If $G$ is a compact topological group, how to show that a finite index subgroup of $G$ is open?

Question

If $G$ is a compact topological group, how to show that a finite index subgroup of $G$ is open ? I really don't know where/how to start...

PS : I precise that by "compact" I mean that it is hausdorff and that any recovering of $G$ by open sets has a finite sub-recovering (sorry if I wasn't unclear)

Answer 1

This is false, so don't feel bad you cannot prove it.

One of the standard non-finite counterexamples is below, it's mostly to illustrate that the problem is more systematic than just the trivial, finite ones. In particular, it starts with a Hausdorff group rather than what some might consider cheating by endowing the original group with a non-Hausdorff topology. From this it constructs a non-closed--but still finite-index--subgroup, so that the quotient space has a non-Hausdorff topology (which is equivalent to the original subgroup being non-closed).

Consider the finite group $A=\Bbb Z/2\Bbb Z$, endowed with the discrete topology. (Note that there are four topologies on a $A$, and that only two of them (the discrete one and the coarsest one) are turning it into a topological group, and only one of them (the discrete one) turns it into a compact topological group.)

Construct the group

$$G=A^\Bbb Z=\prod_{n\in\Bbb Z} A$$

This is compact by Tychonoff's theorem, and manifestly a group. Now from here we let $\tau$ be the topology generated by an ultrafilter, $\mathcal{F}$, containing the Fréchet filter. Let $H\le G$ be defined by the fact that Let $\pi_n(g)=g_n$ be the $n^{th}$ coordinate map, and

$$h\in H\iff |\{n\in\Bbb Z : \pi_n(h)=1\}|\in\{0,\infty\}.$$

$H$ is also dense in $G$ because the Fréchet filter is cofinite in the powerset of $\Bbb Z$, but clearly also $H\ne G$. Then distinguish two special elements, $\mathbf{0},\mathbf{1}\in G$, defined by the rules

$$\begin{cases}\pi_n(\mathbf{0}) =0 & n\in\Bbb Z \\ \pi_n(\mathbf{1})=1 & n\in\Bbb Z\end{cases}$$

Now let $g\in G$ be arbitrary and define sets $S_0, S_1$ by the rule

$$S_i = \{n\in\Bbb Z : g_i=n\}$$

So that $\Bbb Z=S_0\coprod S_1$ since $g_i\in\{0,1\}$. Then we see one of the $S_i\in\mathcal{F}$ by maximality. So $g\equiv x\mod H$ for some $x\in\{\mathbf{0},\mathbf{1}\}$, i.e. $[G:H]\le 2$. Since $H\ne G$, clearly $[G:H]=2$, but $H$ is dense, hence cannot be closed, and since closed + finite index = open, it must be that $H$ is not open, despite being finite index.

The key problem: we chose a non-Hausdorff topology on the quotient, $G/H$, i.e. we took only the trivial topology $\tau=\{\varnothing, G/H\}$. It is classical that $G/H$ Hausdorff iff $H$ is closed (provided $G$ is Hausdorff), so this was the natural approach to producing a non-closed subgroup of finite index. The algebra still allows two cosets, but the topology doesn't allow you to force them apart.

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Edit: An even easier counterexample is just $G=\Bbb Z/2\Bbb Z$ with the trivial topology. Then $\{e\}\le G$ is not open, but clearly has finite index.

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Edit 2: Since the op is using the convention that "compact" means "Hausdorff + finite subcovers" I moved the first edit to the bottom.

Answer 2

As I gave a wrong answer I will try to amend myself. Denote by (P) the following property for a topological group : all finite index subgroups are open. Recall that a profinite group is a topological group isomorphic (in the category of topological groups) to a projective limit of finite groups endowed with discrete topologies. Since a product is but a projective limit, Adam's answer shows in particular that (P) is in general not satisfied for profinite groups. I will give an everyday life counter-example.

Let $L$ be a Galois extension of $K$, but not necessarily a finite extension, and $G$ be the Galois group $\mathrm{Gal}(L/K)$. Put the dicrete topology on $L$, the product topology on $L^L$, and the induced topology on $G$. Let $A$ be the set of finite subextensions of $L/K$. For $\sigma\in G$ and $E\in A$, set $U_{\sigma}(E) := $ the subset of $G$ consisting of elements $\tau$ having same restriction on $E$ that $\sigma$. One can show that $U_{\sigma}(E)$ is a filter basis of $\sigma$ for this topology, and that the restrictions $G \rightarrow \mathrm{Gal}(L'/K)$ are continuous for any subextension $L'/ K$ that is Galois, but not necessarily finite. Moreover, one can show that $G$ is compact and totally disconnected. Actually, if $(L_i)_i$ is a filtered familly of galois subextensions such that $L = \cup_i L_i$, you have $G \simeq \varprojlim_i \mathrm{Gal}(L_i / K)$. (As you can choose all $L_i$'s to be finite over $K$, you got in particular that $G$ is profinite.)

From now we will considerer the extension $\overline{\mathbf{Q}} / \mathbf{Q}$ where $\overline{\mathbf{Q}}$ is the algebraic closure of $\mathbf{Q}$ in $\mathbf{C}$. Let $E = \mathbf{Q}\left[ \{\sqrt{-1}\}\cup\{\sqrt{p}\;|\;p\textrm{ prime}\} \right]$ and $$G := \mathrm{Gal}(E/\mathbf{Q}) \simeq \varprojlim \mathrm{Gal}\left(\mathbf{Q}\left[\sqrt{-1}, \sqrt{2}, \ldots, \sqrt{p}\right] / \mathbf{Q}\right) $$ which is a subgroup of $\mathrm{Gal}(\overline{\mathbf{Q}} / \mathbf{Q})$ and let $H$ be the subgroup of $G$ consisting in elements permuting only finitely many elements among $\sqrt{-1}$ and all the $\sqrt{p}$ for $p$-prime. The subgroup $H$ is dense in $G$ as by definition of $\mathrm{Gal}(\overline{\mathbf{Q}} / \mathbf{Q})$'s topology and $G$'s induced topology every open of $G$ contains obviously an element of $H$. Now $H$ is normal and the group $G / H$ is an $\mathbf{F_2}$-vector space in an obvious manner. Now fix a $d\in\mathbf{N}^{*}$ and let $E$ be a sub-$\mathbf{F_2}$-vector space of $G / H$ of codimension $d$. (Possible by Zorn's lemma.) The inverse image of $E$ in $G$ is a subgroup $K$ of $G$ containing $H$ and of index $\textrm{Card}\left(\mathbf{F_2}^d\right) = 2^d$. The subgroup $K$ is not open because were it open he would be closed which would be a contradiction with the density of $H$ in $G$. Now $K$ is not open, and its inverse image in $\mathrm{Gal}(\overline{\mathbf{Q}} / \mathbf{Q})$ is not open and is of finite index. Indeed, if the inverse image were open, its fixed field would be a nontrivial extension $F$ of $\mathbf{Q})$ contained in $E$ but then $F$ would be fixed by $K$, which is dense... The group $\mathrm{Gal}(\overline{\mathbf{Q}} / \mathbf{Q})$ has not the property (P).

Remark 1. I wrote Were it open he would be closed. For a subgroup $H$ of finite index of a topological group (not necessarily quasi-compact nor compact), being closed or being open is equivalent. Indeed. Make $H$ act of $G$ by left translations and, as $H$ is of finite index, let $g_1,\ldots,g_n$ be a set whose classes modulo $H$ partition $G$ in classes, and suppose that $g_1$'s class is $H$. You can therefore write the disjoint union $G = H \cup \left( \cup_{i=2}^n g_i H\right)$. Note that the $g\mapsto x g$ are homeomorphisms so that a $xH$ is closed (resp. open) if and only if $H$ is. Now, if $H$ is closed, the $g_i H$'s are also, so that the finite union of closed $\cup_{i=2}^n g_i H$ also is, and its complement in $G$, which is but $H$, is open. The same argument applies mutatis mutandis to show that if $G$ is open, it is closed. Note that in this case, our proof would have worked also if there wouldn't have been finitely many $g_i$'s, so that in fact, any open subgroup of a topological group is closed.

Remark 2. The group of $p$-adic integers $\mathbf{Z}_p$ is a profinite group in which all subgroups of finite index are closed and open. You can show this by finding explicitly all these subgroups, but how could you show it "conceptually"?

Answer 3

Consider $$\mathfrak{C}=\prod_{n\in \mathbb{N}_+}\mathbb{Z}/2$$ the set of all sequences $a=(a_n)_{n\ge 1}$ with $a_n \in \{0,1\}\simeq \mathbb{Z}/2$. From Cantor we know $\mathfrak{C}$ is uncountable. Moreover, $\mathfrak{C}$ is an abelian group with the addition $$(a_n) + (b_n) = (a_n+b_n) \mod 2$$

Define the map $\rho : \mathfrak{C} \to [0,1]$ $$\rho((a_n)) = \sum_n \frac{a_n}{2^n}$$ We observe that $$\rho(a) + \rho(b) - \rho(a+b) \ge 0 $$ indeed, the difference is $2 \cdot \sum_{n, a_n=b_n=1} \frac{1}{2^n}$ We define the metric $d$ on $\mathfrak{C}$ by $$d(a,b) = \rho(a-b) ( = \rho(a+b))$$

Intuitively, $d(a,b)$ small means $a$ and $b$ coincide on a large finite set. Indeed, for $a \in \mathfrak{C}$ we have the implications

$$ \rho(a) < \frac{1}{2^N} \implies a_n =0 \text{ for all } 1 \le n \le N \implies \rho(a) \le \frac{1}{2^N} $$

With the topology given by $d$ ( which, by the way, is the product topology from $\mathbb{Z}/2$ with the discrete topology) $\mathfrak{C}$ becomes a compact abelian topological group ( compactness is essentially due to Cantor - and this implies $[0,1]$ is compact)

Inside $\mathfrak{C}$ we have $\mathfrak{C}_0$ , the subgroup of all sequences of $0$,$1$ which have only finitely many nonzero components.

We'll show that $\mathfrak{C}$ has a subgroup of index $2$ containing $\mathfrak{C}_0$. Now, $\mathfrak{C}$ is a a vector space over the field $\mathbb{Z}/2$, and the subgroups of $\mathfrak{C}$ are exactly its subspaces. So we need to show that there exists a subspace of codimension $1$ containing $\mathfrak{C}_0$. For this, extend the standard basis $(e_n)_{n \in \mathbb{N}_+}$ to a basis $(e_i)_{i \in I}$ ( note that $I$ is uncountable since $\mathfrak{C}$ is). Take $i \in I \backslash \mathbb{N}_+$ and define $$\mathfrak{C}' = \text{span}(e_j)_{j \in I, j \ne i}$$ a subgroup of index $2$ containing $\mathfrak{C}_0$. Note that $\mathfrak{C}_0$ is dense in $\mathfrak{C}$ and so $\mathfrak{C}'$ is dense.

This provides a counterexample to the question posted. However, the result holds for compact Lie groups. It follows from the following fact:

Let $G$ a compact connected Lie group and $H$ a subgroup of finite index of $G$. Then $H=G$. From this, using $H \cap G_e$, follows that if $H$ is a subgroup of finite index of a compact Lie group then $H \supset G_e$, the connected component of the identity $e$.

To show that $H=G$ in the case $G$ compact connected, we record the fact that for every $N\ge 1$ the map $$g \mapsto g^N$$ is surjective. Let now $H \subset G$, $H$ of finite index. The normal core $H'$ of $H$ equals $$\bigcap_{g \in G/H} g H g^{-1}$$ and is again of some finite index $N$. The group $G/H'$ is of order $N$ and therefore $N$-torsion. Therefore, for all $g \in G$ we have $g^N \in H'$. But every element is an $N$-th power. We conclude $H'=G$.

Obs: The fact that the power map $g \mapsto g^N$ is surjective for compact connected Lie groups follows from the fact that $G$ connected is a union of tori, and checking the statement for tori (that is easy already).