Existence of non-split sequence

Question

Let $G$ be an abelian group such that $G$ contains non-zero elements of finite order.

Why there exists some short exact non-split sequence:

$0 \rightarrow \mathbb{Z} \rightarrow H \rightarrow G \rightarrow 0$

for some abelian group $H$?

Answer 1

By assumption $G$ has a subgroup $D$ that is nontrivial and can be embedded in to $\mathbb Q/\mathbb Z$. By Zorn's lemma, there exists a maximal such $D$. Then $D=Q/\mathbb Z$ with $\mathbb Z<Q\le \mathbb Q$. Show that $D$ is a direct summand, i.e., $G=G_1\oplus D$ for some group $G_1$. This gives us a short exact seqeunce $$ 0\to\mathbb Z\to G_1\oplus Q\to \underbrace{G_1\oplus (Q/\mathbb Z)}_{=G}\to 0 $$ where the maps are $x\mapsto (0,x)$ and $(x,y)\mapsto (x,y+\mathbb Z)$. In this form, the non-splitting should be apparent from $\mathbb Z<Q $.

Answer 2

I'd be happier with a constructive solution, but this works in standard setting: If $x\in G$ is of order $n$, then ${\mathbb Z}/n{\mathbb Z}\hookrightarrow G$ via $\overline{1}\mapsto x$. By the divisibility/injectivity of ${\mathbb Q}/{\mathbb Z}$, the embedding ${\mathbb Z}/n{\mathbb Z}\hookrightarrow {\mathbb Q}/{\mathbb Z}$, $\overline{1}\mapsto\overline{\tfrac{1}{n}}$ extends to a morphism of abelian groups $\varphi: G\to {\mathbb Q}/{\mathbb Z}$. Moreover, this morphism does not lift along the projection ${\mathbb Q}\to{\mathbb Q}/{\mathbb Z}$ since the torsion element $g$ needs to be mapped to a torsion element, and ${\mathbb Q}$ is torsion free. Hence, the pullback of the short exact sequence $0\to {\mathbb Z}\to{\mathbb Q}\to{\mathbb Q}/{\mathbb Z}\to 0$ along $\varphi$ gives an exact sequence of the form $0\to{\mathbb Z}\to H\to G\to 0$.