prove that $(n \Bbb Z /k \Bbb Z)/(m \Bbb Z /k \Bbb Z) \cong n\Bbb Z / m \Bbb Z$ I need to prove thae following - given $n,m,k \in \Bbb N$ such that $n|m , m|k$ prove that $(n \Bbb Z /k \Bbb Z)/(m \Bbb Z /k \Bbb Z) \cong \Bbb Z / \frac mn \Bbb Z$ What I tried and what missing - I know that $k \Bbb Z , m \Bbb Z$ are normal in $n \Bbb Z$ i don't know how to prove it. I also found that using the 3rd Isomorphism theorm - $(n \Bbb Z /k \Bbb Z)/(m \Bbb Z /k \Bbb Z) \cong n \Bbb Z / m \Bbb Z$ and this is already close .. now im stuck . any help will be appreciated

Hint: for $n=1$ we have the isomorphism $(\mathbb{Z}/k\mathbb{Z})/(m\mathbb{Z}/k\mathbb{Z})\simeq \mathbb{Z}/m\mathbb{Z}$ via $(l+k\mathbb{Z})+m\mathbb{Z}\mapsto l+m\mathbb{Z}$. Now generalize this map to $n\ge 2$.