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?
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$\begingroup$ Do rational numbers mod $1$ contain $\Bbb Z$, which is cyclic? $\endgroup$– J. W. TannerCommented Nov 20, 2020 at 15:46
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2$\begingroup$ I see you edited to insert finite now $\endgroup$– J. W. TannerCommented Nov 20, 2020 at 15:54
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$\begingroup$ I have edited it thanks to your comment. I didn’t think about how it would have been preceived. $\endgroup$– razivoCommented Nov 20, 2020 at 16:05
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3 Answers
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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 $\mathbb{Q}/\mathbb{Z}$.
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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.
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3$\begingroup$ 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. $\endgroup$ Commented Nov 20, 2020 at 16:01
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7$\begingroup$ @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 $0$. $\endgroup$ Commented Nov 20, 2020 at 18:28
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10$\begingroup$ @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. $\endgroup$ Commented Nov 20, 2020 at 18:35
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7$\begingroup$ @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. $\endgroup$ Commented Nov 20, 2020 at 19:28
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14$\begingroup$ @JohnDouma The groups $\{0\}$ under addition and $\{1\}$ under multiplication have empty intersection. $\endgroup$ Commented Nov 20, 2020 at 19:32
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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.
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2$\begingroup$ The group you mentioned is isomorphic to what razivo suggests. $\endgroup$ Commented Nov 20, 2020 at 16:19