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What is the smallest group $S$(in terms of subgroups) for which every finite cyclic group is a subgroup? I know that $\mathbb Q/\mathbb Z$ contain all finite cyclic groups as subgroups, is there a way to prove it doesn’t contain a non-isomorphic subgroup who contains all of them? Can any of you help?

razivo
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3 Answers3

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As Mike points out, "smallest" is not a precise terminology. However, since we already have a model $\mathbb{Q}/\mathbb{Z}$, one can prove what razivo suggests at least, i.e. there's no proper subgroups of $\mathbb{Q}/\mathbb{Z}$ satisfying the desired property. That's basically the existing question:Q/Z has a unique subgroup of order n for any positive integer n?. In the process, one actually shows each element of $\mathbb{Q}/\mathbb{Z}$ lies in some cyclic group of order $n$ (Note: $n$ might not be unique though). Thus if any element is taken away, there must be some cyclic group hurt.

PS. The torsion subgroup of $\mathbb{C}^\times$, which lhf suggests, is isomorphic to $\mathbb{Q}/\mathbb{Z}$ through the exponential map.

EDIT: I would like to present a possibly "fair" way to understand "smallest" in this context. For example, we hope that the smallest group containing $\mathbb{Z}/3\mathbb{Z}, \mathbb{Z}/6\mathbb{Z}$ and $\mathbb{Z}/8\mathbb{Z}$ should be $\mathbb{Z}/24\mathbb{Z}$. When we add more groups, for example, $\mathbb{Z}/5\mathbb{Z}$ and $\mathbb{Z}/10\mathbb{Z}$, we could embed all the groups mentioned above into $\mathbb{Z}/120\mathbb{Z}$, which is again smallest. Now it's inspiring to consider the diagram consisting of all cyclic groups $\mathbb{Z}/n\mathbb{Z}$, with arrows the emdeddings from $\mathbb{Z}/d\mathbb{Z}$ into $\mathbb{Z}/n\mathbb{Z}$ if and only if $d|n$. The colimit of this (directed) diagram seems to be the precise characterisation of "smallest" in our sense. Not surprisingly, one can prove this colimit is exactly $\mathbb{Q}/\mathbb{Z}$.

A convenient proof may use the disjoint-union characterisation of directed colimits, based on which we notice: once we identify $\mathbb{Z}/n\mathbb{Z}$ as $\frac{1}{n}\mathbb{Q}/\mathbb{Z}$, all the equivalent relations required in the characterisation of colimits (see the link above) coincide with those in $\mathbb{Q}/\mathbb{Z}$. For example, $4$ in $\mathbb{Z}/10\mathbb{Z}$ is equivalent with $6$ in $\mathbb{Z}/15\mathbb{Z}$, since they both correspond to $12$ in $\mathbb{Z}/30\mathbb{Z}$. This corresponds exactly to the fact $\frac{4}{10} = \frac{6}{15}$ in $\mathbb{Q}/\mathbb{Z}$ (since they both equal to $\frac{12}{30}$, by elementary arithmetic). Thus the directed colimit coincides with $\mathbb{Q}/\mathbb{Z}$.

Rigid AOE2
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This is a tough question because of the subjectivity of the word "smallest," but I can at least give you a smart-ass correct answer :) The group you're looking for will be the limit of the full subcategory of the category of abelian groups that consists only the (finite) cyclic groups. This isn't helpful except that it tells you such a groups exists since limits must exist in this category. Only considering finite cyclic groups you get something isomorphic to $\mathbf{Q}/\mathbf{Z}$ as folks have pointed out.

Idk how to describe the one where you consider the free cyclic group $\mathbf{Z}$ though. I'll leave this answer CW for anyone who wants to try to flush out details about this group.

Mike Pierce
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    There is no subjectivity. Smallest in mathematics is clear. In this case it means the intersection of all the groups that contain the cyclic groups. – John Douma Nov 20 '20 at 16:01
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    Still not sure what you mean by the limit in the full subcategory, which does not exist clearly. Or could you please explain why we use limits to indicate "smallest" here? – Rigid AOE2 Nov 20 '20 at 17:01
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    @JohnDouma I am not convinced that makes sense. What exactly does "the cyclic groups" mean here? – Derek Holt Nov 20 '20 at 17:30
  • @DerekHolt I am pretty sure you know that when we are asked to find the smallest group that has a property it is the intersection of all groups with that property. – John Douma Nov 20 '20 at 17:36
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    @JohnDouma Taking the intersection may not work generally. We may consider $G = \Pi_{n=1}^{\infty} \mathbb{Z}/n\mathbb{Z}$. By letting the first $i-1$ components be $0$, we obtain a subgroup $G_{i} = \Pi_{n=1}^{i-1} {0} \times \Pi_{n=i}^{\infty} \mathbb{Z}/n\mathbb{Z}$. It's clear that any $G_{i}$ satisfies the desired property of containing all finite cyclic groups. However their intersection is $0$, since any component of the elements in the intersection will be $0$ as $n$ is large enough. In particular, the intersection of all subgroups with that property is $0$. – Rigid AOE2 Nov 20 '20 at 18:28
  • @LawNgai It may not work but the definition of smallest is perfectly clear. That was my point. – John Douma Nov 20 '20 at 18:33
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    @JohnDouma The problem is that there is no canonical choice of an ambient group, where all the subgroups live, so that we can talk about taking intersection. – Rigid AOE2 Nov 20 '20 at 18:35
  • @LawNgai There is no problem. This is how smallest is defined. – John Douma Nov 20 '20 at 18:41
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    @JohnDouma I am afraid that there is a problem. You wrote "the intersection of all the groups that contain the cyclic groups", and I have no idea what you mean by "the cyclic groups". In general, the intersection of two groups is not even a group - it could be the empty set for example. – Derek Holt Nov 20 '20 at 19:28
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    @JohnDouma The groups ${0}$ under addition and ${1}$ under multiplication have empty intersection. – Derek Holt Nov 20 '20 at 19:32
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    I'm so happy that I get notified of this discussion ;) – Mike Pierce Nov 20 '20 at 20:43
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    @DerekHolt: It's not. It's how you define "smallest subgroup $H$ of $G$ containing a set $S$". But your definition does not work when there's no universe you're restricting to, or when there's not a particular set $S$ but instead some categorization of structure (isomorphism classes). Both problems are present here. – R.. GitHub STOP HELPING ICE Nov 21 '20 at 02:33
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The smallest group that contains (copies of) all finite cyclic groups is the torsion subgroup of $\mathbb C^\times$, that is, the elements of finite order. It's the set of all complex roots of unity.

lhf
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