$A_p:=\{x\in A: p^kx=0~ \text{for some}~k\}$ is a sylow p-subgroup without using the structural theorem of abelian groups Let $A$ be an abelian group and for each prime number $n$, $A_p:=\{x\in A: p^kx=0~ \text{for some}~k\}$.
I wonder how to prove that $A_p$ is a sylow subgroup without using the structural theorem of abelian groups.
I can show $A_p$ has the order a power of $p$ by Cauchy's theorem, but I don't know how to show the maximality

It depends on what you mean "prove that $A_p$ is a $p$-Sylow" (actually, the $p$-Sylow).
If you already know Sylow's theorems (here what's important is the existence of a Sylow subgroup), then it's really easy : assume $S$ is a $p$-subgroup of $A$. Then $|S|$ is a prime power so by Lagrange's theorem, its elements have order dividing $p^k$ for some $k$, hence $S\subset A_p$. By taking $S$ a (the) $p$-Sylow of $A$, we see that $A_p$ is indeed a $p$-Sylow.
If you don't know Sylow's theorems, then it's a bit more complicated (but not that much). Since $A$ is abelian, $A_p$ is normal in $A$ and so you may consider $A/A_p$. If the exponent of $p$ in $|A_p|$ weren't maximal, then $p\mid |A/A_p|$, hence by Cauchy's theorem there is $x$ of order $p$ in $A/A_p$. Let $\pi$ denote the canonical morphism $A\to A/A_p$, and consider $L=\pi^{-1}(\langle x \rangle)\subset A$. Then $0\to A_p \to L \to \langle x\rangle \to 0$ is exact, which implies that $|L|$ is a power of $p$ (to be specific, $|A_p|\times p$), and so its elements have order a power of $p$: therefore they are in $A_p$: $L\subset A_p$, which is a contradiction.  Therefore, the exponent of $p$ in $|A_p|$ is maximal and $A_p$ is indeed a (the) $p$-Sylow of $A$.
