First of all, I am aware of the First Isomorphism Theorem but I am not sure how to use it/if it is useful here
$G$ is an abelian group and $f:G\rightarrow\mathbb{Z}$ is a surjective group homomorphism. I need to show $G\cong \ker(f) \times \mathbb{Z}$. I need to do so by the following: choose some $s\in G$ with $f(s)=1\in\mathbb{Z}$, and use $s$ to find a homomorphism from $G$ to $\ker(f)\times\mathbb{Z}$ and show this is bijective.
I am not really sure how $s$ would help me find this 'second' homomorphism so I haven't really made any progress. My only attempt has been to do with that homomorphisms preserve the identity, but since $1$ is not the identity of $\mathbb{Z}$ I don't think is the right way to think about the problem. I am also not sure how $G$ being abelian helps.
Let for $m\in\mathbb{Z}$, $A_m:=\{g\in G:f(g)=m\}$. Since $f$ is a subjective homomorphism, $A_m$ is non-empty for every $m\in\mathbb{Z}$.
Let $s\in G$ be such that $f(s)=1$. Such $s$ exists due to subjectivity of $f$. Then by properties of homomorphis, $f(ms)=m$ $\forall$ $m\in\mathbb{Z}$. Define the following map, $\varphi: G\longrightarrow \mathrm{ker}(f)\times \mathbb{Z}$ by the following rule, $$\varphi(g):= (g-f(g)s,f(g))\quad (\forall g\in G)$$ It's easy to verify that $\varphi$ is a group homomorphism. Then we show that $\varphi$ is also a bijection in order to establish isomorphism.
$$\varphi(g_1)=\varphi(g_2)\implies g_1-f(g_1)s = g_2-f(g_2)s\implies g_1=g_2$$ Hence $\varphi$ is injective.
Let $(k,m)$ be in $\mathrm{ker}(f)\times\mathbb{Z}$. We have to choose a $g$ in $A_m$ such that $k=g-f(g)s=g-ms$. Obviously we see that $g=k+ms$ works. Then $\varphi$ is subjective also. Hence $\varphi$ is bijective or $G\cong \mathrm{ker}(f)\times\mathbb{Z}$.