Properties of a Hausdorff topological group ("proof" verification).

Question

Problem. The centralizer $C_{G}(S)$ of a subset $S$ of a group $G$ is defined by $$C_{G}(S) = \{g \in G \mid sg = gs\mathrm{\,for\,all\,}s \in S\}$$ and the normalizer $N_{G}(H)$ of a subgroup $H$ of $G$ is defined by $$N_{G}(H) = \{g \in G \mid g^{-1}hg \in H\mathrm{\,and\,}ghg^{-1} \in H\mathrm{\,for\,all\,}h \in H\}.$$ Suppose that $G$ is a Hausdorff topological group.

(a) Prove that the centralizer of each subset of $G$ is closed and the normalizer of each closed subgroup is closed.

(b) Prove that each abelian closed subgroup $A$ is contained in a maximal closed abelian subgroup.

(c) Prove that if there is a series $$1 = G_{0} \triangleleft G_{1} \triangleleft \cdots \triangleleft G_{n} = G$$ of subgroups such that $G_{i}/G_{i-1}$ is abelian for each $i = 1,...,n$, then there is such series consisting of closed subgroups.

My attempt.

(a) OK

(b) First, I prove the lemma:

Lemma 1. Suppose that $G$ is a Hausdorff topological group. If $H$ is a subgroup of $G$, then $\overline{H}$ is a subgroup of $G$. If $H$ is abelian, $\overline{H}$ is abelian.

So, I take $\mathcal{F} = \{A \mid A \subset G; A = \overline{A}; A\mathrm{\,abelian}\}$ and $\mathcal{C}$ a chain in $\mathcal{F}$. If $\mathcal{C}$ has no an increasing sequence of elements, then $\mathcal{C}$ has a maximal element. Otherwise, take $A$ as an union of all elements of the chain. Then $A$ is an abelian subgroup and, by the Lema 1, $\overline{A}$ is closed. Therefore, using Zorn's Lema, $\mathcal{F}$ has a maximal element.

PS. The author consider the family $\mathcal{F}$ partially ordered.

(c) I proved the more some lemmas.

Lemma 2. If $H$ is a subgroup of $G$ and $H \triangleleft G$, then $\overline{H} \triangleleft \overline{G}$.

Lemma 3. If $G$ is a group and $N \triangleleft G$, then $\overline{G}/\overline{N} \subset \overline{G/N}$.

Thus, is enough take the closures of each element of the series.

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This is just sketch of my proof. I would like to verify if the ideas are correct and make sense. If I wrote something stupid, I would like some hints to correct.

I'm a little insecure about the item (c). I thought that I needed to use the item (a).

Answer 1

For the Zorn argument use the poset $\{H: A \subseteq H, H \text{closed Abelian subgroup of G } \}$ and we just need to note that the closure of the union of any chain is an upperbound (no cases). Then Zorn applies straight away.

Lemma 2 is easily seen to hold using nets, e.g.

Could you prove your lemma 3? You need another formulation: if $G_1 \triangleleft G_2$ and both are subgroups of $G$, then $\overline{G_1}/{\overline{G_2}} \subseteq \overline{G_1/{G_2}}$. The closure of $G$ has no point in your statement when it is the whole group already. If this lemma 3 holds where do you take the closure on the right? Taking the closures in the sequence is I think the right idea. Lemma 3 might not even be needed; you only need that the quotients of the closures are still Abelian.

Answer 2

Friend from the Portuguese speaking world, I am somewhat unsure about your claim in lemma 3, which if I may be pedantic is not actually rigorously formulated (as Henno Brandsma has already pointed out). I gather you meant something of the following kind:

Claim: If $G$ is a Hausdorff topological group and $K \trianglelefteq H \leqslant G$ then the quotient $\overline{H}/\overline{K} \leqslant \mathrm{N}_G(\overline{K})/\overline{K}$ can somehow be canonically embedded in the quotient $\overline{H/K} \leqslant \mathrm{N}_{G}(K)/K$

where the closure in the latter quotient is understood with respect to the ambient topology of the quotient $\mathrm{N}_G(K)/K$.

Let us wait to see if the wiser ones among us could clarify whether such a proposition has any chances to hold in general and in the mean while let us make the following argument to establish the result in c).

Lemma 1. If $(G, \cdot, \mathscr{T})$ is an arbitrary topological group and $K, H \leqslant G$, then the following holds for commutators: $$[\overline{K}, \overline{H}] \leqslant \overline{[K, H]}$$

Proof: Consider the commutator bracket map $$[\ ,\ ]: G \times G \to G\\ (x,y) \mapsto [x,y]=x^{-1}y^{-1}xy$$ When equipping $G \times G$ with the direct product (''direct square'' so to speak) topology, this map is clearly continuous. Hence we have that the inverse image set $$[\ ,\ ]^{-1}(\overline{[K, H]})=M$$ is closed, as the preimage through a continuous map of a closed subset. On the other hand, $K \times H \subseteq M$ by definition, so we infer that $$\overline{K \times H}=\overline{K} \times \overline{H} \subseteq M$$ This means that you have the following for direct images: $$[\ ,\ ](\overline{K} \times \overline{H}) \subseteq \overline{[K,H]}$$ As the right-hand side in the above inclusion is a subgroup, it will include the subgroup generated by the left-hand side, in other words: $$[\overline{K}, \overline{H}]=\langle [\ ,\ ](\overline{K} \times \overline{H}) \rangle \leqslant \overline{[K,H]}$$ and we are done. $\Box$

Lemma 2. If $(G, \cdot, \mathscr{T})$ is an arbitrary topological group and $H=\overline{H} \leqslant G$ then the normalizer $\mathrm{N}_G(H)$ is also closed (i.e. normalizers of closed subgroups are closed).

Proof: For arbitrary $x \in G$ define the map $$\theta_x: G \to G,\ \theta_x(t)=txt^{-1}$$ Such a map is clearly continuous (in more formal parlance we would say $\theta_x \in \mathrm{End}_{\mathrm{Top}}(G, \mathscr{T})$, the latter being the set of all endomorphisms of the topological space $(G, \mathscr{T})$). Hence, if we introduce $$T_x=\theta_x^{-1}(H)$$ we infer that $T_x$ is closed for any $x \in G$, as $H$ is closed.

By definition, we can write $$\mathrm{N}_G(H)= \bigcap_{x \in H} (T_x \cap T_x^{-1})$$ where for arbitrary subset $X \subseteq G$ we define the symmetric $X^{-1}=\{x^{-1}\}_{x \in X}$. As the symmetry map $$^{-1}: G \to G,\ ^{-1}(x)=x^{-1}$$ is an automorphism of the space $(G, \mathscr{T})$, it will send closed subsets to closed subsets. Consequently the normalizer $\mathrm{N}_G(H)$ will be closed itself, as an intersection of closed subsets. $\Box$

Lemma 3. If $(G, \cdot, \mathscr{T})$ is a topological group and $K \trianglelefteq H \leqslant G$ are such that the quotient $H/K$ is abelian, then $\overline{K} \trianglelefteq \overline{H}$ and the quotient $\overline{H}/\overline{K}$ is also abelian.

Proof: By lemma 2, $\mathrm{N}_G(\overline{K})$ is closed; as it is generally valid that $\mathrm{N}_G(K) \leqslant \mathrm{N}_G(\overline{K})$, from the chain $$K \leqslant H \leqslant \mathrm{N}_G(K)$$ we obtain by taking closures $$\overline{K} \leqslant \overline{H} \leqslant \overline{\mathrm{N}_G(K)} \leqslant \mathrm{N}_G(\overline{K})$$ so it is indeed the case that $\overline{K} \trianglelefteq \overline{H}$. Claiming that $H/K$ is abelian is equivalent to claiming $$\mathrm{D}(H)=[H, H] \leqslant K$$ hence by taking closures and applying lemma 1 we infer $$\mathrm{D}(\overline{H})=[\overline{H}, \overline{H}] \leqslant \overline{[H, H]}=\overline{\mathrm{D}(H)} \leqslant \overline{K}$$ by which we conclude that $\overline{H}/\overline{K}$ is also abelian. $\Box$