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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.

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    $\begingroup$ 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 $B$. $\endgroup$ – YCor Apr 4 at 4:15
  • $\begingroup$ @YCor Do you know some reference with the proof of that? $\endgroup$ – Lucas Corrêa Apr 4 at 4:23
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    $\begingroup$ Yes, where's the confusion? $\endgroup$ – YCor Apr 4 at 7:12
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    $\begingroup$ "Let $T$ be a family ..." In particular, $T$ can be empty. Do you want $T$ to contain all unions of cosets $Kg$? $\endgroup$ – punctured dusk Apr 7 at 8:51
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    $\begingroup$ @rabota I will check the book, but I think that is "$T$ be the family". $\endgroup$ – Lucas Corrêa Apr 7 at 8:59
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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.

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

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