Let $m, n \geq 1$, prove that $[\mathbb{Z}/m \mathbb{Z} : n\left( \mathbb{Z}/m \mathbb{Z} \right)] = \text{gcd}(m, n)$
When writing out the index as the cardinality of the quotient space, this looks like the third isomorphism theorem, but the $n$ doesn't allow this to work.
Also we could try to prove this with the first isomorphism theorem, constructing a homomorphism $\phi : \mathbb{Z} \rightarrow n\left( \mathbb{Z}/m \mathbb{Z} \right)$ with kernel: $\frac{m}{\text{gcd}(m, n)} \mathbb{Z}$. Then $\frac{m}{\text{gcd}(m, n)} = \lvert \mathbb{Z} / \frac{m}{\text{gcd}(m, n)} \mathbb{Z} \rvert = \lvert n\left( \mathbb{Z}/m \mathbb{Z} \right) \rvert$, and thus by Lagrange: $[\mathbb{Z}/m \mathbb{Z} : n\left( \mathbb{Z}/m \mathbb{Z} \right)] = \frac{m}{\frac{m}{\text{gcd}(m, n)}} = \text{gcd}(m, n)$.
But none of these approaches seem to work. Any hint on where to start?