Is this an equivalent way of stating the classification theorem of finite abelian groups?

Question

Let $(G,\cdot)$ be a finite abelian group with $n$ elements. I know the following version of the classification theorem of finite abelian groups : there $\exists$ the integers $d_1, d_2, ... d_k\ge 2$ such that $d_1 | d_2 | ... |d_k$ and $G \cong \mathbb{Z_{d_1}} \times \mathbb{Z_{d_2}}\times...\times \mathbb{Z_{d_k}}$ .
I was wondering if this also holds : if $p_1^{m_1}\cdot p_2^{m_2}\cdot ... \cdot p_l^{m_l}$ is the prime factor decomposition of $n$, then $G \cong \mathbb{Z_{p_1^{m_1}}}\times \mathbb{Z_{p_2^{m_2}}}\times ... \times \mathbb{Z_{p_l^{m_l}}}$. I came up with this idea while reading an article (https://www.msri.org/people/members/chillar/files/autabeliangrps.pdf this one, Theorem 1.1. more precisely), but I could neiher find this result anywhere nor prove it(this is why I think it may be false, so any counterexample is welcome).

Answer 1

That is not true in general. Take for example $G=\mathbb{Z}_2\times\mathbb{Z}_2$. Then $n=4$ but $G$ is not isomorphic to $\mathbb{Z}_4$ (can you see why?)

Answer 2

Something like what you're after is indeed true. But as written it's incorrect. The prime decomposition won't work, all by itself, because of the fact that, for instance, $\Bbb Z_{p^2}\not\cong\Bbb Z_p×\Bbb Z_p$.

To rectify it, take, instead of the prime decomposition, an appropriate decomposition with some of the $p_i\,\bf{possibly\, being\, repeated}$.

Note that, if you just use the prime factor decomposition, you just get the cyclic group of order $n$.

The correct statement is that any finite abelian group can be written as a product of cyclic groups of prime power order. So there are indeed two ways of stating the theorem.