Proving a Subgroup of a Solvable Group is Solvable

In summary, the conversation discusses the definition of a solvable group and the need to show that a subgroup of a solvable group is also solvable. The conversation also touches on the use of composition series and subnormal series, and the importance of having abelian factor groups in these series. Additionally, there is a discussion on the simplicity of abelian groups and how it relates to showing that distinct groups from a composition series for a subgroup are simple. Finally, the conversation suggests alternative methods for finding a composition series for a subgroup with abelian factor groups.
  • #1
samkolb
37
0
I'm supposed to show that a subgroup of a solvable group is solvable.

(I am using the Fraleigh Abstract Algebra book and the given definition of a solvable group is a group which has a COMPOSITION series in which each of the factor groups is abelian. In other books I have looked at a solvable group is defined as one which has a SUBNORMAL series in which all the factor groups are abelian.)

Let K be a subgroup of a solvable group G. Let

{e}=Ho<H1<...<Hn=G be a composition series for G in which all the factor groups are abelian.

I showed that the distinct groups from among (K & Hi) (&= set intersection)
form a subnormal series for K.

Then I used the second isomorphism to show that (K & Hi)/(K & Hi-1) is isomprphic to Hi-1(K & Hi)/Hi-1.

I then showed that Hi-1(K & Hi) is a subgroup of Hi. Since Hi/Hi-1 is abelian, Hi-1(K & Hi)/Hi-1 is abelian, and so (K & Hi)/(K & Hi-1) is abelain.

What I can't show is that the distinct groups among (K & Hi) from a composition series for K. That is, I can't show that (K & Hi)/(K & Hi-1) is simple.

Please help!
 
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  • #2
samkolb said:
(I am using the Fraleigh Abstract Algebra book and the given definition of a solvable group is a group which has a COMPOSITION series in which each of the factor groups is abelian. In other books I have looked at a solvable group is defined as one which has a SUBNORMAL series in which all the factor groups are abelian.)

These are the same: factor groups only exist for normal subgroups.

Let K be a subgroup of a solvable group G. Let

{e}=Ho<H1<...<Hn=G be a composition series for G in which all the factor groups are abelian.

I showed that the distinct groups from among (K & Hi) (&= set intersection)
form a subnormal series for K.

Then I used the second isomorphism to show that (K & Hi)/(K & Hi-1) is isomprphic to Hi-1(K & Hi)/Hi-1.

please could you tex this clearly so people can understand what you mean?


What I can't show is that the distinct groups among (K & Hi) from a composition series for K. That is, I can't show that (K & Hi)/(K & Hi-1) is simple.

Surely you mean abelian, not simple?
 
  • #3
Thank you for responding to my question.

As I said in the preface to my question, I need to find a composition series for K in which each of the factor groups is abelian. So I need each of the factor groups (K & Hi)/(K & Hi-1) to be both abelian and simple.
 
  • #4
samkolb said:
What I can't show is that the distinct groups among (K & Hi) from a composition series for K. That is, I can't show that (K & Hi)/(K & Hi-1) is simple.

Why do you need to show that? [itex]K\cap H_{i}/K\cap H_{i-1}[/itex] is isomorphic to a subgroup of [itex]H_i/H_{i-1}[/itex], so it is abelian, and this is enough to show that K is simple.

Actually, if K is a strict subgroup of G, you will have [itex]H_i\cap K=H_{i-1}\cap K[/itex] for at least one i. So, [itex]K\cap H_i[/itex] will not be a composition series -- you would have to remove some of them from the series.
 
  • #5
I've got 2 further comments on your question.

First, you don't need to use a composition series. Any subnormal series for which the quotients [itex]H_i/H_{i-1}[/itex] are Abelian will do. Then, you can eliminate any [itex]H_i[/itex] for which [itex]H_i=H_{i-1}[/itex]. Then, if any of the quotients are not simple you can enlarge the series, until you reach a maximum one. This will be a composition series and the quotients will still be abelian.

Second, the only simple abelian groups are the cyclic groups of prime order. Any subgroup will be either the trivial group, or the whole group, so still simple. So, the way you have set up the problem it will be the case that you get a composition series once you eliminate any terms with [itex]K\cap H_i=K\cap H_{i-1}[/itex].
 
  • #6
gel:

Thank you for your comments. They were very helpful.
 

1. How do you define a subgroup of a solvable group?

A subgroup of a solvable group is a subset of the original group that is itself a group. This means that it has a binary operation that is closed, associative, and has an identity element. In addition, the subgroup must contain the inverse of each of its elements.

2. What is the significance of proving that a subgroup of a solvable group is solvable?

Proving that a subgroup of a solvable group is solvable is important because it helps to understand the structure and properties of the original group. It also allows for the application of solvable group theory to the subgroup, which can lead to further insights and applications in other areas of mathematics.

3. What is the process for proving that a subgroup of a solvable group is solvable?

The process for proving that a subgroup of a solvable group is solvable involves showing that the subgroup itself is a solvable group. This can be done by demonstrating that the subgroup has a normal series with abelian factors, or by showing that it has a composition series with cyclic factors. These two methods are equivalent and can be used interchangeably.

4. Can a subgroup of a solvable group be solvable if the original group is not solvable?

No, a subgroup of a solvable group cannot be solvable if the original group is not solvable. This is because the subgroup inherits its solvability from the original group. If the original group is not solvable, then it is not possible for any of its subgroups to be solvable.

5. Are there any applications of proving a subgroup of a solvable group is solvable?

Yes, there are various applications of proving a subgroup of a solvable group is solvable. One example is in the study of finite groups, where solvable groups play an important role. Additionally, solvable groups have applications in other areas of mathematics, such as algebraic geometry and number theory. Knowing that a subgroup of a solvable group is solvable can also help in solving certain mathematical problems and equations.

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