# Can a countable group have uncountably many subgroups?

If $G$ is a countable group,can it have an uncountable number of distinct subgroups?

Let $V$ be a vector space of dimension $\aleph_0$ over a countable field $F$ (so $V$ is countable) and let $B$ be a basis for $V$ over $F$. Then every subset of $B$ spans a different subspace of $V$, so $V$ has $2^{\aleph_0}$ different subspaces, and its additive group has $2^{\aleph_0}$ different subgroups.

• OK. More interesting question: Can a countable group have precisely $\omega_1$ subgroups? (Without assuming $\mathsf{CH}$, of course.) – Andrés E. Caicedo Oct 13 '14 at 6:25
• @AndresCaicedo Hmm. Isn't the set of all subgroups of $G$ a closed set in the Tychonoff topology of $2^G$? Then, if $G$ is countable, the set of all subgroups of $G$ is homeomorphic to a closed set of real numbers, so it's either countable or has the cardinality of the continuum? Right? So, if $G$ is countable, then it can't have exactly $\omega_1$ subgroups, unless CH holds. – bof Oct 13 '14 at 6:33
• Yes, that's the only argument I know of. I'm half wondering if one can provide a combinatorial argument for the existence of continuum many subgroups that is not just this descriptive set theoretic approach in disguise. – Andrés E. Caicedo Oct 13 '14 at 6:42
• @AndresCaicedo Of course the "topological" argument has nothing to do with groups, it applies to any countable algebraic structure with finitary operations. I suppose there are some nontrivial settings where the existence of continuum many subalgebras can be proved by a "combinatorial" argument (whatever that means)? – bof Oct 13 '14 at 7:12

Consider the direct sum of countably many $\mathbb{Z}/2\mathbb{Z}$ groups, which I'll denote by $$G = \displaystyle \bigoplus_{n = 1}^\infty \left( \mathbb{Z} / 2\mathbb{Z} \right)_n$$ and where the index is to keep track of each copy of $\mathbb{Z}/2\mathbb{Z}$. A set of subgroups of $G$ are formed by including or excluding the $n$th copy of $\mathbb{Z}/2\mathbb{Z}$ (but as Slade corrected me in the comments, this does not give all the subgroups). Nonetheless, any infinite binary sequence yields a distinct subgroup by including those indices that are $1$ in the sequence, and so we have an injection from $2^\mathbb{N}$ into the set of subgroups of $G$. Thus the set of subgroups is uncountable.

• Ah, if you use direct sum, I believe this works. – Slade Oct 13 '14 at 4:38
• @Slade: yes, you're right on all accounts. It does work if I use the direct sum, and I'll edit that in – davidlowryduda Oct 13 '14 at 4:47
• Sorry,I am not getting the bijective map.It would be of great help if someone explains it in more details for me – Learnmore Oct 13 '14 at 4:51
• @learningmaths If you were referring to the original (as I just edited), it's because it wasn't actually bijective. Sorry about that. – davidlowryduda Oct 13 '14 at 4:52
• @mixedmath Doesn't $G$ have an uncountable number of elements? – Prince Kumar Dec 1 '16 at 5:55

One more example, using a very familiar group, the additive group $\mathbb Q$ of rational numbers. For any set $S$ of primes, consider the subgroup of $\mathbb Q$ consisting of those numbers that can be written as fractions (integer over integer) in which all the prime divisors of the denominator belong to $S$. (So, for example, when $S$ is empty, this subgroup is $\mathbb Z$, and when $S$ is the set of all primes, this subgroup is all of $\mathbb Q$.) There are continuum many choices for $S$, and each one leads to a different subgroup of $\mathbb Q$.

A finitely presented example: the free group $\mathbb{F}_2$ of rank $2$. Indeed, it contains the free group $\mathbb{F}_{\infty}$ of countable infinite rank. Let $\{x_1,x_2,\dots\}$ be a free basis for such a subgroup. For any sequence $\mathfrak{n} = (n_i)$ of positive integers, let $S( \mathfrak{n})$ denote the free subgroup generated by $\{x_{n_1},x_{n_2}, \dots\}$. Finally, $\{S (\mathfrak{n}) \mid \mathfrak{n} \}$ defines an uncountable family of pairwise distinct subgroups.

More generally, it may be noticed that any SQ-universal group has uncountably many normal subgroups. Let $G$ be such a group. It is clear that a countable group has countably many 2-generated subgroups, and because there exist uncountably many non-isomorphic such groups, $G$ must have uncountably many quotients. A fortiori, $G$ has uncountably many normal subgroups.

• For example, $\{a^iba^i:i\in\mathbb{Z}\}$ forms a basis for a copy of $\mathbb{F}_{\infty}$ in $\mathbb{F}_2$ (the derived subgroup $[\mathbb{F}_2, \mathbb{F}_2]$ is also isomorphic to $\mathbb{F}_{\infty}$, but the proof is less obvious, and would have to think what the basis would be!). – user1729 Oct 13 '14 at 10:42
• In fact, it seems to be natural to deduce that the rank of $[\mathbb{F}_2,\mathbb{F}_2]$ is infinite from algebraic topology. See for example chiasme.wordpress.com/2013/10/24/… Furthermore, an explicit basis may be found. – Seirios Oct 13 '14 at 11:34
• @Serios Yes, I agree, algebraic topology wins the day! – user1729 Oct 13 '14 at 12:09
• Algebraic geometry proof about the derived subgroup by user1729: math.stackexchange.com/a/983990/111377 – evgeny Feb 19 '16 at 16:44

In a similar vein to Andreas' answer, consider the additive group $\mathbb{Z}[X]$ of integer-coefficient polynomials (so with finitely many terms). This group is countable, but for each subset of the naturals $S \subset \mathbb{N}$ we get a subgroup $\langle \left\{ x^n \vert n \in S \right\} \rangle$ and these are all distinct.

• It is a good example, essentially $\mathbb{Z}^{(\mathbb{N})}$, that is, the multiplicative structure is not involved. – Orest Bucicovschi Oct 13 '14 at 18:25
• Indeed, the groups are familiar and I find monomials $x^n$ easier to think about than $\langle \frac{1}{p_n} \rangle$, where $p_n$ is the $n$th prime. – yatima2975 Oct 14 '14 at 9:28

The set of all bijection from $$\mathbb{N}$$ to $$\mathbb{N}$$ that forms a group and I think it has uncountably many subgroups.

For each $(b_n) \in \prod_{n\in \mathbb{N}}(\mathbb{Z}/2)$ consider the map from $\oplus _{n\in \mathbb{N}}(\mathbb{Z}/2)$ to $\mathbb{Z}/2$ $$(a_n) \mapsto \sum_n a_n \cdot b_n$$

take the kernel of this map; we get $2^{\aleph_0}$ subgroups of index $2$ of $\oplus _{n\in \mathbb{N}}(\mathbb{Z}/2)$.

@user1729: thanks! for correcting a previous statement about the number of subgroups of finite index of a countable group -- it may well be uncountable!

The example of @seirios: is a finitely generated group having uncountably many subgroups. However, a finitely generated group has finitely many subgroups of a given index $n$. Indeed, one can reduce to normal subgroups ( the normal core has index $N \le n!$) and then notice that there are finitely many morphisms from the groups to the finitely many groups of order $N$. Thus a finitely generated group has countably many subgroups of finite index.

• I do not understand how this answers the question. You have only shown that there are (finite or) countably many finite-index, normal subgroups of a finitely generated group (note: finitely generated, not countable, as mixedmath's answer demonstrates). Which is not awfully relevant... – user1729 Oct 14 '14 at 8:22
• Your now answer has some issues - not every $(b_n)\in\prod(\mathbb{Z}/2\mathbb{Z})$ corresponds to an element of the direct sum. This is because the direct sum contains only those elements with finite support (it is not the Cartesian product). Also, this is just a re-hashing of mixedmath's answer. If I were you, I would just delete everything above "The example of @seirios..." – user1729 Oct 17 '14 at 10:39
• @user1729: The kernel of the functional given by $(b_n)\ne (0)$ is a subgroup of index $2$. – Orest Bucicovschi Oct 17 '14 at 11:29