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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.

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    $\begingroup$ It doesn't fail for $m=1$ or $n=1$. $\endgroup$ – egreg Sep 6 '17 at 14:55
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    $\begingroup$ What makes you think the proposition fails for $m,n=1$? Then we have $$|\left<m\right>/\left<nm\right>|=|\left<1\right>/\left<1\right>| = |\mathbb Z/\mathbb Z|=|\{e\}| = 1 = m$$ just as claimed. $\endgroup$ – hmakholm left over Monica Sep 6 '17 at 14:56
  • $\begingroup$ Hint: $\langle n\rangle/\langle nm\rangle=\{0+\langle nm\rangle,n+\langle nm\rangle,2n+\langle nm\rangle,\dots,n(m-1)+\langle nm\rangle\}$. $\endgroup$ – Michael Burr Sep 6 '17 at 14:57
  • $\begingroup$ If does fail if either $m$ or $n$ is $0$, though. (Except, by coincidence, for $(n,m)=(0,1)$). $\endgroup$ – hmakholm left over Monica Sep 6 '17 at 14:58
  • $\begingroup$ If $n=1$, then $\langle n\rangle$ is not a proper subgroup, but it is still a subgroup. $\endgroup$ – Michael Burr Sep 6 '17 at 14:58
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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 $\ge1$.

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