0
$\begingroup$

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.

$\endgroup$
3
  • 3
    $\begingroup$ 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. $\endgroup$
    – Derek Holt
    Apr 19, 2020 at 10:39
  • $\begingroup$ 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? $\endgroup$
    – Guenterino
    Apr 19, 2020 at 13:51
  • $\begingroup$ @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? $\endgroup$
    – Guenterino
    Apr 21, 2020 at 6:12

0

Reset to default

You must log in to answer this question.