I don't know if it is appropriate to ask these kind of stuff in here.
I know the Chinese remainder theorem that if gcd(m,n)=1:
$\mathbb{Z}/mn\mathbb{Z}$ isomorphic to $\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$
and for finite type abelian groups:
$G$ is isomorphic to $ \mathbb{Z}^r\times \mathbb{Z}/a_1\mathbb{Z} \times ... \mathbb{Z}/a_{s} \mathbb{Z} $ with $r,s,a_1...a_s$ uniques anda $a_{i+1} | a_{i}$
The problem is: Is there a theorem such that for any finite abelian group $G$ is isomorphic to $\mathbb{Z}/p_{n_1} ^{m_1} \times ... \mathbb{Z}/p_{n_i} ^{m_i}$ where $p_{n_i}$ are prime numbers?
With something like $p_{n_1} ... p_{n_i} = |G|$. (That's the main property I want to get)
If yes, could you send me a paper or link with something to read about it? Maybe give me the conditions and the theorem?
Would you tell me anything near about it that you know?
I couldn't find this result where I searched for it, I'm pretty sure I've seen people doing it before. (I'm a beginner in group theory take it easy please.)
The decomposition $\mathbb{Z}/n\mathbb{Z}\cong\mathbb{Z}/p_1^{n_1}\mathbb{Z}\times\cdots\times\mathbb{Z}/p_k^{n_k}\mathbb{Z}$ when $n=p_1^{n_1}\cdots p_k^{n_k}$ follows immediately from the Chinese Remainder Theorem and induction on the number of prime factors in the factorization of $n$.
When $n=p_1^{n_1}p_2^{n_2}$, then $\mathbb{Z}/n\mathbb{Z}\cong\mathbb{Z}/p_1^{n_1}\mathbb{Z}\times\mathbb{Z}/p_2^{n_2}\mathbb{Z}$, by the Chinese Remainder Theorem, since $\gcd(p_1,p_2)=1$. When $n=p_1^{n_1}p_2^{n_2}\cdots p_k^{n_k}$, $$ \mathbb{Z}/n\mathbb{Z}\cong \mathbb{Z}/p_1^{n_1}\mathbb{Z}\times \mathbb{Z}/m\mathbb{Z} $$ where $m=p_2^{n_2}\cdots p_k^{n_k}$, since $\gcd(p_1,m)=1$. Now, $m$ has fewer prime factors than $n$, so $$ \mathbb{Z}/m\mathbb{Z}\cong\mathbb{Z}/p_2^{n_2}\mathbb{Z}\times\cdots\times\mathbb{Z}/p_k^{n_k}\mathbb{Z} $$ by induction.
A proof of this fact should appear in almost any introductory group theory book. Look for the fundamental theorem of finite abelian groups.