Order of cyclic subgroup

Question

This should be easy, but I keep getting stuck on it.

Suppose $D$ is a cyclic group of order $m$ (written additively) and then let $D[n] = \{x \in D : nx = 0\}$.

I'm trying to show $D[n]$ has order $d$, where $d = hcf(m,n)$. I can see that $D[n]$ should be generated by $\frac{m}{d}$ and this has the required order, but how to show that this is the generator? I'm sure I need to use Bezouts lemma somewhere but can't quite see it.

Thanks

Answer 1

You don't need Bézout's lemma. Let $g$ be a generator of $D$. Then

$$k\cdot g \in D[n] \iff (nk)\cdot g = 0 \iff m\mid nk.$$

Now just write $n = \nu d$ and $m = \mu d$ and you see that

$$m\mid nk \iff \mu \mid \nu k \iff \mu \mid k,$$

since $\gcd(\mu,\nu) = 1$.