# Isomorphism between this subgroup of complex numbers and all finitely generated abelian groups ?

Every cyclic (abelian) group of infinite order is isomorphic to $G=(\mathbb{Z},+)$. Is there a corresponding set of groups $S_G=\{G\}$ such that every finitely generated abelian group of infinite order is isomorphic to at least one group in $S_G$?

More precisely,

1. Let $A$ be the set of all finitely generated abelian groups of infinite order, then is there an $S_G$ so that $\exists$ a mapping $f: A \to S_G$ such that $f(A_1)=G\Leftrightarrow A_1 \approx G$?
2. Can $S_G,f$ be chosen such that $f$ is an isomorphism?

Here's an idea I had for a possible $S_G$:

$I_N=\{1,...,N\}$, integer $m$ and $p_i$ being the $i$th prime, I posit that each group, with binary operation of multiplication, of the form $$G_{m,N,M}=\{ re^{i \theta}| r=\prod_{i \in I_N}p_i^{a_i} , \theta \in \{\frac{1}{m}\prod_{i \in I_M}p_i^{b_i} \mod 1\}$$ Given that $m,N,M$ are free to vary, I think $S_G$ is sufficiently big for an $f$ to exist. Is this provable, and can a very similar $G_{m,N,M}$ be contructed with fewer free parameters (i.e. here, $m,N,M$)?

• Can't you just use the classification theorem? The question you posted solves your problem perfectly. – Ian Coley Mar 17 '14 at 18:27
• @IanColey wasn't aware of this, quite possibly. – Meow Mar 17 '14 at 18:28

## 2 Answers

Take $$S_G=\left\{\mathbb Z^r\times \prod_{i=1}^n \mathbb Z/p_i^{a_i}\mathbb Z:n,r,a_i\in\mathbb Z^+,p_i\text{ prime}\right\}.$$ This is using the classification theorem for finitely generated abelian groups.

There is a classification theorem for finitely generated abelian groups. See http://en.wikipedia.org/wiki/Finitely-generated_abelian_group

The statement is that if $G$ is finitely generated and abelian, then $G \cong \mathbb{Z}^n \oplus \mathbb{Z}_{m_1} \oplus \cdots \oplus \mathbb{Z}_{m_\ell}$ where $n \in \mathbb{N}$ and each $m_i$ is a prime power.