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Let $(G,+)$ be an (abelian) group whose cardinality is $\mathcal{k}$. What is the size of the space of all subgroups of $G$?

This question suggests that it is possible the answer is $2^{\mathcal{k}}$.

Anyway, my question is concerned only with the case $G=\Bbb{Q}^n$, where $n\ge1$. And I hope (with few chances) that the cardinality of the set of all its subgroups is exactly $\aleph_0$.

Edit: In the previous version of the question, I wrote $G=\Bbb{Q}$.

LBJFS
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  • See https://math.stackexchange.com/questions/104571/how-to-find-all-subgroups-of-mathbbq for most of the answer to this question. – Eric Towers Dec 29 '19 at 13:32
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    Does this answer your question? How to find all subgroups of $(\mathbb{Q},+)$ – Chain Markov Dec 29 '19 at 13:34
  • I have changed the question a little bit. Anyway, the accepted answer gives a reference and I have to check the paper. Thank you to both of you :) – LBJFS Dec 29 '19 at 13:40
  • If I understand correctly Theorem 1 of the reference, the set of all subgroups of $\Bbb{Q}$ has the same cardinality as $\Bbb{N}^{\Bbb{N}}$, i.e. the cardinality of continuum $2^{\aleph_0}$. – LBJFS Dec 29 '19 at 14:31
  • So, for any positive integer $n$, we have that $2^{\aleph_0}$ is a lower bound for the size of the set of subgroups of $\Bbb{Q}^n$, and hence it equals $\mathcal{c}$. Am I right? – LBJFS Dec 29 '19 at 14:36

1 Answers1

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Given that the cardinality is infinite, I think that the answer is $2^k$. The cardinality of an infinitely generated group is less or equal to the cardinality of the set of it's generators. Then you can define an injective function from $2^S$ to the set of subgroups of $G$ where $S$ is a minimal generating set, such that a vector in $2^S$ is sent to the subgroup of $G$ generated by the elements whose entries in the vector is $1$. For example if $S=\{ g_1, g_2,g_3,g_4 \}$, then $(1,0,0,1)\mapsto \langle g_1,g_4\rangle$.

If this argument is correct then $\vert 2^S\vert \leq \vert\mathcal{S}(G) \vert$, where $\mathcal{S}(G)$ is the set of subgroups of $G$. Since $\vert S\vert\geq k$, then $\vert \mathcal{S}(G)\vert \geq 2^k$.

I am using the answer I obtained in this thread: Cardinality of a generated group

Keen-ameteur
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  • In this thread, https://math.stackexchange.com/questions/257175/countable-group-uncountably-many-distinct-subgroup, doesn't Nate Eldrege's answer show that $\mathbb{Q}$ has uncountably many additive subgroups? – Keen-ameteur Dec 29 '19 at 15:47