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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:

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

  2. 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.

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Can you prove that an infinite abelian group must have an infinite proper subgroup? then can you see how to use this to rule out a finite composition series? – Gerry Myerson Sep 2 '12 at 12:49
Can you rule out the case where there is an infinite cyclic subgroup? – i. m. soloveichik Sep 2 '12 at 13:16
@GerryMyerson: I consider $\mathbb Z_2\times\mathbb Z$ for what you noted and see $\{1\}\times\mathbb Z$ is its infinte subgroup. I am thinking of that "rule out"... – Babak S. Sep 2 '12 at 13:20
@GerryMyerson: I think that, since we need $G_{i+1}$ be a maximal normal subgroup of $G_i$ in any strictly arranged composition series; we will encounter a contradiction if we regard that infinite subgroup. Right? – Babak S. Sep 2 '12 at 13:38
@BabakSorouh You can deduce that $G$ must be finitely generated and apply the fundamental theorem. – user38268 Sep 2 '12 at 14:58
up vote 5 down vote accepted

A non-trivial simple Abelian group is cyclic of prime order. A composition series must have finite length. This should suffice.

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Sorry Geoff for so many asking but did you use this fact that if $G$, a finite group, has a normal series with factor groups $H_0,H_1,...H_n$ then $|G|=\prod|H_i|$? – Babak S. Sep 2 '12 at 14:53
Yes, if a group has a composition series with finite composition factors, then its order is the product of the order of the composition factors. – Geoff Robinson Sep 2 '12 at 15:39
An infinite group which has a composition series is ${\rm PSL}(2,\mathbb{C}),$ which is simple and infinite. – Geoff Robinson Sep 2 '12 at 17:09
Thanks for your kind help about 2. Thanks for that again. I think it would be a bit difficult to me to find it. – Babak S. Sep 2 '12 at 17:27

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. $\hspace{6in} \square$

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If $n$ is positive then there is an infinite direct summand. There are not many cases to consider where this is not a proper subgroup ... – Mark Bennet Sep 2 '12 at 13:40
@MarkBennet Yes I see it now. – user38268 Sep 2 '12 at 13:50
@BenjaLim: Honestly, I was thinking about the fundamental theorem of finitely generated abelian groups but I wonder why J.J.Rose brought this question in the chapter 5 while he would discuss about abelian groups in chapter 10!!! :-) – Babak S. Sep 2 '12 at 15:04

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