# Why isn't there interest in nontrivial, nondiscrete topologies on finite groups?

A topology on a group is required to be compatible with the group structure (multiplication must be a continuous map $G\times G\to G$ and inversion must be continuous). I've only ever seen the discrete topology referenced on finite groups, however (as e.g. subgroups of continuous matrix groups).

Why isn't there any interest in nontrivial, nondiscrete topologies on finite groups? Is it because they are easy to classify, their study reduces to other areas of mathematics like graph theory, or that they haven't seen any external purpose elsewhere?

Surely these topologies exist. For a subgroup $H\le G$ we can set all cosets of $H$ to be a base. We can speak of generating (group-compatible) topologies from subsets $X\subseteq G$; let the topology $\tau_G$ be the collection of all left and right translates of a given collection of subsets, as well as their unions and intersections (which will obviously be finite), and then their translates, etc. (this process will surely terminate because it acts like a monotone function being applied to the double power set $\mathcal{P}(\mathcal{P}(G))$.)

Then we can ask for sufficient and necessary conditions for a family of subsets to generate the discrete topology. Other potentially interesting questions and answers: for a given family or handful of finite groups, we can ask for classification of their admissible topologies. So I for one think there are potentially interesting questions about nontrivial nondiscrete finite group topologies, but there does not seem to be any theory about it $-$ is there something I'm missing?

• A finite group with compatible hausdorff topology (or even lesser separation) is discrete. Oct 18 '12 at 21:42
• As you have noticed in your own answer, the group topologies on a finite group $G$ are only the pullback of the Hausdorff topologies of its quotients; hence they are indexed by normal subgroups $G$. This is because the closure of $\{1\}$ is a normal subgroup, and because a topological group with closed $\{1\}$ is Hausdorff.
– YCor
Apr 28 '19 at 11:41

Recall that a topology is completely determined by the closure relation. Let $G$ be a finite group with some topology and let $G_0$ be the closure of the identity. Then $G_0$ is a normal subgroup. Moreover, the closure of any finite subset of $G$ is the union of the corresponding translates of $G_0$ (since this is a closed subset contained in the closure). Consequently, the topology on $G$ is completely determined by $G_0$. In other words, studying nontrivial topologies on a finite group is equivalent to studying normal subgroups.

In general, if $G$ is a topological group and $G_0$ the closure of the identity, then $G_0$ is a normal subgroup and $G/G_0$ is the universal Hausdorff topological group into which $G$ maps. (Recall that as Hagen mentions in the comments, for finite spaces Hausdorff is equivalent to discrete.) So the study of topological groups reduces almost immediately to the study of Hausdorff topological groups.

• I am having issues proving that $G_0$ is a subgroup. I have shown it is a normal subset and that it completely determines the group topology, however. There are indications that connectedness might offer a proof, but I am not very familiar with connectedness and I don't know if the closure and connected component of the trivial subgroup are necessarily the same. Hint?
– blue
Oct 19 '12 at 2:31
• @sea: $G_0$ is the closure of every element of $G_0$. Oct 19 '12 at 2:50
• In a topological group (even if not $T_1$) the closure of any subgroup is a subgroup. $G_0$ is the closure of the subgroup $\{e\}$. Oct 19 '12 at 3:19
• Thanks; I was able to finish fleshing out the details with the hint.
– blue
Oct 19 '12 at 5:03
• Let me flesh out the conclusion that general topological groups reduce to Hausdorff ones. The fact that $G_0$ is the closure of any of its points means that $G_0$ has the indiscrete topology, as do all of its cosets, the fibers of the quotient map $G \to G/G_0$. That is, the points of $G_0$ are topologically indistinguishable, and $G/G_0$ is not only the universal Hausdorff (or $T_2$) quotient of $G$, but also the much milder Kolmogorov (or $T_0$) quotient. Hence a set is open in $G$ iff (1) it is closed under multiplication by $G_0$ and (2) its image is open in $G/G_0$. Mar 17 '16 at 12:59

Evidently I was missing some things. All is well now, though.

In response to Qiaochu's answer I wanted to flesh out the details. Let $(G,\tau_G)$ be a finite topological group, $1:=\{1_G\}$ the trivial subgroup, $\mathrm{cl}(\cdot)$ the topological closure operator, and $\mathrm{ncr}(\cdot)$ the group-theoretic normal core operator. So $\mathrm{cl}(A)=\bigcap \{F~\textrm{closed}:A\subseteq F\}$ and $\mathrm{ncr}(A)=\bigcap_{g\in G}g^{-1}Ag$.

Lemma 1. Let $(X,\tau_X)$ be a topological space. The pairs $A,\,\mathrm{cl}_X(A)$ determine $\tau_X$ and vice-versa.

Proof. The vice-versa direction is clear by $\mathrm{cl}$'s intersection formula. Conversely, a subset $A\subseteq X$ is closed if and only if $\mathrm{cl}(A)=A$ (if $A$ is closed then $A\subseteq \bigcap F\subseteq A\implies \mathrm{cl}(A)=A$, and conversely if equality holds then $A$ is an intersection of closed sets hence is closed). Hence we can determine the open sets of $X$ precisely in terms of closure: $A\subseteq X$ is open iff $X\setminus A=\mathrm{cl}(X\setminus A)$.

Lemma 2. The topology $\tau_G$ on $G$ is determined by $S:=\mathrm{cl}(1)$.

Proof. Let $X\subseteq G$ be a subset. Any closed set $F$ containing $X$ contains any singleton subset of $X$, so also contains the closure $\mathrm{cl}(x)$ for all $x\in X$, hence will contain the finite union $Y:=\cup_{x\in X}\mathrm{cl}(x)$; since the union is finite, $Y$ is also closed. Note $X\subseteq Y$ since $x\in\mathrm{cl}(x)$ always. So we conclude that $\mathrm{cl}(X)=Y$ because $Y$ exhibits the universal property of the closure of $X$. Moreover,

$$\begin{array}{cl} \mathrm{cl}(x) & = \bigcap \{F~\textrm{closed}:x\in F\} \\ & = \bigcap x\{x^{-1}F~\textrm{closed}:1_G\in x^{-1}F\} \\ & =x\bigcap\{E~\textrm{closed}:1_G\in E\} \\ & = x\,\mathrm{cl}(1)=xS \end{array}$$ $$\therefore~~ \mathrm{cl}(X)=\bigcup_{x\in X}\mathrm{cl}(x)=\bigcup_{x\in X}xS=XS.$$

All closures are therefore uniquely determined by $S$, and hence so with $\tau_G$ by Lemma 1.

Lemma 3. $S=\mathrm{cl}(1)\trianglelefteq G$ is a normal subgroup.

Proof of normality. Since left and right translation are continuous and $S$ is closed containing $1_G$, any conjugate $g^{-1}Sg$ must be closed and contain $1_G$ hence $S\subseteq g^{-1}Sg=S^g$ for all $g$. Therefore $$S\subseteq \bigcap_{g\in G}S^g=\mathrm{ncr}(S) \subseteq S\implies S=\mathrm{ncr}(S).$$ Since the group-theoretic normal core is normal, $S$ is a normal subset of $G$.

Proof of subgroup. Note that $S$ is a closed set containing $x$ for any $x\in S$, hence $xS\subseteq\mathrm{cl}(x)\subseteq S$, and since $x^{-1}S\subseteq S\implies S\subseteq xS$ by left-multiplication, $xS=S$ for all $x\in S$, which gives closure under multiplication as well as inverses (since $1_G\in S=xS\implies 1_G=xy$ for some $y\in S$).

Theorem. The only nontrivial topologies on a finite group are lifted from discrete topologies on factor groups. That is, a topology on $G$ must have as base the coset space $G/N$ for a $N\trianglelefteq G$.

Proof. Let $\tau_G$ be a topology on $G$ and let $N=S=\mathrm{cl}_G(1)$. A subset $X\subseteq G$ is open iff $G\setminus X$ is closed iff $G\setminus X=\mathrm{cl}(G\setminus X)=(G\setminus X)S=\cup_{y\in G\setminus X}yN$ is a union of left cosets of $N$ (and indeed since $N$ is closed and translation is continuous, any union of left cosets of $N$ is closed). But since cosets partition $G$, if $G\setminus X$ is a union of cosets, so is $X$. Hence $G/N$ is a base for $\tau_G$.

Remark. Suppose $(G,\tau_G)$ is Hausdorff. Then for each nonidentity element $g\in G$, there are disjoint open sets $1_G\in U_g$ and $g\in V_g$. Then $1_G\in\cap_{g\ne 1_G}U_g$ is open and cannot contain a non-identity element hence $\{1_G\}$ is open, and subsequently by continuity any singleton and by union any subset is open, so $\tau_G=\mathcal{P}(G)$ is in fact the discrete topology.