$G/H \cong \mathbb{Z}$ implies existence of $K \leq G$ s.t. $K \cap H = 1$ and $HK = G$

Question

Let $H \lhd G$ and $G/H \cong \mathbb{Z}$. The claim is that there is a subgroup $K \leq G$ such that $K \cap H = 1$ and $HK = G$.

Is the following correct?

Take $a$ to be any element in a generator of $G/H$. Then $K = \left< a \right>$ will work: Clearly $K \cap H = 1$ and if $g \in G$, then $Hg = Ha^m$ for some $m$ so $g \in HK$.

It's not clear to me how $G/H \cong \mathbb{Z}$ is used. What stops this working for a general normal $H$? If $H$ isn't infinite cyclic, is the group $K$ then messed up? Would it even work for finite cyclic?

Answer 1

You use $K\cong\mathbb{Z}$ when you write "Clearly, $K\cap H=1$".

For example, if $G$ is torsion-free and $G/H$ is finite, then every non-trivial subgroup $K$ has $K\cap H\neq1$. Therefore, it doesn't work for finite cyclic groups.

The more general result here is that every map to a free group $G\twoheadrightarrow F$ splits. The proof here is harder.

Answer 2

Every short exact sequence of the form

$$1\to H\to G\to\mathbb{Z}\to 1$$

must be split, because $\mathbb{Z}$ is not only free-abelian but also free among all groups (see the comments). Therefore if $f:G\to\mathbb{Z}$ is an epimorphism and $f(g)=1$ for some $g\in G$, then $h:\mathbb{Z}\to G$ given by $h(1)=g$ is the partial inverse. So the sequence (right)-splits.

It follows that $G$ is a semidirect product of $H$ and $\Bbb Z$. So we can chose $K\cong \Bbb Z$ with $K\cap H=1$ and $HK=G$.