are their finite intersections $$\tag2\bigcap_{r=1}^m U_{i_r,a_r}$$ and ultimately the arbitrary (though in this example: wlog. finite) unions of sets of form $(2)$. In order to be dense, our desired subgroup must intersect every non-empty open set, or equivalently must intersect every set of form $(2)$.
If $H$ is of index $p^n$ in $\Bbb F_p^I$, then clearly $\Bbb F_p^I/H\cong \Bbb F_p^n$, which is a nice vector space, and $H$ can be described as kernal of a homomoprhism to a finite-dimensional vector space. Let's start with the basic case $n=1$. What kind of onto homomorphisms $\Bbb F_p^I\to \Bbb F_p$ can we think of? The simplest ones are projections to some coordinate, $(x_\iota)_{\iota\in I}\mapsto x_{\iota_0}$. That would make $ H=\{\,(x_\iota)\mid x_{\iota_0}=0\,\}$, so $H\cap U_{\iota_0;1}=\emptyset$, and thus $H$ is unfortunately not dense. We need a different kind of homomorphism!
On the powerset $\mathcal P(I)$, define an equivalence relation $A\sim B\iff |A\mathop\Delta B|<\infty$. Then every equivalence class has only countably many members, hence $J:=\mathcal P(I)/{\sim}$ is as uncountable as $\mathcal P(I)$ is. For each $j\in J$ except the one with $\emptyset\in j$, we choose a representative $A_j\in j$ and define $e_j=(e_{j;\iota})_{\iota\in I}$ as $$e_{j;\iota}=\begin{cases}1&\iota\in A_j\\0&\iota\notin A_j\end{cases} $$ Additionally, we have for each $i\in I$ a vector $f_i$ given by $$f_{i;\iota}=\begin{cases}1&\iota=i\\0&\iota\ne i\end{cases} $$ Then these $e_j$, $f_i$ are linearly independent: If $$\tag3\sum_{j\in J} a_je_{j}+\sum_{i\in I}b_if_i=0,$$ where only finitely many coefficients are non-zero, then for each $j$ with $a_j\ne 0$, the summand $a_je_j$ is non-zero in infinitely many components where none of the other summands of the first sum is non-zero. As the second sum is non-zero only in finitely many components, at least one component remains where the left hand side of $(3)$ is non-zero, contradiction. Therefore $a_j=0$ for all $j\in J$. But for the second sum it is directly clear that $b_i=0$ for all $i\in I$, as was to be shown.
Thus (using the Axiom of Choice a second time) we can extend the collection of all $e_j$ and all $f_i$ to a basis $\{e_j\}_{j\in J}\cup \{f_i\}_{i\in I}\cup \{h_k\}_{k\in K}$ of $\Bbb F_p^I$. Now pick distinct $j_1,\ldots, j_n\in J$ and let $H$ be the kernel of $\Bbb F_p^I\to \Bbb F_p^n$, $$\sum_{j\in J}a_je_j+\sum_{i\in I}b_if_i+\sum_{k\in K}c_kh_k\mapsto (a_{j_1},\ldots, a_{j_n}). $$ Then $H$ is of index $p^n$ and intersects every set of form $(2)$ because we can pick any $x=\sum a_je_i+\sum b_if_i+\sum c_kh_k\in H$ and adjust the coefficient $b_i$ to achieve any desired value in the $i$th component, and repeat this for the indices $i_r$, $1\le r\le m$ given in $(2)$.
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 $p^n$.
