# Finitely generated abelian groups clarification

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?

Thanks!

• "1. Each abelian group $G$..." and then $G$ is never mentioned again, but suddenly we have an $A$. That should be $A$. Commented Aug 2, 2022 at 16:59
• 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 $g^3$. Commented Aug 2, 2022 at 17:00

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, $$\Bbb Z^n/\rm{ker}\phi\cong G$$.