I read about finitely generated abelian groups and would be grateful for a clarification.

My lecturer noted these statments:

  1. Each abelian group $G$ such that $a_1,a_2,\cdots,a_n\in A$ are been chosen, exists a unique homomorphism $\phi:\mathbb{Z}^n\to A$ such that $\phi(e_i)=a_i$.

  2. Each finitely generated abelian groups isomerphic to quotient of $\mathbb{Z}^n$ for some $n\in \mathbb{N}$.

The proof of the above statment based on the the first isomorphism theorem.

I try to build my own example:

Let $G=\langle g\rangle$ be a cyclic (and abelian) group, such that $|G|=4$.

The generators of $G$ are $\{g,g^3\}$.

Denote 2 homomorphism ($n=1$):

$\phi_1:\mathbb{Z}^1 \to G$ such that $\phi_1(1)=g$.

$\phi_2:\mathbb{Z}^1 \to G$ such that $\phi_2(1)=g^3$

There is no a contradiction of the uniqueness part of statement 1 since I choose other generator of $G$?

Using statement 2, $\mathbb{Z}^1/4\mathbb{Z}^1\cong G$ since $\ker\phi=4\mathbb{Z}^1$?

Is my example correct?


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    $\begingroup$ "1. Each abelian group $G$..." and then $G$ is never mentioned again, but suddenly we have an $A$. That should be $A$. $\endgroup$ Commented Aug 2, 2022 at 16:59
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    $\begingroup$ There is no contradiction to uniqueness because the uniqueness is dependent on your choice of $a_i$. $\phi_1$ is the unique map that sends $1$ to $g$. $\phi_2$ is the unique map that sends $a$ to $g^3$. $\endgroup$ Commented Aug 2, 2022 at 17:00

2 Answers 2


Let's rewrite the two statements:

  1. for each abelian group $A$ and every choice of $a_1,a_2,\dots,a_n\in A$ there exists a unique homomorphism $\phi\colon\mathbb{Z}^n\to A$ such that $\phi(e_i)=a_i$

  2. each finitely generated abelian group $A$ is isomorphic to a quotient of $\mathbb{Z}^n$, for some $n\in\mathbb{N}$.

Is there a contradiction with $G=\langle g\rangle$ being cyclic of order $4$ and the existence of two distinct homomorphisms $\phi_1\colon\mathbb{Z}^1\to G$ and $\phi_2\colon\mathbb{Z}^1\to G$, where $\phi_1(1)=g$ and $\phi_2(1)=g^3$?

No, because the two homomorphisms map $1$ to different elements of $G$.

In both cases, the homomorphism is surjective and by the homomorphism theorem $G\cong\mathbb{Z}/\ker\phi_1$, so we must have $\ker\phi_1=4\mathbb{Z}$ by the structure of subgroups of $\mathbb{Z}$. Similarly, $\ker\phi_2=4\mathbb{Z}$, but this is again not a contradiction.


Your lecturer is referring to the universal property of the free abelian group $\Bbb Z^n$.

For $2)$, $\Bbb Z^n$ is the free abelian group with basis $S=\{a_1,a_2,\dots, a_n\}$, where $G=\langle a_1,a_2,\dots, a_n\rangle $. The homomorphism $\phi:\Bbb Z^n\to G$ (from $1)$) is surjective. Hence by the first isomorphism theorem, $\Bbb Z^n/\rm{ker}\phi\cong G$.


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