Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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"... –  B. 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? –  B. S. Sep 2 '12 at 13:38
1  
@BabakSorouh You can deduce that $G$ must be finitely generated and apply the fundamental theorem. –  user38268 Sep 2 '12 at 14:58
add comment

2 Answers 2

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.

share|improve this answer
    
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 $|G|=\prod|H_i|$? –  B. S. Sep 2 '12 at 14:53
1  
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
1  
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. –  B. S. Sep 2 '12 at 17:27
add comment

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. $\hspace{6in} \square$

share|improve this answer
    
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!!! :-) –  B. S. Sep 2 '12 at 15:04
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.