prove: A finitely generated abelian group can not be isomorphic to a proper quotient group of itself.

Question

Let $G$ be a finitely generated abelian group. Then prove that it is not isomorphic to $\frac{G}{N}$, for every subgroup $N\neq\langle 1\rangle$.

Answer 1

Suppose $f:G\to G/N$ is an isomorphism and let $p:G\to G/N$ be the projection.

Let $N_1=N$, $N_2=p^{-1}(f(N_1))$, $N_3=p^{-1}(f(N_2))$ and so on. This defines a increasing sequence $$N_1\subseteq N_2\subseteq N_3\subseteq\cdots$$ of subgroups of $G$. You can check that if $N$ is non-trivial, then the sequence is strictly increasing.

This is absurd, since $G$ is a noetherian $\mathbb Z$-module.

Later. Alternatively, let $N$ be a subgroup of $G$. We have a short exact sequence $$0\to N\to G\to G/N\to 0$$ so for the ranks of the groups we have $\newcommand\rk{\operatorname{rk}}\rk G=\rk N+\rk G/N$. If $G\cong G/N$ we must have $\rk N=0$, so $N$ is torsion. But then $N$ is contained in the torsion subgroup $t(G)$ of $G$, and clearly the torsion subgroup $t(G/N)$ has less elements than $t(G)$ unless $N=0$. It follows that $G\not\cong G/N$ unless $N=0$.