Are all the finite abelian groups isomorphic to the pruduct of some $\mathbb{Z}_n$s and some $\mathbb{F}_q$s?

Question

Are all the finite abelian groups isomorphic to the pruduct of some $(\mathbb{Z}_n,+)$s and some $(\mathbb{F}_q,+)$s? where $\mathbb{Z}_n$ denotes the residue class ring of order $n$, $\mathbb{F}_q$ denotes the finite field of order $q$.

This problem seems similar with the problem( Every finite group is isomorphic to the Galois group of some polynomial ) and has connections with the Inverse Galois Problem( https://en.wikipedia.org/wiki/Inverse_Galois_problem ).

In fact, I want to go through all the finite abelian groups while the $\mathbb{Z}_n$ and $\mathbb{F}_q$ have been gone through.

Thanks for any advise.

Answer 1

This is a special case of the fundamental theorem of finitely generated abelian groups.

I am not sure why you are talking about ring and field structure in a question regarding classification of groups.

Long story short, any abelian group can be written as the product of cyclic groups of order $p^k$ where $p$ is a prime. (You can think of cyclic groups in terms of residues if you like.)

Answer 2

The fundamental theorem of finitely generated abelian groups states that if $G$ is a finitely generated abelian group, then $$ G\cong\Bbb Z^r\oplus\left(\bigoplus_{i = 1}^m\Bbb Z/(n_i)\right), $$ where $r\geq 0$ and the $n_i$ are powers of primes (not necessarily distinct). Any finite abelian group is finitely generated, and since $r > 0$ implies $G$ is infinite, you actually have that any finite abelian group $G$ is of the form $$ G\cong\bigoplus_{i = 1}^m\Bbb Z/(n_i). $$

So, the answer to your question is "yes." Moreover, the groups you use in your description are redundant: if $q = p^n$ is a prime power, then $(\Bbb F_q,+)\cong\left(\Bbb Z/(p)\right)^n$. So $(\Bbb F_q,+)$ is already a product of $\Bbb Z/(p)$'s.