I tried to solve this question by the First Isomorphism Theorem but without a success.

Let $m,n$ be natural numbers such that $m \mid n$. Letting $d=n/m$, prove that $m\mathbb{Z}/n\mathbb{Z}$ isomorphic to $\mathbb{Z}/d\mathbb{Z}$.

I tried to find the right homomorphism to prove it but I didn't succeed.

  • $\begingroup$ Sal's answer below works fine. Let me add that morphisms from $\mathbb Z$ are determined completely by where $1$ goes. So whenever you need to prove something involving $\mathbb Z$ try to start with $\mathbb Z$ as the domain of a morphism. Then you can do what Sal does below, or define $\mathbb Z\to m\mathbb Z/n\mathbb Z$, $1\mapsto m$ and see if this is well defined and surjective. $\endgroup$ Dec 3, 2014 at 6:19

1 Answer 1


Consider the homomorphism $f\colon \mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}$ defined by $f(x)=mx+n\mathbb{Z}$.

Show that $\ker(f)=d\mathbb{Z}$ and that $\mathrm{im}(f)=m\mathbb{Z}/n\mathbb{Z}$, so the 1st isomorphism theorem implies $$\mathbb{Z}/d\mathbb{Z}=\mathbb{Z}/\ker(f)\cong\mathrm{im}(f)=m\mathbb{Z}/n\mathbb{Z}$$


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