Let B_{p} be the subgroup of ℚ/ℤ generated by the classes of the elements of the form \frac{a}{p^{i}}, a ∈ ℤ, i ≥ 0 where p is a fixed prime. Show that

ℚ/ℤ = ⨁_{p prime} B_{p}.

Proof. Let p_{j} be a prime number, j ∈ ℕ. Let φ_{j} : B_{p_{j}} → ℚ/ℤ be the natural inclusion map, i.e.,

\overline{\frac{a}{p^{i}}} ↦ \overline{\frac{a}{p^{i}}}

where ^{−} denotes the class of the element. By the UMP there exists a unique group homomorphism φ : ⊕_{p prime}B_{p} → ℚ/ℤ given by

(\overline{\frac{a_{1}}{p_{1}^{i_{1}}}} , \overline{\frac{a_{2}}{p_{2}^{i_{2}}}} , …, \overline{\frac{a_{s}}{p_{s}^{i_{s}}}}) ↦ \overline{\frac{a_{1}}{p_{1}^{i_{1}}}} + \overline{\frac{a_{2}}{p_{2}^{i_{2}}}} + ⋯ + \overline{\frac{a_{s}}{p_{s}^{i_{s}}}} = \overline{\frac{a_{1}}{p_{1}^{i_{1}}} + \frac{a_{2}}{p_{2}^{i_{2}}} + ⋯ + \frac{a_{s}}{p_{s}^{i_{s}}}}

Claim: φ is a group isomorphism.

We begin by showing that φ is one-to-one. Suppose

\overline{\frac{a_{1}}{p_{1}^{i_{1}}} + \frac{a_{2}}{p_{2}^{i_{2}}} + ⋯ + \frac{a_{s}}{p_{s}^{i_{s}}}} = 0

Then

\overline{\frac{p_{2}^{i_{2}}p_{3}^{i_{3}} ⋯ p_{s}^{i_{s}}a_{1} + p_{1}^{i_{1}}p_{3}^{i_{3}} ⋯ p_{s}^{i_{s}}a_{2} + ⋯ + p_{1}^{i_{1}}p_{2}^{i_{2}} ⋯ p_{s−1}^{i_{s−1}}a_{s}}{p_{1}^{i_{1}}p_{2}^{i_{2}} ⋯ p_{s}^{i_{s}}}} = 0,

or equivalently,

p_{1}^{i_{1}}p_{2}^{i_{2}} ⋯ p_{s}^{i_{s}}|p_{2}^{i_{2}}p_{3}^{i_{3}} ⋯ p_{s}^{i_{s}}a_{1} + p_{1}^{i_{1}}p_{3}^{i_{3}} ⋯ p_{s}^{i_{s}}a_{2} + ⋯ + p_{1}^{i_{1}}p_{2}^{i_{2}} ⋯ p_{s−1}^{i_{s−1}}a_{s}.

Since p_{1}^{i_{1}}|p_{1}^{i_{1}}p_{2}^{i_{2}} ⋯ p_{s}^{i_{s}}, we have

p_{2}^{i_{2}}p_{3}^{i_{3}} ⋯ p_{s}^{i_{s}}a_{1} ≡ 0 (mod p_{1}^{i_{1}}).

From the above expression, we see that a_{1} ≡ 0 (mod p_{1}^{i_{1}}), or \overline{\frac{a_{1}}{p_{1}^{i_{1}}}} = 0 in B_{p_{1}}. A similar

argument will show that \overline{\frac{a_{j}}{p_{j}^{j_{1}}}} = 0 in B_{p_{j}} for j = 2, …, s. We have now shown that

(\overline{\frac{a_{1}}{p_{1}^{i_{1}}}} , \overline{\frac{a_{2}}{p_{2}^{i_{2}}}} , …, \overline{\frac{a_{s}}{p_{s}^{i_{s}}}}) = (0, 0, …, 0)

so the map is one-to-one.

The proof that φ is onto follows from the lemma below.

Lemma 1. Let \frac{a}{b} ∈ ℚ and suppose that b = p_{1}^{k_{1}} ⋯ p_{s}^{k_{s}} is the primary decomposition of b. Then we can find a_{1}, …, a_{k} such that

\frac{a}{b} = \frac{a_{1}}{p_{1}^{k_{1}}} + ⋯ + \frac{a_{s}}{p_{s}^{k_{s}}} .

Proof. Suppose p_{1}^{k_{1}}|b but no higher power of p_{1} divides b. Hence b = p_{1}^{k_{1}}b′. We observe that gcd(p_{1}^{k_{1}}, b′) = 1. Thus, by using the Euclidean algorithm we can find x and y in ℤ such that

p_{1}^{k_{1}}x + b′y = 1.

Thus p_{1}^{k_{1}}(ax) + b′(ay) = a. An easy calculation will show that

\frac{ax}{b′} + \frac{ay}{p_{1}^{k_{1}}} = \frac{a}{p_{1}^{k_{1}}b′} = \frac{a}{b}.

Set a_{1} = ay and a′ = ax. Now repeat the above argument for \frac{a′}{b′} by using one of the prime factors of b′.

Let \overline{\frac{a}{b}} ∈ ℚ/ℤ. By the above lemma, we have

\overline{\frac{a}{b}} = \overline{\frac{a_{1}}{p_{1}^{k_{1}}}} + ⋯ + \overline{\frac{a_{s}}{p_{s}^{k_{s}}}} .

It is then clear that (\overline{\frac{a_{1}}{p_{1}^{k_{1}}}} , …, \overline{\frac{a_{s}}{p_{s}^{ks}}}) maps to this element.