Finitely generated abelian groups clarification I read about finitely generated abelian groups and would be grateful for a clarification.
My lecturer noted these statments:

*

*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 $\phi(e_i)=a_i$.


*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?
Thanks!
 A: 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,  $\Bbb Z^n/\rm{ker}\phi\cong G$.
A: Let's rewrite the two statements:

*

*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 $\phi(e_i)=a_i$


*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.
