1

I am to show that for all $m,n \in \mathbb{Z}$, the following isomorphism:

$\dfrac{(m,n)\mathbb{Z}}{m\mathbb{Z}} \cong \dfrac{n\mathbb{Z}}{[m,n]\mathbb{Z}}$

where $(m,n) = gcd(m,n)$ and $[m,n] = lcm(m,n)$.

I really have no idea where to start here. I tried using one of the isomorphism theorems as a starting point, but it hasn't yielded any obvious results.

This is a HW assignment so I only need a hint or some initial guidance of how to start this proof. Thank you.

Calculus08
  • 783
  • Can you prove, for instance, that $(m,n)\cdot [m,n]=mn$? – Milo Brandt Feb 09 '15 at 02:42
  • This is very easy, but i'm not sure how this helps. :( – Calculus08 Feb 09 '15 at 02:43
  • Well, it's suggestive that if you let $k=\frac{n}{(m,n)}$, then multiplying numerator and denominator of the lefthand quotient by $k$ gives the righthand quotient. This, at least, establishes that $x\mapsto x\cdot k$ is an isomorphism of the additive groups. (If you're looking at multiplicative structure as well, there's probably some more thinking, but that's a good starting place) – Milo Brandt Feb 09 '15 at 02:49
  • I figured out a very easy way to show this isomorphism if you see the answer below. Thank you for the help! :) – Calculus08 Feb 09 '15 at 02:52

1 Answers1

5

I have found the answer. Using the fact that $n\mathbb{Z} + m\mathbb{Z} = (m,n)\mathbb{Z}$ and $n\mathbb{Z} \cap m\mathbb{Z} = [n,m]\mathbb{Z}$, this follows from the 2nd Iso. Theorem.

Pedro
  • 122,002
Calculus08
  • 783