A group with a particular topology that is a topological group

Question

Suppose that $G$ is an additive abelian group and select a collection of subgroups $\{H_\alpha\}_{\alpha\in A}$.

Now consider the topology on $G$ generated by the family $$\mathcal F =\{H_\alpha+g\}_{\alpha\in A,g\in G}$$

In other words $\mathcal F$ is a subbasis for the topology. Why do we have that $G$ is a topological group with this topology?

Answer 1

It is necessary and sufficient to check that the neighbourhood system of the identity (so $(H_\alpha)_{\alpha \in A}$ obeys the following axioms :

1. $\forall H_\alpha \exists H_\beta , H_\gamma : H_\gamma + H_\beta \subset H_\alpha$

2. $\forall H_\alpha: \exists H_\beta: -H_\beta \subseteq H_\alpha$.

for continuity of $+$ and $-$. This is quite clear, as we can just take $H_\gamma = H_\beta = H_\alpha$ everywhere,as we have subgroups.

To get a valid topology: we need to check that

3. $\forall H_\alpha, H_\beta \exists H_\gamma: H_\gamma \subset H_\alpha \cap H_\beta$,

and this is clear as the $H_\alpha$ generate the base of intersections that we can use here.

Answer 2

You have to show that the product $m:G\times G\rightarrow G$ and the inverse $i:G\rightarrow G$ is continue.

$i^{-1}(H_{\alpha}+g)=H_{\alpha}-g$ this implies that $i$ is continuous.

$m^{-1}(H_{\alpha}+g)=\bigcup_{x\in G} (H_{\alpha}+x)\times (H_{\alpha}+g-x)$ this implies that $m$ is continuous. To show the last equality, remark that $\bigcup_{x\in G} (H_{\alpha}+x)\times (H_{\alpha}+g-x)\subset H_{\alpha}+g$ since for every $h,h'\in H_{\alpha}, h+x+g+h'-x=h+h'+g\in H_{\alpha}+g$. Conversely, let $(y,y')\in m^{-1}(H_{\alpha}+g)$, $y+y'=g+h$. This implies that $y'=g-(g-y')$ and $y=g+h-y'=h+(g-y')$. Write $x=g-y'$, we deduce that $(y,y')\in (H_{\alpha}+x)\times (H_{\alpha}+g-x)$

Answer 3

Whenever you have a nonempty family $\mathcal{U}$ of subgroups of $G$ which is closed under finite intersections, you get a group topology on $G$ by declaring that a subset is open if and only if it is the union of subsets of the form $g+H$, for some $g\in G$ and $H\in\mathcal{U}$.

Closure under arbitrary unions is obvious; the empty set is the union of the empty family and $G=\bigcup_{g\in G}(g+H)$, where $H$ is any member of $\mathcal{U}$. As for finite intersections, consider $$ \Bigl(\bigcup_{\lambda}(x_\lambda+H_\lambda)\Bigr) \cap \Bigl(\bigcup_{\mu}(y_\mu+K_\lambda)\Bigr)= \bigcup_{\lambda,\mu}\bigl((x_\lambda+H_\lambda)\cap(y_\mu+K_\mu)\bigr) $$ Then we only have to show that, for $H,K\in\mathcal{U}$, $x,y\in G$, the intersection $$ (x+H)\cap(y+K) $$ satisfies our definition of open set. There are two cases: either the intersection is empty, or it is $z+(H\cap K)$, for some $z\in G$; prove that if the intersection is not empty and $z\in(x+H)\cap(y+K)$, then $z+(H\cap K)=(x+H)\cap(y+K)$.

This is clearly a group topology: there is a basis of neighborhoods of zero satisfying $H+H=H$ and $-H=H$.

Your topology can be obtained in the above way by taking finite intersections of the members of the given family of subgroups.