# $| \ker{\phi} |<\infty \iff [G: \phi(G)]<\infty$ for group homomorphism $\phi$. [closed]

Let $$G$$ be a countably infinite abelian group.

If $$\phi:G \to G$$ is a group homomorphism is it true that either both $$| \ker{\phi} |$$ and $$[G: \phi(G)]$$ are finite or both infinite?

If so, is there some additional relation between $$| \ker{\phi} |$$ and $$[G: \phi(G)]$$?

• Yes there is an additional relation - the first isomorphim theorem en.wikipedia.org/wiki/Fundamental_theorem_on_homomorphisms . Commented Jun 28 at 9:33
• I am well aware of that. It only gives that $G / \ker{\phi} \cong \phi(G)$, which is NOT necessarily a relation between the cardinality of $\ker{\phi}$ and the index of $\phi(G)$ in $G$. Right? Does the first isomorphism theorem even imply a positive answer to my original question?
– User
Commented Jun 28 at 9:39
• @Desperado That's not true. Consider $\mathbb{Z}$ and the map $n \mapsto 2n$. I think it only holds for finite groups, but I am not asking about these.
– User
Commented Jun 28 at 9:51
• My mistake, I misread the question. Good question, and indeed the answer is no as the answer below explains. Commented Jun 29 at 11:51

Consider $$G=\Bbb{Q}/\Bbb{Z}$$. It is known that $$\Bbb{Q}/\Bbb{Z} = \bigoplus_p \Bbb{Z}_{p^\infty}$$, where $$\Bbb{Z}_{p^\infty}$$ is the abelian group generated by elements $$x_1,\dots,x_n,\dots$$ such that $$px_1=0$$ and $$px_{n+1}=x_n$$. Consider the map $$\Bbb{Z}_{p^\infty}\xrightarrow{\cdot p}\Bbb{Z}_{p^\infty}$$. On generators, this maps $$x_{n+1}\mapsto x_n$$, thus is clearly surjective, and it has kernel generated by $$x_1$$, i.e. $$\Bbb{Z}/p\Bbb{Z}$$.
Now let $$\phi$$ be the direct sum of these maps. It has kernel $$\bigoplus_p \Bbb{Z}/p\Bbb{Z}$$, which is clearly infinite, and it is surjective, thus $$[G:\phi(G)] = 1$$, giving a counter-example.
As a sidenote: for finitely generated abelian groups the answer is yes. The proof basically boils down to the fact that a finitely generated group is infinite iff it has positive rank, and rank is dimension when tensoring with $$\Bbb{Q}$$, which preserves exactness, and then you can use rank-nullity for vector spaces.