prove that $(n \Bbb Z /k \Bbb Z)/(m \Bbb Z /k \Bbb Z) \cong n\Bbb Z / m \Bbb Z$

Question

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

Answer 1

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 $n\ge 2$.

Answer 2

You can consider the following homomorphism: $$ \phi:\frac{<n>}{<k>} \longrightarrow \frac{<n>}{<m>} $$ which send

$$ np+<k> \longmapsto np+<m> $$ where $p\in\mathbb{Z}$.

This a homomorphism of groups, and in particular $\phi$ is surjective. Now if you show that $ker\phi=\frac{<m>}{<k>}$ by the fundamental theorem of isomorphism you have finished.

But now $$np+<m>=<m>;$$ so is zero in the quotient; iff $np \in <m> $ iff $ p=ml+<k>$ iff $p\in \frac{<m>}{<n>}$. Intact if $p=ml+<k>$ then $np=nml+n<k>$; obviously $nml\in<m>$ and $<k>$ is a subgroup of $<m>$ since $m|k$.