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

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).

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?)

• I like how Chris answered after with a similar example...
– user567763
Commented Feb 29, 2020 at 18:43

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.