Exercise 4D, 4).

From 3), we only need to show that for every J ⊆ ℤ_{m} and any ℤ_{m}-module homomorphism ϕ : J → ℤ_{m}, the following diagram:

0 → J →^{i} ℤ_{m}
ϕ ↓
ℤ_{m}

can be completed with a map χ such that χ ∘ i = ϕ.

Note that J = ⟨d⟩, where d | m, so ϕ is determined by ϕ(d). Also, 0 = ϕ(\frac{m}{d} · d) = \frac{m}{d}ϕ(d), so ϕ(d) = rd, some r ∈ ℤ_{m}.

Similarly, i(d) = sd, some s ∈ ℤ_{m}. i is 1-1, so i is onto ⟨d⟩. Thus, i(td) = d, for some t ∈ ℤ_{m}.

Now, i(std) = si(td) = sd = i(d) ⇒ std = d( by 1 − 1). Also, ϕ(td) = trd, so we define χ by χ(1) = tr. Then for any l ∈ ℤ_{m}, χ(i(ld)) = χ(li(d)) = lχ(i(d)) = lχ(sd) = lsχ(d) = lstrd = lr(std) = lrd = r(ld) = ϕ(ld).

Thus, χ ∘ i = ϕ.