I'm learning abstract algebra, specifically group theory, and need help with the following problem:
Let $m, n \in \mathbb{N}$, with $n, m \geq 2$. For every natural number $k$, we denote $\langle k \rangle$ the subgroup of $(\mathbb{Z}, +)$ generated by $k$. Show that $|\langle n \rangle / \langle nm \rangle| = m$.
I'm sorry for my lack of effort but I was not able to do much with this one. Is is also unclear to me at the moment why the proposition fails for $m,n = 1$. Any help would be appreciated.
Consider the homomorphisms \begin{align} f\colon\langle n\rangle &\to \mathbb{Z} & g\colon\mathbb{Z} &\to \mathbb{Z}/\langle m\rangle \\ x&\mapsto x/n & x&\mapsto x+\langle m\rangle \end{align} What is the kernel of $g\circ f$? Then note that $g\circ f$ is surjective and apply the homomorphism theorem.
Actually there is no restriction about $m$ and $n$; they just need to be $\ge1$.