Index of cyclic subgroup $\langle h \rangle$ in a subgroup $H$ with finite index in a finitely generated group $G$

Question

Let $\phi: H\to \mathbb{Z}$ be a homomorphism with finite kernel.

Let $h \in H$ such that $\phi(h) \not=0$.

Can someone help me understanding why exactly the cyclic group $\langle h\rangle$ has finite index in $H$ ?

Answer 1

The image $\phi(H)$ is cyclic group generated by $\phi(t)$ for some $t\in H$. Let $K$ be the finite kernel. Every element $z\in H$ satisfies $\phi(z)=\phi(t)^k$ for some $k$. Hence $z=t^k s$, $s\in K$. Hence $H=\langle t\rangle K$. Your element $h$ is then equal to $t^cs'$ for some integer $c$ and $s'\in K$. Therefore $\langle h\rangle K=\langle t^c\rangle K$ so $\langle h\rangle$ has finite index in $H$.

Answer 2

By first isomorphism theorem $\frac{H}{\operatorname{Ker}\phi}\cong \mathbb Z$. Since $h\notin \operatorname{Ker}\phi$, we get $\bar h\in \frac{H}{\operatorname{Ker}\phi} $ generates a non-zero subgroup of $\frac{H}{\operatorname{Ker}\phi}$. Since any non-zero subgorup of $\mathbb Z$ has finite index we get $\langle \bar h \rangle$ has finite index in $\frac{H}{\operatorname{Ker}\phi}$. Thus $\exists \ x_1,x_2\dots,x_n$ such that given $x\in H$, we can write $x=sx_jh^k$ where $s\in \operatorname{Ker}\phi$ for some $j,k$. Since the set $\{sx_j | s\in \operatorname{Ker}\phi ,j=1,2,\dots,n \}$ is finite, we get $\langle h \rangle $ has finite index in $H$.