# How Likely is a Finite Group to be Hausdorff?

This is the third in a series of posts on some fun little toy problems of mine. Today’s question is one I came across while trying to come up with an example of a topological group which is not Hausdorff. Anyone familiar with topological groups knows that such examples are scarce in the literature. This is not without reason: any $\mathrm{T}_0$ group is actually Hausdorff, so non-Hausdorff topological groups are particularly pathological. But how scarce really are the non-Hausdorff topological groups? In other words, how likely is a topological group to be Hausdorff?

See Is Every Set Fieldable? and Classification of Abelian Lie Groups for the previous two posts on this series!

We begin by noting that, as stated, our question is ill-posed. First, we should clarify that we are working with a fixed group $G$ and considering all possible topologies which turn it into a topological group — which we call “$G$-topologies”. That is to say, our question is actually: given $G$ and some random $G$-topology, how likely is it to be Housdorff? However — and this is more important — the meaning of “some random topology” is still unclear.

To make the question precise, we focus on the case where $G$ is finite. This is because a finite set $X$ can only be endowed with a finite number of distinct topologies — indeed, the power set $\mathcal{P}(X)$ is finite. Counting how many topologies there are in terms of the size of $X$ is a famous open problem and these numbers have only being computed for very small $X$. Here is a plot of (the logarithm of) the first 19 of these numbers:

See the excellent page on these computation at the OEIS website!

What is relevant to us is that this implies a finite group $G$ admits only a finite number of $G$-topologies. We may thus consider a random variable $\tau \in \{ G \text{-topologies} \}$ with uniform distribution and ask: can we compute $\mathbb{P}(G\ \text{is Hausdorff}) = \mathbb{P}(G\ \text{is Hausdorff under}\ \tau)$? This is the question we will attempt to answer in this article. Since any discrete group is Hausdorff, we certainly know

What is perhaps surprising is that $\mathbb{P}(G\ \text{is Hausdorff})$ is precisely $1/\#\{G\text{-topologies}\}$. In other words, the odds of $G$ being Hausdorff are as slim as they could possibly be! Indeed, it is a well known fact from the theory of finite topological spaces that any a finite Hausdorff space is discrete, so the only finite Hausdorff groups are the discrete ones. All it’s left is to compute $\# \{G\text{-topologies}\}$. To do so, we investigate the closure $\overline{ \{ 1 \} }^\tau$ of $1 \in G$ under a given $G$-topology $\tau$.

Since the translations by elements of $G$ are all homeomorphism, $\overline{\{g\}}^\tau = g \cdot \overline{\{1\}}^\tau$ for all $g \in G$. It is then easy to see that

for any $n, m \in \overline{\{1\}}^\tau$. Similarly, since the inversion $g \mapsto g^{-1}$ is a homeomorphism, we find $n^{-1} \in \overline{\{1^{-1}\}}^\tau = \overline{\{1\}}^\tau$ for any $n \in \overline{\{1\}}^\tau$. Finally, $g \cdot \overline{\{1\}}^\tau \cdot g^{-1} = \overline{\{g 1 g^{-1}\}}^\tau = \overline{\{1\}}^\tau$ for all $g \in G$ and $n \in \overline{\{1\}}^\tau$. This goes to show $\overline{\{1\}}^\tau$ is a normal subgroup of $G$.

There is thus a natural map

between the space of $G$-topologies and the set of normal subgroups of $G$. That’s all well and good, but how does any of this helps us to count $G$-topologies? Well, the point is that, perhaps surprisingly, this map is actually a bijection.

To see that our map is injective, we remark that the collection of subsets of $G$ which are closed under a given $G$-topology $\tau$ is actually determined by $N_\tau$. Indeed,

for any $F \subset G$ which is closed under $\tau$. Conversely, since $G$ is finite, any union of $N_\tau$-cosets is closed under $\tau$ — a finite union of closed sets is closed. All in all, the subsets which are closed under $\tau$ are precisely the unions of $N_\tau$-cosets.

This last equation also goes to show that our map is surjective: given some $N \triangleright G$, the set $\tau = \left\{ \bigcup_{g \in X} g \cdot N : X \subset G \right\}$ is a natural candidate for a $G$-topology such that $N_\tau = N$. It is clear that $\tau$ is a topology and $\overline{\{1\}}^\tau = N$. With enough effort, one can also show that if $m : G \times G \to G$ and $i : G \to G$ are the group multiplication and inversion, respectively, then

which goes to show that $\tau$ is indeed a $G$-topology.

Finally, $\#\{G\text{-topologies}\} = \#\{N \triangleright G\}$ and thus

We conclude this article with a table of the values of $\mathbb{P}(G\ \text{is Hausdorff})$ for some noteworthy finite groups $G$, which were computed with the help of the Group Explorer library. I would like to thank my dear friend Eduardo Sodré for his help with the problem. I really hope the ride was as interesting to you as it was us both! 😁

$G$
$\mathbb{P}(G\ \text{is Hausdorff})$

$Q_4$

$1/6$

$K_4$

$1/5$

$S_4$

$1/4$

$S_n$ for $n \ge 5$

$1/3$

$D_n$ for even $n$

$\frac{1}{3 + \tau(n)}$

$D_n$ for odd $n$

$\frac{1}{1 + \tau(n)}$