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For a group $G$, define a collection of subsets $\tau_G$ of $G$ by declaring that a subset $H$ is in $\tau_G$ if

  • $H$ is empty, i.e. $H=\emptyset$, or
  • $H$ is a subgroup of $G$.

As first an example: $(\mathbb{Z}_{3},+)$ is a group. Then $\tau_{\mathbb{Z}_3} = \{\emptyset,\{0\},\mathbb{Z}_{3}\}$ is a topology on $\mathbb{Z}_3$. If $n$ is a prime power, then one can similarly show that $\tau_{\mathbb{Z}_n}$ is a topology on $\mathbb{Z}_n$.

For what groups $G$ is the set $\tau_G$ a topology on $G$?

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    $\begingroup$ Because union of open sets must be open, you need that the union of any two subgroups is a subgroup. If $H$ and $K$ are subgroups of $G$ and $H\cup K$ is a subgroup of $G$, then either $H\subseteq K$ or $K\subseteq H$. Thus any group where this construction works must have a subgroup lattice that is a path. The group $\mathbb{Z}_n$ where $n$ is a prime power satisfies this. I'm not sure if any other groups do. $\endgroup$ – Adam Lowrance Dec 10 '17 at 13:28
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    $\begingroup$ For the abelian case see this question and for the general case see this question. $\endgroup$ – J.-E. Pin Dec 10 '17 at 14:54
  • $\begingroup$ Thank you Adam on pointing out my typo on the original question. The empty set was missing on my notes too. $\endgroup$ – user496118 Dec 10 '17 at 17:06
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The intersection of two subgroups is always a subgroup, but the union of two subgroups is a subgroup if and only if one subgroup is a subset of the other. Since unions of open sets are open in a topology, it follows that $\tau_G$ is a topology if and only if its subgroup lattice is a chain $\{e\} \subset H_1 \subset H_2 \subset \cdots \subset H_n=G$.

Let $a, b\in G$. Since all subgroups are nested, either $\langle a \rangle \subset \langle b \rangle$ or $\langle b \rangle \subset \langle a \rangle$. In either case, we can conclude that $ab=ba$, and hence $G$ is abelian.

Claim: If $G$ is finite and its subgroup lattice is a chain, then $G$ is cyclic.

Proof of Claim: Suppose that $G$ is not cyclic and finite. Choose $a\in G$ of maximum order. Since $G$ is not cyclic, there is a $b\in G\setminus \langle a \rangle$. Since $b\notin \langle a \rangle$, it follows that $\langle b \rangle$ is not a subset of $\langle a \rangle$. If $\langle a \rangle$ is properly contained in $\langle b \rangle$, then $|a| < |b|$ contradicting that $a$ is an element of maximum order. Thus the subgroup lattice of $G$ is not a chain.

As noted in the question, the group $\mathbb{Z}_{p^k}$ has a subgroup lattice that is chain: $$\langle 0 \rangle \subset \langle p^{k-1}\rangle \subset \langle p^{k-2} \rangle \subset \cdots \subset \langle p \rangle \subset \langle 1 \rangle=\mathbb{Z}_{p^k}.$$ If $n$ has two distinct prime factors $p$ and $q$, then the subgroups $\langle n/p\rangle$ and $\langle n/q\rangle$ of $\mathbb{Z}_n$ do not contain one another. This completes the finite order case.

The remainder of this response is taken nearly verbatim from the second response here. Suppose $G$ is infinite. $G$ cannot contain elements of both finite and infinite order since the cyclic groups generated by those elements would have to be nested. If $G$ has no nontrivial elements of finite order, then $G$ contains two elements $a$ and $b$ of infinite order such that $\langle a \rangle$ is not a subset of $\langle b \rangle$ and vice versa. If $G$ only contains elements of finite order, but there exists elements of prime orders $p$ and $q$ for $p\neq q$, then you can use those elements to generate subgroups that are not subsets of one another.

What remains is the case where every element is of order $p^k$ for a fixed prime $p$, that is $G$ is an infinite $p$-group. For any $k$, the elements of order dividing $p^k$ is a cyclic subgroup. Every element of $G$ must be contained in such a subgroup. Thus $G$ is a direct limit of $\mathbb{Z}_{p^k}$ for $k\geq 1$. More concretely, $G$ is the subgroup of the unit complex numbers consisting of elements whose order is a power of a fixed prime $p$: $$G= \{z \in \mathbb{C}^*~:~ z=e^{2\pi i/p^k}~\text{for some $k$}\}.$$

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