Let $M$ be a module over a commutative and unital ring $R$. A finite presentation for $M$ is an exact sequence $$F\xrightarrow{\alpha} E\xrightarrow{\Phi} M\rightarrow 0,$$ where $E$ and $F$ are free $R$-modules with finite bases.

A presentation matrix for $M$ is defined as a presentation matrix for $\alpha$ with respect to bases of $F$ and $E$.

In the lecture we had an example without proof and I am not sure why this holds:

Consider a finite abelian group $G$, i.e. $G$ is a $\mathbb{Z}$-module. It has a presentation matrix $A$ with $\det A=|G|$.

What is the idea for proving this, i.e. what can I choose for $F,E,\alpha,\Phi$?

Solution. We have $G\cong\bigoplus_{i=1}^n\mathbb{Z}/p_i^{k_i}\mathbb{Z}$ for primes $p_i$. Besides we have an homomorphism $$\Phi:\bigoplus_{i=1}^n\mathbb{Z}\rightarrow\bigoplus_{i=1}^n\mathbb{Z}/p_i^{k_i}\mathbb{Z},\quad (m_1,...,m_n)\mapsto ([m_1],...,[m_n]).$$

Then we get $\ker \Phi=\bigoplus_{i=1}^np_i^{k_i}\mathbb{Z}$. If we define $$\alpha:\bigoplus_{i=1}^n\mathbb{Z}\rightarrow\bigoplus_{i=1}^n\mathbb{Z},\quad (m_1,...,m_n)\mapsto (p_1^{k_1}m_1,...,p_n^{k_n}m_n)$$ then we get $\text{im } \alpha=\ker\Phi$. Clearly, $\Phi$ is surjective, hence we get an exact sequence. The determinant respective the standard basis is equal to $\prod_i p_i^{k_i}=|G|$.

  • $\begingroup$ You can simplify the problem significantly by using the structure theorem of finitely generated Abelian groups, if you have that available. $\endgroup$
    – VJP
    May 9, 2017 at 21:08
  • $\begingroup$ I do and I have already thought about this! But without any results. Can you give more details? $\endgroup$
    – user444847
    May 9, 2017 at 21:11
  • $\begingroup$ First, decompose $M$ to a sum of cyclic groups. Then, for $E$ take a free Abelian group or order equal to the number of cyclic summands of $M$. $\Phi$ will hopefully be obvious now and then finish off by figuring out the kernel of $\Phi$ to get suitable choices for $F$ and $\alpha$. $\endgroup$
    – VJP
    May 9, 2017 at 21:18
  • $\begingroup$ I have added a solution to my question, may you please check? One question: The structure theorem stated that the group is iso to $\mathbb{Z}^r\oplus ...$, Why can we use $r=0$? Because otherwise $|G|$ would be infinite? $\endgroup$
    – user444847
    May 9, 2017 at 21:55
  • $\begingroup$ Yes, exactly. The group is finite, if and only if the free part is of rank $0$, i.e. $r = 0$. $\endgroup$
    – VJP
    May 9, 2017 at 22:00

1 Answer 1


The building blocks of abelian groups are the cyclic subgroups of order power of a prime. So we start with that. Observe the exact sequence $p^nZ\rightarrow Z\rightarrow Z/p^nZ\rightarrow 0$. Once u get this part just use direct sums repeatedly till you cover the whole group. Also note that $p^nZ$ is a free module. As for the determinant in the elementary case we have it is equal to $p^n$


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