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### Problem 4

Suppose that $k$ is a divisor of $n$. Prove that $\mathbb{Z}_n / \langle k \rangle \simeq \mathbb{Z}_k$.

**Solution**

Suppose that $k$ is a divisor of $n$, with $\mathbb{Z}_n / \langle k \rangle$.

Then $\mid k \mid = \dfrac{n}{gcd(n,k)}$.

$\hspace{40pt} =\dfrac{n}{k}$, since $k$ divides $n$ and $k$ is the greatest divisor of itself.

Consequently, $\mid \mathbb{Z}_n / \langle k \rangle \mid = \dfrac{n}{\bigg( \dfrac{n}{k} \bigg) } = k$.

Now, by Theorem 15.9, $\mathbb{Z}_n / \langle k \rangle$ is cyclic.

Furthermore, $\mathbb{Z}_n / \langle k \rangle$ is abelian, since every cyclic group is abelian.

Now by the Fundamental Theorem of Finitely Generated Abelian Groups, any cyclic group of order $k$ must be isomorphic to $\mathbb{Z}_k$.

Therefore, $\mathbb{Z}_n / \langle k \rangle \simeq \mathbb{Z}_k$.

$\hspace{100 pt} \clubsuit$