Let $G$ be a subgroup of a finite product of finite cyclic groups. Is it easy to prove that $G$ is itself a finite product of cyclic groups without appealing to the structure theorem of finite abelian groups?
Some context, to avoid this topic to be closed. I am looking for a proof of the structure theorem for a finite abelian groups $G$. I have the following fun argument that shows that shows that $G$ is contained in a product of finite cyclic groups: if $G$ has order $n$, it embeds in the permutations group on $n$ elements, itself embedding in the group of invertible matrices over $\mathbb{C}$. This way, one may interpret elements of $G$ as $n$ commuting endomorphisms of a dimension $n$ vector space, all annihilated by the simple roots polynomial $X^n-1$. In particular, they are codiagonalizable, with eigenvalues being roots of the unity!