# A group allowing exactly 7 group topologies

Is there a group $G$ allowing exactly 7 group topologies on $G$: $\mathcal T_{\text{trivial}}, \mathcal T_{\text{discrete}}, \mathcal T_1, \mathcal T_2,\mathcal T_3,\mathcal T_4, \mathcal T_5$ with

$$\mathcal T_i\nsubseteq \mathcal T_j,~~ \mathcal T_j\nsubseteq \mathcal T_i$$ for every distinct $i,j\in \{1,2,3,4,5\}$?

• The question above is a rephrased form of this question: Is there a group with the lattice of all group topologies on it order-isomorphic to the lattice $M_5$?
• It can be proved that $G$ cannot be finite. For a similar question with $p+3$ group topologies (instead of $7$) where $p$ is prime, there is a simple proof for existence of such group. This is why the number 7 is selected for this question.

• There is an infinite group with exactly 2 group topologies which may be used in constructing $G$. This is why the question above can be interesting.

• why do you want to know? – Will Jagy Aug 17 '14 at 20:54
• @WillJagy: I'm captured by a few aliens. they say I'll be released only if I can give them a proof if such $G$ exists. – Minimus Heximus Aug 17 '14 at 21:00
• "This is why they say 7". Who is they? – Steven Landsburg Aug 17 '14 at 23:09
• @Noah S: A topology on a finite group is the pullback of the discrete topology on the quotient by the closed and normal subgroup given by intersecting of all closed subsets containing the identity. So topologies are the same as normal subgroups; the non-inclusivity condition is asking that any two nontrivial normal subgroups intersect only in the identity, so $G$ is the product of any two. If there are at least 3 nontrivial such subgroups, they must all be the same simple group $H$ and $G = H \times H$, and has $2 + |Aut(H)|$ nontrivial normal subgroups. – Vivek Shende Aug 18 '14 at 6:46
• @VivekShende If $G$ is infinite, then $G \times G \to G$ is not continuous for the cofinite topology. – Dan Petersen Aug 18 '14 at 19:33

Edit (Oct 12, 2015):

My original answer contained an error, as noticed by user47958. I now record only those parts of what I wrote which might be relevant to a future complete solution. (What I am writing here is NOT a complete solution.)

I order topologies so that $\mathcal T\subseteq \mathcal T'$ means that $\mathcal T'$ is finer than $\mathcal T$. I let $M_n$ denote the lattice of height three with $n$ atoms.

Background info:

1. If $N$ is any normal subgroup of a group $G$, then the set $\mathcal T_N$ consisting of those subsets of $G$ that are unions of cosets of $N$ is a group topology on $G$. The discrete and indiscrete topologies have the form $\mathcal T_N$ for $N =\{e\}$ or $N=G$ respectively. If $M\neq N$, then $\mathcal T_M\neq \mathcal T_N$. Since $\mathcal T_M\vee \mathcal T_N = \mathcal T_{M\cap N}$ and $\mathcal T_M\wedge \mathcal T_N = \mathcal T_{MN}$, the lattice of normal subgroups of $G$ is dually embedded in the lattice of group topologies via $N\mapsto \mathcal T_N$.

2. For any group topology $\mathcal T$ on $G$ the intersection of the neighborhoods of the identity in $\mathcal T$ is a normal subgroup $N\lhd G$. Any set in $\mathcal T$ is a union of cosets of $N$. The topology $\mathcal T_N$ from Item 1 contains the starting topology $\mathcal T$. In particular, if $\mathcal T$ is coatom in the lattice of group topologies, and the intersection $N$ is nontrivial, then $\mathcal T = \mathcal T_N$. (Briefly, non-Hausdorff coatoms in the lattice of group topologies have the form $\mathcal T_N$ for some normal subgroup $N$.)

3. The intersection $N$ from Item 2 equals $\{e\}$ iff $\mathcal T$ is $T_0$ (equivalently Hausdorff, equivalently Tychonoff).

4. If $G$ is finite, then all group topologies on $G$ have the form $\mathcal T_N$. The lattice of group topologies is therefore dually isomorphic to the lattice of normal subgroups of $G$.

5. If a group $G$ has at least three pairwise complementary normal subgroups, then $G$ is finite and in fact $G\cong \mathbb Z_p\times \mathbb Z_p$ for some prime $p$. In this case the normal subgroup lattice of $G$ is $M_{p+1}$.

The problem asks if there is a topological group whose lattice of group topologies is $M_5$. Since $5$ is not a prime plus one, such a group cannot be finite. In fact, no more than two of the (co)atoms in the topology lattice can have the form $\mathcal T_N$ for $N\lhd G$. This means that (i) $G$ has at most two proper nontrivial normal subgroups, each of which is an atom and a coatom in the normal subgroup lattice of $G$. Therefore (ii) at most two of the coatoms in the group topology lattice can be non-Hausdorff. Hence (iii) at least three of the atoms in the lattice of group topologies on $G$ are pairwise complementary Hausdorff topologies. Item (i) splits the full problem into these cases:

Case I. $G$ is simple and has 5 pairwise complementary proper nontrivial Hausdorff group topologies.

Case II. $G$ has exactly one nontrivial proper normal subgroup $N$, $G/N$ has no nontrivial proper group topology, but $G$ has four pairwise complementary Hausdorff topologies that are nontrivial and proper.

Case III. $G=S\times T$ is a product of two simple groups, where neither $S$ nor $T$ has any nontrivial proper group topology, but $G$ has three pairwise complementary Hausdorff topologies that are nontrivial and proper.

I have learned that there is no known group that has even one pair of complementary proper Hausdorff group topologies. No abelian group can have complementary proper Hausdorff group topologies, as proved in

Relationship between the meet and join operators in the lattice of group topologies, Bradd Clark and Victor Schneider, Proceedings AMS 98(4) (1986) 681-682.

The general problem of whether there can exist complementary proper Hausdorff group topologies is posed as Question 7.1(a) in

On transversal group topologies, Dikran Dikranjan, Mikhail Tkachenko, Ivan Yaschenko, Topology and its Applications 153 (2005) 786-817.

Here is a construction of a group topology that might be applicable to derive properties of groups in Cases I, II or III. It is a generalization of the $\mathcal T_N$ topology.

Let $H$ be a subgroup of $G$. Let $S_H$ be the set of subgroups of $G$ that are finite intersections of conjugates of $H$. The groups in $S_H$ form a system of neighborhoods of $e$ for a group topology $\mathcal T_H$ on $G$. The topology $\mathcal T_H$ is nontrivial if $H$ is proper and is proper if $\{e\}\notin S_H$. $\mathcal T_H$ is Hausdorff if $H$ is core-free.

Here is an example that hints about how this construction might be applicable to the problem. Suppose that $G = A_{\omega}$ is the alternating group on $\omega$, $E\subseteq \omega$ is (say) the set of even natural numbers, and $H$ is the stabilizer of $E$. Every group in $S_H$ is the stabilizer of some set that is the symmetric difference of $E$ and a finite subset of $\omega$. Therefore $\mathcal T_H$ is a nontrivial proper Hausdorff group topology that is tied to the set $E$. You can produce lots more Hausdorff group topologies on $A_{\omega}$ by varying the set $E$.

For this construction to fail to produce nontrivial proper Hausdorff group topologies, it must be that whenever $H\leq G$ is proper and core-free, then $\{e\}$ is the intersection of finitely many conjugates of $H$.

• I'm surprised that someone has answered that question so I'm commenting before reading your reasoning for the answer "No". I will be released after a year! – Minimus Heximus Aug 15 '15 at 22:11
• That's a really nice answer! You must have thought about this question for quite a while... – user62017 Aug 20 '15 at 8:00
• "At most one of the atoms in the lattice of compatible topologies can be Hausdorff (by Claim 3)". How do you deduce it from Claim 3? Claim 3 is talking about intersection of topologies. We don't have any assumptions about intersection. – Minimus Heximus Oct 10 '15 at 8:20