Let's analyze each statement about the set of rational numbers \(\mathbb{Q}\) under addition, considering its properties as a group \((\mathbb{Q}, +)\).

Subgroups of Rational Numbers \(\mathbb{Q}\)

Statement 1: \(\mathbb{Q}\) has countably many subgroups

This statement claims that the group \((\mathbb{Q}, +)\) has a countable number of subgroups. To verify this, we need to understand the structure of subgroups of \(\mathbb{Q}\).

Consider subgroups of the form \(G_S = \{ x \in \mathbb{Q} \mid \text{the denominator of } x \text{ (in lowest terms) is only divisible by primes in } S \}\) for different sets of prime numbers \(S\). For example:

• If \(S = \emptyset\), then the denominator can only be 1, so \(G_\emptyset = \mathbb{Z}\).
• If \(S = \{2\}\), then \(G_{\{2\}} = \{ a/2^k \mid a \in \mathbb{Z}, k \ge 0 \}\), the dyadic rationals.
• If \(S = \{p \text{ | } p \text{ is prime}\}\), then \(G_S = \mathbb{Q}\).

For any two distinct sets of primes \(S_1 \ne S_2\), the corresponding subgroups \(G_{S_1}\) and \(G_{S_2}\) are distinct. For instance, if \(p \in S_1\) but \(p \notin S_2\), then \(1/p \in G_{S_1}\) but \(1/p \notin G_{S_2}\) (assuming \(p\) is the only prime factor in the denominator). There are uncountably many distinct sets of prime numbers (the power set of the set of primes, which is infinite, is uncountable). While these \(G_S\) don't represent *all* subgroups of \(\mathbb{Q}\), they show there's an uncountable family of distinct subgroups.

In fact, it is a known result that the group \((\mathbb{Q}, +)\) has uncountably many subgroups. Therefore, statement 1 is false.

Subsets of Rational Numbers \(\mathbb{Q}\)

Statement 2: \(\mathbb{Q}\) has uncountably many subsets

The set of rational numbers \(\mathbb{Q}\) is countably infinite. This means its cardinality is \(\aleph_0\).

The number of subsets of any set \(A\) is given by the cardinality of its power set, denoted by \(|\mathcal{P}(A)|\) or \(2^{|A|}\).

For \(\mathbb{Q}\), the number of subsets is \(|\mathcal{P}(\mathbb{Q})| = 2^{|\mathbb{Q}|} = 2^{\aleph_0}\).

By Cantor's theorem, for any set \(A\), \(|A| < |\mathcal{P}(A)|\). Thus, \(\aleph_0 < 2^{\aleph_0}\).

The cardinality \(2^{\aleph_0}\) is the cardinality of the continuum, which is uncountable. For example, it is the same cardinality as the set of real numbers \(\mathbb{R}\).

Therefore, \(\mathbb{Q}\) has uncountably many subsets. Statement 2 is true.

Finitely Generated Subgroups of \(\mathbb{Q}\)

Statement 3: Every finitely generated subgroup of \(\mathbb{Q}\) is cyclic

Let \(H\) be a finitely generated subgroup of \((\mathbb{Q}, +)\). By definition, this means \(H\) is generated by a finite set of rational numbers, say \(\{q_1, q_2, \dots, q_n\}\), where \(q_i \in \mathbb{Q}\).

Any element in \(H\) can be written as a linear combination with integer coefficients:

\(h = k_1 q_1 + k_2 q_2 + \dots + k_n q_n\)

where \(k_i \in \mathbb{Z}\).

Let each \(q_i = \frac{a_i}{b_i}\) where \(a_i, b_i \in \mathbb{Z}\) and \(b_i \ne 0\). We can find a common denominator for all \(q_i\). Let \(L = \text{lcm}(|b_1|, |b_2|, \dots, |b_n|)\). Then each \(q_i\) can be written as \(q_i = \frac{a_i'}{L}\) for some integer \(a_i'\).

So, an element \(h \in H\) is of the form:

\(h = k_1 \frac{a_1'}{L} + k_2 \frac{a_2'}{L} + \dots + k_n \frac{a_n'}{L} = \frac{k_1 a_1' + k_2 a_2' + \dots + k_n a_n'}{L}\)

Let \(a' = \text{gcd}(a_1', a_2', \dots, a_n')\). Then each \(a_i' = a' \cdot c_i\) for some integers \(c_i\), and \(\text{gcd}(c_1, c_2, \dots, c_n) = 1\).

So, \(h = \frac{k_1 a' c_1 + \dots + k_n a' c_n}{L} = \frac{a'(k_1 c_1 + \dots + k_n c_n)}{L}\).

Let \(q = \frac{a'}{L}\). Any element in \(H\) is of the form \(\frac{a' \cdot m}{L} = m \cdot q\), where \(m = k_1 c_1 + \dots + k_n c_n\) is an integer.

Since \(\text{gcd}(c_1, \dots, c_n) = 1\), by Bezout's identity, there exist integers \(x_1, \dots, x_n\) such that \(x_1 c_1 + \dots + x_n c_n = 1\). This implies that any integer \(m\) can be expressed in the form \(k_1 c_1 + \dots + k_n c_n\) by choosing \(k_i = m x_i\).

Thus, the set of all possible values for \(m\) is precisely the set of all integers \(\mathbb{Z}\).

So, \(H = \{ m \cdot q \mid m \in \mathbb{Z} \}\), which is the cyclic subgroup generated by \(q = \frac{a'}{L}\).

Therefore, every finitely generated subgroup of \(\mathbb{Q}\) is cyclic. Statement 3 is true.

Isomorphism of \(\mathbb{Q}\) with \(\mathbb{Q} \times \mathbb{Q}\)

Statement 4: \(\mathbb{Q}\) is isomorphic to \(\mathbb{Q}\) × \(\mathbb{Q}\) as groups

This statement claims that the group \((\mathbb{Q}, +)\) is isomorphic to the direct product group \((\mathbb{Q} \times \mathbb{Q}, +)\), where addition is component-wise: \((a,b) + (c,d) = (a+c, b+d)\).

Let's consider the properties of these groups. Both are abelian groups.

We can view \((\mathbb{Q}, +)\) as a vector space over the field \(\mathbb{Q}\). A basis for this vector space is \(\{1\}\), so its dimension is 1.

Similarly, \((\mathbb{Q} \times \mathbb{Q}, +)\) can be viewed as a vector space over \(\mathbb{Q}\) with scalar multiplication defined as \(r(a,b) = (ra, rb)\) for \(r \in \mathbb{Q}\). A basis for this vector space is \(\{(1,0), (0,1)\}\), so its dimension is 2.

Two vector spaces over the same field are isomorphic if and only if they have the same dimension. Since the dimensions are 1 and 2, they are not isomorphic as vector spaces over \(\mathbb{Q}\).

While a group isomorphism doesn't strictly require preserving scalar multiplication, the vector space structure over \(\mathbb{Q}\) reflects the divisible and torsion-free nature of these groups. If a group isomorphism \(\phi: \mathbb{Q} \to \mathbb{Q} \times \mathbb{Q}\) existed, it would preserve the group structure.

Consider the finitely generated subgroups. As shown in Statement 3, every finitely generated subgroup of \(\mathbb{Q}\) is cyclic (isomorphic to \(\mathbb{Z}\) or \(\{0\}\)).

Now consider \(\mathbb{Q} \times \mathbb{Q}\). The subgroup generated by \(\{(1,0), (0,1)\}\) is \(\{ k(1,0) + m(0,1) \mid k, m \in \mathbb{Z} \} = \{ (k, m) \mid k, m \in \mathbb{Z} \} = \mathbb{Z} \times \mathbb{Z}\). This is a finitely generated subgroup of \(\mathbb{Q} \times \mathbb{Q}\).

The group \(\mathbb{Z} \times \mathbb{Z}\) is not cyclic (e.g., there is no single element \((a,b)\) such that every element \((k,m)\) can be written as \(n(a,b)\) for some integer \(n\), unless \((a,b) = (\pm 1, 0)\), \((0, \pm 1)\) or \((a,b)\) is a generator of one of these, but these only generate subgroups isomorphic to \(\mathbb{Z}\), not \(\mathbb{Z} \times \mathbb{Z}\)).

If \(\mathbb{Q} \cong \mathbb{Q} \times \mathbb{Q}\), then any property preserved by group isomorphisms must hold for both groups. Being "every finitely generated subgroup is cyclic" is such a property. \(\mathbb{Q}\) has this property (Statement 3 is true), but \(\mathbb{Q} \times \mathbb{Q}\) does not (it has \(\mathbb{Z} \times \mathbb{Z}\) as a finitely generated subgroup which is not cyclic).

Therefore, \(\mathbb{Q}\) is not isomorphic to \(\mathbb{Q} \times \mathbb{Q}\) as groups. Statement 4 is false.

Summary of Statements

Based on the analysis:

• Statement 1: \(\mathbb{Q}\) has countably many subgroups - False.
• Statement 2: \(\mathbb{Q}\) has uncountably many subsets - True.
• Statement 3: Every finitely generated subgroup of \(\mathbb{Q}\) is cyclic - True.
• Statement 4: \(\mathbb{Q}\) is isomorphic to \(\mathbb{Q}\) × \(\mathbb{Q}\) as groups - False.

The statements that are true are Statement 2 and Statement 3.