When does an abelian group have a composition series? There is an exercise in the book "An Introduction to the group theory by J.J. Rose" which can also be found as a proposition in "Abstract algebra by T. Hungerford":

Every finite group has a composition series $^*$.

Now I am doing the exercise $5.9$ of the first above book:


*

*An abelian group has a composition series iff it is finite.


*Give an example of an infinite group which has a composition series.

About 1. :  Since $(*)$; one side can be carried out. For other side; what would be happened if we assumed the group was infinite? In fact, if an abelian group is infinite; it cannot have a composition series with finite length? Is this our contradiction? I see this by considering $\mathbb Z_{p^\infty}$ but cannot see the right way. Thanks.
 A: A non-trivial simple Abelian group is cyclic of prime order. A composition series must have finite length. This should suffice.
A: I think you can proceed with your problem by reducing to the case for (1) that your abelian group must be finitely generated. For suppose that your abelian group $G$ has a composition series. Then it would follow (I think) that all chains in $G$ are bounded in length, consequently $G$ satisfies the ascending chain condition and descending chain condition (as a $\Bbb{Z}$ - module) and so is finitely generated .  Now by the fundamental theorem of finitely generated abelian groups, we get that
$$G \cong \Bbb{Z}^n \oplus \Bbb{Z}_{p_1} \oplus \ldots \oplus \Bbb{Z}_{p_n}$$
for some prime numbers $p_1,\ldots,p_n$. Now if $n > 0$, you have a copy of $\Bbb{Z}$ sitting inside of $G$ that gives rise to a descending chain of subgroups inside of $G$ that does not terminate, contradicting $G$ being Artinian. It follows $n =0$ and consequently $G$ is a finite abelian group.
$\hspace{6in} \square$
