Characterization of a collection of sets containing identity in a group so that it generates a topology?

Question

Start with a group $G$ without topology, we want to generate a topology on the group by a collection of sets containing identity.

To be more specific, let $\mathcal{S}_e$ be a collection of sets containing identity $e$. Let $\mathcal{S}_x:= \{L_x(U) : x\in G, U \in \mathcal{S}_e\}$, where $L_x$ is the left multiplication by $x$. What I want to ask is under what conditions (if such conditions exist at all) of $\mathcal{S}_e$, $\mathcal{S}= \bigcup_{x\in G} \mathcal{S}_x $ is a base so that it generates a topology for $G$.

The same question can be asked in the manner of "subbase" if we do finite intersection first. Define $\mathcal{B}_e:= \{U : U \text{ is a finite intersection of sets in } S_e\}$. $\mathcal{B}_x:= \{L_x(U) : x\in G, U \in \mathcal{B}_e\}$. Under what conditions (if such conditions exist at all) of $\mathcal{S}_e$, $\mathcal{B} = \bigcup_{x\in G} \mathcal{B}_x$ is a base so that it generates a topology for $G$.

Note here if $\mathcal{S}$ or $\mathcal{B}$ is a base then the topology it generates is automatically translation-invariant. To simplify the question, I do not require $G$ is a topological group under this topology so we do not need to worry about taking inverse is continuous or not.

I think A random collection doesn't work, in either case. A counterexample in my mind is $G = Z_5$, if we define $\mathcal{S}_0 = \{4,0,1\}.$ $\mathcal{B}=\mathcal{S}$ = {{4,0,1},{0,1,2},{1,2,3},{2,3,4},{3,4,0}}. It is obviously not a base for any topology.

Answer 1

A requirement for B to be a base is
for all x,y in G, U,V in B$_e$,
(a in xU $\cap$ yV implies xU $\cap$ yV in L$_a$).

"we want to generate ..." Please speak for yourself only, I have no desire to do that.

After expressing the desire "I want to ask ..." you still have not asked your question. Do not beat around the bush, be brave and directly ask your question "What conditions are needed for S to be a base of some topology for G"?

Answer 2

A classic basic theorem that characterises local bases of $e$: a non-empty collection of $\mathcal{V}$ of subsets of $G$ is a local base at $e$ iff

1. $\forall U \in \mathcal{V}: e \in U$ (obviously)
2. $\forall U,V \in \mathcal{V}: \exists W \in \mathcal{V}: W \subseteq U \cap V$ (filterbase property)
3. $\forall U \in \mathcal{V}: \exists V \in \mathcal{V}: VV^{-1} \subseteq U$
4. $\forall U \in \mathcal{V} \forall x \in U: \exists V \in \mathcal{V}: xV \subseteq U$
5. $\forall U \in \mathcal{V}: \forall a \in G: \exists W \in \mathcal{V}: aWa^{-1} \subseteq U$.

And to characterise Hausdorff topological groups replace 1 by 1':

1'. $\bigcap \mathcal{V} = \{e\}$.

Under these circumstances $\{xV, Vx: x \in G, V \in \mathcal{V}\}$ forms a base for the topology on $G$ that makes it into a (Hausdorff) topological group.

I quoted this version from Grupos Topológicos, by Tkachenko et al. (in Spanish, bought it in Mexico), but I believe Bourbaki has a version of this too, probably many other books as well.