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?
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 $o(\operatorname{ker} \phi ) = p^n$.
Best Answer
Here is @halrankard's comment that answers my question. I post it here to remove this question from unanswered list. All credits are given to @halrankard.