TITLE: Splitting of exact sequence of groups when middle group has split subgroup.
QUESTION [5 upvotes]: I am trying to show that a short exact sequence of abelian groups splits. I have a short exact sequence,
$$0\rightarrow \mathbb{Z} \rightarrow G \rightarrow \mathbb{Z}_2 \rightarrow 0$$
and I know that  $\mathbb Z \oplus \mathbb Z_2$ is a subgroup of $G$. Do I have a splitting by sending the generator of $\mathbb Z_2$ to the generator of $\mathbb Z_2$ in $G$?

REPLY [4 votes]: If you have a sequence
$$0 \to F \to G \to H \to 0$$
then $H$ is a quotient of $G$. In general there will be no map $H \to G$ giving a section of $G\to H$. However, if such a section exists then the sequence is indeed split on the right (by definition), and therefore split. The important thing is not only the existence of an injection $H\to G$, but the existence of one which is a section to the given map $G\to H$.
In your case, since $G$ contains an element $\epsilon$ of order $2$, there is a unique injection $H\to G$ taking the generator of $H$ to $\epsilon$. So the question is: is this injection necessarily a section of $G \to H$? In other words, does $G\to H$ necessarily map $\epsilon$ to the generator of $H$? Well, if it doesn't, then $\epsilon$ is in the kernel, hence in the image of $\mathbf Z$. But $\mathbf Z$ has no element of finite order. Therefore, the map you have described is necessarily a section, and the sequence splits.