Can a countable group have uncountably many 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.

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.

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 $\mathbb Q$.