Since $f:G\to \Z$ is surjective, there exists an element $a\in G$ such that
\[f(a)=1.\]
Let $H=\langle a \rangle$ be the subgroup of $G$ generated by the element $a$.

We show that $G\cong \ker(f)\times H$.
To prove this isomorphism, it suffices to prove the following three conditions.

The subgroups $\ker(f)$ and $H$ are normal in $G$.

The intersection is trivial: $\ker(f) \cap H=\{e\}$, where $e$ is the identity element of $G$.

Every element of $G$ is a product of elements of $\ker(f)$ and $H$. That is, $G=\ker(f)H$.

The first condition follows immediately since the group $G$ is abelian, hence all the subgroups of $G$ are normal.

To check condition 2, let $x\in \ker(f) \cap H$.
Then $x=a^n$ for some $n\in \Z$ and we have
\begin{align*}
0&=f(x) && \text{since $x \in \ker(f)$}\\
&=f(a^n)\\
&=nf(a) && \text{since $f$ is a homomorphism.}\\
&=n &&\text{since $f(a)=1$}.
\end{align*}
Thus, as a result we have $x=a^0=e$, and hence $\ker(f) \cap H=\{e\}$.
So condition 2 is met.

To prove condition 3, let $b$ be an arbitrary element in $G$.
Let $n=f(b) \in \Z$. Then we have
\[f(b)=n=f(a^n),\]
and thus we have
\[f(ba^{-n})=0.\]
It follows that $ba^{-n}\in \ker(f)$.
So there exists $z\in \ker(f)$ such that $ba^{-n}=z$.
Therefore we have
\begin{align*}
b=za^n\in \ker(f)H.
\end{align*}
This implies that $G=\ker(f)H$.

We have proved all the conditions, hence we obtain
\[G\cong \ker(f)\times H.\]
Since $H$ is a cyclic group of infinite order, it is isomorphic to $\Z$.
(If $H$ has a finite order, then there exists a positive integer $n$ such that $a^n=e$. Then we have
\begin{align*}
0=f(e)=f(a^n)=nf(a)=n,
\end{align*}
and this contradicts the positivity of $n$.)

Combining these isomorphisms, we have
\[G\cong \ker(f)\times \Z,\]
as required.

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