A simple proof of Sylow theorem for abelian groups

In my attempt to prove

Let $G$ be an abelian group and $H,K$ its subgroups such that $o(H) = m$ and $o(K) = n$. Then $G$ has a subgroup of order $operatorname{lcm}(m,n)$.

, the crucial argument is

Sylow theorem for abelian groups: Let $G$ be an abelian group, $p$ a prime number, and $n$ a natural number. If $o(G)$ is divisible by $p^n$, then $G$ has a subgroup of order $p^n$.

IMHO, my proof is quite different from those I’ve seen around. Could you please verify if my attempt is fine or contains logical mistakes?

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My attempt:

First, we need an auxiliary result

Cauchy’s theorem for abelian group (CTFAG): Let $G$ be an abelian group. If the order of $G$ is divisible by a prime number $p$, then $G$ has an element of order $p$.

Then we define recursively a finite sequence of group epimorphisms $(phi_k)_{1 le k le n}$ as follows. Let $G_0 := G$. By CTFAG, there exists $x_k in G_k$ such that $o(x_k) = p$. Let $G_{k+1} =G_k / langle x_k rangle$. Because $G_k$ is abelian, the map $phi_{k+1}: G_k to G_{k+1}, quad y mapsto y langle x_k rangle$ is a group epimorphism. Moreover, $G_{k+1}$ is also abelian and $o(G_{k+1}) = o(G_{k})/p$.

It follows that $o(G_n) = o(G) / p^n$ and that $phi := phi_n circ cdots circ phi_1$ is a group epimorphism from $G$ to $G_n$. By first isomorphism theorem, we have $G / operatorname{ker} phi cong G_n$ and thus $o(G / operatorname{ker} phi ) = o(G_n)$. Consequently, $o(G) = o(G_n) o(operatorname{ker} phi) = o(G) o(operatorname{ker} phi ) / p^n$. Finally, we have $o(operatorname{ker} phi ) = p^n$.