We examine topologies on finite groups, and determine which topologies contain proper subgroups and how many
We will examine finite topological spaces in which the finite set in question is a finite group. We have two primary questions of interest: Given a finite group, which topologies, if any, will contain proper subgroups as open sets and how many? We offer only a partial answer to these questions, focusing on the special case in which the finite group in question is of prime order.
Consider the trivial group $G = \{e\}$. There is only one topology on this group, the trivial topology $\tau = \{\emptyset, G\}$, which clearly contains no proper subgroups as open sets. Next, consider the group $G = \{e, a\}$, where $a^2 = e$, so $G \cong Z_2$. There are four topologies on this group: $$\tau_1 = \{\emptyset, G\}$$ $$\tau_2 = \{\emptyset, \{e\}, G\}$$ $$\tau_3 = \{\emptyset, \{a\}, G\}$$ $$\tau_4 = \{\emptyset, \{e\}, \{a\}, G\}$$ Observe that $\tau_2$ and $\tau_4$ contain exactly one proper subgroup, namely $\{e\}$. The following theorem allows us to generalize this observation to any group of prime order.
Now, consider the case where $G\cong Z_3$. The number of possible topologies increases to 29 in this case, so rather than list them all out, we make an observation. By Lagrange's Theorem, for any group of prime order, the only proper subgroup is the trivial subgroup. Thus, the questions of interest become determining which topologies contain $\{e\}$ as an open set and how many. The following theorem tells us that for a group $G$ of prime order p, the number of topologies on $G$ that contain a proper subgroup is equal to the total number of topologies on a set with $p-1$ elements.
Let $G$ be a finite group of prime order, and $T_G$ be the set of all topologies on G that contain $\{e\}$ as an open set. Let $X = G - \{e\}$, and $T_X$ be the set of all topologies on X. Then there exists a bijection $\phi : T_G \to T_X $, so $|T_G|=|T_X|$.
Suppose $T \in T_G$. Define $\phi(T) = \{U \cap X | U \in T \}$. This is the subspace topology on $X$, thus is in $T_X$. To show that $\phi$ is injective, suppose $T1, T2 \in T_G$ and $\phi(T1)=\phi(T2)$. We must show $T1=T2$. Suppose $U1 \in T1$. Then $U1 \cap X = U2 \cap X$ for some $U2 \in T2$.
If $e \in U1, U1 \cap X = U1 - \{e\}$. If $U2$ also contains e, then $U2 \cap X = U2 - \{e\}$, so $U1=U2$. If $e \notin U2$, then $U2 \cap X = U2$, so $U1 \cap X = U1 - \{e\} = U2$ and $U1 = U2 \cup \{ e\} \in T2$ by closure of unions and the assumption that all topologies in $T_G$ contain $\{e\}$.
If $e \notin U1, U1 \cap X = U1$. If $e \notin U2$, then $U2 \cap \{e\} = U2$, so $U1 = U2$. In the final case, we are trying to show that if $e \notin U1$ and $e \in U2$, then $U1 \in T2$. We consider the contrapositive statement, so suppose $U1 \notin T2$. We must show $e \in U1$ or $e \notin U2$. If $e \notin U2$, we're done. So suppose $e \in U2$. Then we must have $e \in U1$, since otherwise, $U1 \cap X = U1$ and $U2 \cap X = U2$, so $U1=U2$ and $U1 \in T2$, a contradiction. Thus $U1 \in T2$ and $T1 \subset T2$. A similar proof shows that $T2 \subset T1$, so $T1 = T2$ and $\phi$ is injective.
To show $\phi$ is surjective, suppose $Y \in T_X$. Let $T = \emptyset \cup \{U \cup \{ e \} | U \in Y \}$. We claim $T \in T_G$ and $\phi(T) = Y$. Observe, $\emptyset, G, \{e\} \in T$. Also, since $Y$ is a topology, it is closed under union and finite intersection, thus so is $T$, since we union each set of $Y$ with $\{e\}$. So $T \in T_G$. Also, $\phi(T) = Y$ by the definition of $\phi$. Thus, $\phi$ is surjective and we conclude that $\phi$ is a bijection. $ \qquad \blacksquare $
Using the above theorem, we can return to the case where $|G| = 3$, and conclude that the number of topologies on $G$ which contain a proper subgroup as an open set, $T_o$, is equal to total number of topologies on a set with 2 elements, $T(2) = 4$. Similarly, $T_o(5) = T(4) = 355$, and $T_o(7) = T(6) = 209527$. [1]
We have provided a partial classification to the question of which topologies on a finite group will contain a proper subgroup and how many. For a group $G$ of prime order p, the number of topologies containing a proper subgroup as an open set is equal to $T(p-1)$, where $T$ returns the total number of topologies on a set with n elements. Naturally, we would like a result that applies to arbitrary finite groups. Other related questions arise when considering arbitrary groups, such as how the number changes for larger groups with more than one isomorphism class? Also, is there algorithm for determining which topologies will contain a proper subgroup beyond brute-force computation? As an extension, we could also ask if we can form a basis for a topology entirely from proper subgroups.
[1] https://oeis.org/A000798
[2] Munkres, J. R. (2000). Topology (2nd ed.). Prentice Hall, Inc.
[3] Gallian, J. (2021). Contemporary Abstract Algebra (10th ed.). Chapman and Hall/CRC