I'm trying to show that if $G=\{a/p^n\in\mathbb{Q}:a\in\mathbb{Z},n\geq0\}$ for a fixed prime $p$, then the quotient $G/\mathbb{Z}$ is Artinian.
One minor detail I need to finish is $$ \langle 1/p^n+\mathbb{Z}\rangle\supset\langle 1/p^m+\mathbb{Z}\rangle\implies n\geq m $$
This is one of the last details in the example on planetmath.
This should be minor, but I can't figure it out. If $1/p^m+\mathbb{Z}\in\langle 1/p^n+\mathbb{Z}\rangle$, then $a/p^n-1/p^m\in\mathbb{Z}$ for some $a\in\mathbb{Z}$ nonzero. Towards a contradiction assume $n<m$. If $p\mid a$, then we can write this as $b/p^l-1/p^m\in\mathbb{Z}$ for $(b,p)=1$ and $l<m$, which is essentially the same case. This would imply $$ \frac{bp^{m-l}-1}{p^m}\in\mathbb{Z} $$ which doesn't look true. Is there an obvious contradiction to be had here?
$a/p^n-1/p^m\in\mathbb{Z}$ and $m>n$ means $$ \frac{ap^{m-n}-1}{p^m}\in\mathbb{Z} $$ Since $m>0$, this implies that $p$ divides $ap^{m-n}-1$ and, since $m-n>0$, we conclude that $p$ divides $-1$, a contradiction.