# Topology “generated by” normal subgroups (Topological groups).

Let $$G$$ be a group and $$L$$ be a non-empty family of normal subgroups such that if $$K_{1},K_{2} \in L$$ and $$K_{3}$$ is a normal subgroup containing $$K_{1} \cap K_{2}$$ then $$K_{3} \in L$$. Let $$T$$ be a family of unions of sets of cosets $$Kg$$ with $$K \in L,g \in G$$. Show that $$T$$ is a topology in $$G$$ and that $$G$$ is a topological group with respect to this topology. Show that $$L$$ is the set of open normal subgroups of $$G$$ with respect to this topology.

This is the first problem of Wilson's book "Profinite Groups". I'm having trouble with this question.

Using YCor's comment, I was able to prove that $$T$$ is a topology with basis:

$$B = \left\{C_{I,J} \mid C_{I,J} = \{K_{i}g_{j}\}_{i \in I, j \in J}\right\}.$$

Now, to prove that $$G$$ is topological group, I need to show that the maps $$\varphi: G \times G \to G$$ given by $$\varphi(x,y) = xy$$ and $$\psi: G \to G$$ given by $$\psi(x) = x^{-1}$$ are continuous.

I'm still not familiar with topological groups (this is the first problem of the book), so I'm having a bit of trouble time with this question.

So, I need to show that if $$U$$ is open in $$G$$, $$\varphi^{-1}(U)$$ is open in $$G \times G$$. But, is enough consider $$U \in B$$. I worked with a simple subset of $$B$$: $$\{Kg\}$$ for some $$K \in L$$ and $$g \in G$$. The sets

$$Kg \times \{1\}1$$ $$K1\times \{1\}g$$ $$Kg^{-n}\times \{1\}g^{n+1}$$ $$Kg^{n+1} \times \{1\}g^{-n}$$

and the "symmetrical" products are maps in $$\{Kg\}$$ and the union of this sets is in $$T$$. If I'm not completely wrong, I think that the general case is similar, but I dont know how to formalize.

Also, for the last part, Im having troubles to show that the set of open normal sets of $$G$$ is cointained in $$L$$. I would like a hint for it.

• If $B$ is a set of subsets of a set $X$ that is stable under taking finite intersections with $X\in B$, then $B$ is a basis for a topology $T_B$ on $X$, namely the set of unions of subsets of $B$. – YCor Apr 4 at 4:15
• @YCor Do you know some reference with the proof of that? – Lucas Corrêa Apr 4 at 4:23
• Yes, where's the confusion? – YCor Apr 4 at 7:12
• "Let $T$ be a family ..." In particular, $T$ can be empty. Do you want $T$ to contain all unions of cosets $Kg$? – punctured dusk Apr 7 at 8:51
• @rabota I will check the book, but I think that is "$T$ be the family". – Lucas Corrêa Apr 7 at 8:59

The assumption on $$L$$ is equivalent to the following two conditions:

(a) If $$K_1, K_2 \in L$$, then $$K_1 \cap K_2 \in L$$. [because $$K_1 \cap K_2$$ is normal]

(b) If $$K \in L$$ and $$N$$ is a normal subgroup such that $$K \subset N$$, then $$N \in L$$. [take $$K_1 = K_2 = K$$]

We shall now prove a few claims.

1) Let $$K \in L$$ and $$g,g' \in G$$ such that $$g \in Kg'$$. Then $$Kg'= Kg$$.

We have $$g = kg'$$ for some $$k \in K$$. Since $$Kk' = K$$ for any $$k' \in K$$, we get $$Kg' = K k^{-1} g = K g$$.

2) Let $$K_1, K_2 \in L$$ and $$g_1,g_2 \in G$$. If $$K_1g_1 \cap K_2g_2 \ne \emptyset$$, then $$K_1g_1 \cap K_2g_2 = (K_1 \cap K_2)g$$ for some $$g \in G$$.

Let $$g \in K_1g_1 \cap K_2g_2$$. Then $$K_1g_1 = K_1g, K_2g_2 = K_2g$$. Hence $$K_1g_1 \cap K_2g_2 = K_1g \cap K_2g = (K_1 \cap K_2)g$$.

3) $$T$$ is a topology on $$G$$.

$$\emptyset \in T$$ (it is the union of the empty family of cosets $$Kg$$.)

$$X \in T$$ ($$X = \bigcup_{g \in G} Kg$$ for any $$K \in L$$.)

$$T$$ contains all unions of members of $$T$$ (obvious from the definition.)

$$T$$ contains the intersection of any two members of $$T$$:

For $$k = 1,2$$ let $$U_k = \bigcup_{i_k \in A_k, j_k \in B_k} K_{k,i_k}g_{k,j_k}$$. Then $$U_1 \cap U_2 = \bigcup_{i_1 \in A_1, j_1 \in B_1,i_2 \in A_2, j_2 \in B_2} K_{1,i_1}g_{1,j_1} \cap K_{2,i_2}g_{2,j_2}$$ which belongs to $$T$$ by 2).

4) For each $$g \in G$$, the set $$L(g) = \{ Kg \mid \ K \in L \}$$ is a base of open neigborhoods of $$g$$.

Let $$U$$ be an open neighborhood of $$g$$. Choose $$K \in L$$ and $$g' \in G$$ such that $$g \in Kg' \subset U$$. By 1) we have $$Kg' = Kg$$.

5) $$\varphi$$ is continuous.

Let $$g_1,g_2 \in G$$ and $$U$$ be an open neighborhood of $$g_1g_2$$. Choose $$K \in L$$ such that $$Kg_1g_2 \subset U$$. The sets $$V_i = Kg_i$$ are open neighborhoods of $$g_i$$. Since $$K$$ is normal, we have $$g_1K = Kg_1$$. We get $$\varphi(V_1,V_2) = Kg_1Kg_2 = KKg_1g_2 = Kg_1g_2 \subset U$$.

6) $$\psi$$ is continuos.

Let $$g \in G$$ and $$U$$ be an open neighborhood of $$g^{-1}$$. Choose $$K \in L$$ such that $$Kg^{-1} \subset U$$. The set $$V = Kg$$ is an open neighborhood of $$g$$. We get $$\psi(V) = (Kg)^{-1} = g^{-1}K^{-1} = g^{-1}K = Kg^{-1} \subset U$$.

7) $$L$$ is the set of open normal subgroups.

Let $$N$$ be an open normal subgroup. It contains the neutral $$e$$ of $$G$$. Hence there exists $$K \in L$$ such $$K = Ke \subset N$$. By assumption on $$L$$ we get $$N \in L$$.

Conversely, if $$K \in L$$, then $$K = Ke$$ is open.

• Awesome! It's a very clear answer. Thank you! – Lucas Corrêa Apr 7 at 17:06