Given $k,m \geq 0$, prove $\langle k \rangle / \langle km \rangle \simeq \mathbb {Z}_m$ where $\langle k \rangle$ is the subgroup generated by $k$ and $k\in \mathbb{Z}$
I think this is an immediate consequence of the fact that
Every finite cyclic group of order n is isomorphic to the additive group of $Z/nZ$, the integers modulo n.
Because $\langle k \rangle / \langle km \rangle $ has order $m$. I still want to give a more explicit proof, defining an isomorphism.
The set $\langle k \rangle / \langle km \rangle $ could be descriibed as $\{[k], [2k], \dots [mk]\}$. Then if we define $\phi:\langle k \rangle / \langle km \rangle \to \mathbb{Z}_m$ by $\phi([xk]) = [x]$ the function $\phi$ is bijective, thus the sets are isomorphic.
I think that the efficient way to do this is to consider a group homomorphism $\Bbb Z\to k\Bbb Z/km\Bbb Z$, which can also be written, as you do, as $\langle k\rangle/\langle km\rangle$.
The homomorphism I have in mind is $n\mapsto[kn]$, where the latter is the coset of $kn$ in $k\Bbb Z/km\Bbb Z$. This homomorphism is clearly onto, and what’s the kernel? Multiples of $m$, i.e. the group $m\Bbb Z$. Thus, by the First Isomorphism Theorem, $\Bbb Z/m\Bbb Z\cong k\Bbb Z/km\Bbb Z$