Kernel of surjective homomorphism isomorphic to $\mathbb{Z}/m\mathbb{Z}$?

Question

Assume $d \mid n$. Is the kernel of the surjective homomorphism $f: \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/d\mathbb{Z}$ given by $[n] \mapsto [n]$ isomorphic to $\mathbb{Z}/m\mathbb{Z}$ where $m = {n\over{d}}$? Is there a way to answer this question without using anything advanced like the isomorphism theorems?

Answer 1

Yes. Since $f$ is surjective, as a result of the first isomorphism theorem, we have $$\frac{\#\mathbb Z/n\mathbb Z}{\#\ker f}={\#\mathbb Z/d\mathbb Z}$$so $\#\ker f =\frac nd$. But $\ker f$ is a subgroup of the cyclic group ${\mathbb Z/n\mathbb Z}$ so must itself be cyclic, and its order $m = \frac nd$.

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Indeed, there is an explicit isomorphism $$\ker f = \{[k]\in \mathbb Z/n\mathbb Z:k\equiv 0 \pmod d\}\to\mathbb Z/m\mathbb Z\\ [k]\mapsto \frac kd \pmod m$$