# Homomorphism between abelian groups

An abelian group $G$ has property $N$ if for every sequence of subgroups of $G$ such that $G_1 \subseteq G_2 \subseteq ... \subseteq G$, then there exists positive integer $N$ such that $G_N=G_M$ for all $M \geq N$. Now let $\phi:G \rightarrow G_1$ be a group homomorphism of abelian groups. Prove that if $G$ has property $N$, then $\ker(\phi)$ has property $N$.

I know that given any sequence $K_1 \subseteq K_2 \subseteq ... \subseteq K \subseteq G$ there is an integer $N$ satisfying the above property since $G$ has property $N$. But how do I show that when we leave out $G$ this is still true? Any hints? Thanks.

## 2 Answers

Consider a sequence of subgroups of $\ker \phi$ say $P_1\subseteq P_2...\subseteq \ker \phi$. Then since $\ker \phi$ is a subgroup of $G$ so each $P_i$ is a subgroup of $G$ and thus we get a sequence of subgroups of $G$ and $G$ has property $N\implies$ we get some integer $K$ such that....

Please complete from here

You don't need to leave out $G$. You can simply use the fact that the sequence stabilizes to see that all but finitely many $K_i$ are the same.