Can a countable group have uncountably many subgroups?

Question

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

Answer 1

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.

Answer 2

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 $\mathbb Q$.

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

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

Answer 7

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.