# How to find the Quotients of Z^n as a group?

I know that for finitely generated abelian groups (and thus also for my case), it is that they are isomorphic to $$\mathbb{Z}^r \times \mathbb{Z}_{p_1^{e_1}} \times \mathbb{Z}_{p_2^{e_2}}\times \cdots \times \mathbb{Z}_{p_k^{e_k}}$$ for some prime numbers $$p_1, \dots p_k$$.

I am now trying to proof that the quotients of $$\mathbb{Z}^n$$ are of precisely this form without using this theorem (the theorem would then follow from the fact that every abelian group is a $$\mathbb{Z}$$-Modul).

My approach so far was to find all subgroups $$G\subset \mathbb{Z}^n$$ and then proof what form $$\mathbb{Z}^n/G$$ needs to have. I believe to know that every subgroup of $$\mathbb{Z}^n$$ is isomorphic to some $$\mathbb{Z}^m$$. But as I'm trying to find quotients this does not help me much since $$\mathbb{Z}/2\mathbb{Z} \neq \mathbb{Z}/3\mathbb{Z}$$, even though both subgroups $$2\mathbb{Z}\cong \mathbb{Z} \cong 3\mathbb{Z}$$. This means I would need to generalize for all subgroups without relying on their isomorphy. I'm stuck here since I am not finding a way around this.

Any help on how to proof this is appreciated.

• Some of the proofs of the Fundamental Theorem of Abelian Groups take this approach anyway. The trick is to prove that there exists a free basis $b_1,\ldots,b_n$ of ${\mathbb Z}^n$ such that the given subgroup is freely generated by multiples $n_ib_i$ (with $n_i \in {\mathbb N}$) of the basis elements. for some of the $b_i$. This is equialent to showing that matrices over ${\mathbb Z}$ can be put into Smith Normal Form using unimodular row and column operations. Apr 19, 2020 at 10:39
• Thank you for your answer. Looking at the free basis seems like a good idea. What I am essentially doing is change of basis like in linear algebra. If I can show that the subgroup is freely generated by multiples $n_i b_i$, then I can make a change of base and thus get the wanted form - is this correct? Apr 19, 2020 at 13:51
• @DerekHolt do you have a reference for a (quotable) source from which I could read up the proof for the existence of such a basis? Apr 21, 2020 at 6:12