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Yesterday I asked a question which turned out to be pretty bad posed due to the fact that I am not really experienced with presentations of groups. However, I was told the following result:

If an abelian group $G$ is finitely generated, then $G$ is finitely presented.

I thought that I know how to prove this, but I keep struggling with it and I am not able to do it. I considered $G$ to be generated by the set $X=\{x_1, x_2, ... , x_n\}$. Then I took the free abelian group on $X$ and I considered the map that sends $X$'s generators to those of the free abelian group. This gave me a homomorphism between $G$ and the free abelian group on $X$. However, I don't know what to do next to get that $G$ has a finite presentation. I also don't know if my approach is fruitful. Sorry if this is well-known, but I could find no references.

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Let $F$ be the free abelian group generated by symbols $x_1,\ldots,x_n$, and let $\phi:F \to G$ be the homomorphism that takes each $x_i$ to the corresponding generator of $G$. Then the kernel $K$ of $\phi$ is generated by a finite subset $R$ of $F$.

Then a presentation of $G$ is $$\langle x_1,\ldots,x_n \mid R \cup C \rangle,$$ where $C$ consists of relations $x_ix_j=x_jx_i$ for all $1 \le i,j \le n$. The relations in $C$ ensure that $G$ is abelian).

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  • $\begingroup$ Thank you very much! I couldn't find this anywhere and it really annoyed me, because it was the missing piece I needed to understand how to prove the classification of finitely generated abelian groups by using the Smith Normal Form. $\endgroup$ – TheZone Apr 23 at 22:37

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