Ring projection from \(\mathbb{Z}/m\mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z}\)

Let \(n,m \in \mathbb{Z}^+\) with \(n \mid m\). Prove that the natural surjective ring projection \(\mathbb{Z}/m\mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z}\) is also surjective on the units: \((\mathbb{Z}/m\mathbb{Z})^\times \rightarrow (\mathbb{Z}/n\mathbb{Z})^\times\).

Let \(\varphi\) be the ring projection. Suppose that \(m = p^a\) and \(n = p^b\) where \(p\) is a prime and \(a,b \in \mathbb{Z}^+ \cup \{0\}\). Clearly if \((x,n) = 1\), then \((x,m) = 1\) as well. Thus \(\varphi\) is surjective on the units.

Now suppose \(m,n\) are any positive integers; let \(m = {p_1}^{a_1} \ldots {p_n}^{a_n}\) and \(n = {p_1}^{b_1} \ldots {p_n}^{b_n}\) be their prime factorization (where \(b_i\) is set to 0 where needed). By the Chinese Remainder Theorem we have \((\mathbb{Z}/m\mathbb{Z})^\times \cong (\mathbb{Z}/{p_1}^{a_1}\mathbb{Z})^\times \times \ldots \times (\mathbb{Z}/{p_n}^{a_n}\mathbb{Z})^\times\) (likewise for \((\mathbb{Z}/n\mathbb{Z})^\times\)). Since the natural projections corresponding to each factor is surjective on the units as shown in the previous paragraph, \(\varphi\) is surjective.