(Cauchy's theorem in the abelian case): If $G$ is a finite abelian group such that $p \ | \ |G|$, where $p$ is a prime, then $\exists g \in G$ such that $o(g) = p$.
Basically the part of the proof that I'm stuck on is the following. After choosing an $h \in G$ for which $h \neq 1_G$ and assuming $o(h) = l$, we form the subgroup $ H = \langle h \rangle$. We can then form the quotient group $G / H$ which has order $$|G/H| = \frac{|G|}{|H|} = \frac{|G|}{l}$$ by Lagrange's theorem. The next part of the proof says that $p$ divides $\frac{|G|}{l}$ because $p$ divided $|G|$, and I can't immediately see why this is the case.
What I think is that this part of the proof must be using some argument from elementary number theory, and I think that the theorem that's being referenced must be (similar to) the following conjecture of mine:
Conjecure: If $p$ is prime and $p \ | \ x$ and $\frac{x}{l} = m$ where $m, l \in \mathbb{Z}$ then $p \ | \ m$.
In trying to prove the above this is as far as I got.
Attempted Proof: Since $p | x$ there exists an $n \in \mathbb{Z}$ such that $pn =x$, Then $$\frac{x}{l } = m \implies \frac{pn}{l} = m \implies pn = ml$$ and since $p$ divides the left hand side of the above equation it follows that the right hand side of the equation must also be divisible by $p$. So $p|ml$ and since $p$ is prime $p|l$ or $p|m$. If $p|m$ we are done, if $p|l$ then...
And that's as far as I got. Is the above conjecture actually a theorem? If so is it the one being used in Cauchy's theorem and if so how could I finish proving the above conjecture?
It is not true in general. Obviously $3|6$ but $3$ doesn't divide $\frac{6}{3}=2$. What is that $l$ that appears in the proof? It can't be just any number.