Let $G$ and $H$ be abelian groups of order $n$. I want to prove that $G$ is isomorphic to $H$ if and only if for every prime $p\mid n$, Sylow $p$-subgroup of $G$ is isomorphic to Sylow $p$-subgroup of $H$.
One direction is obvious.
How do I show that the other direction, i.e. if Sylow $p$-groups are isomorphic then Groups are isomorphic. The goal of this question is to make easier to verify list of finite abelian groups of order $n$ (up to isomorphism).
We can certainly prove a baby version of the structure theorem for finite abelian groups:
Theorem: A finite abelian group $G$ is the direct product of its Sylow $p$-subgroups.
Proof: Let $\#G=n=p_1^{n_1}\dots p_r^{n_r}$. By Euclid/Bezout/..., there exists integers $m_1,\dots,m_r$ such that $n/p_i^{n_i}$ divides $m_i$ and $m_i\equiv 1\pmod{p_i^{n_i}}$. Then $m_1+m_2+\dots+m_r\equiv 1\pmod{p_i^{n_i}}$ for all $i$ and hence $\equiv 1\pmod{n}$. By Cauchy, $g^n=1$ for all $g\in G$ so $g^{m_i}$ has $p$-power order, i.e., lies in the Sylow-$p_i$ of $G$, and $$ g=g^{m_1+m_2+\dots+m_r}=g^{m_1}g^{m_2}\dots g^{m_r} $$ gives the desired result. QED.
