Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Possible Duplicate:
A Nontrivial Subgroup of a Solvable Group

If $H$ is nontrivial normal subgroup of the solvable group $G$, then how can I show that there is a nontrivial subgroup $A\leq H$ such that $A$ is abelian and normal in $G$?

I am looking for hints so that I can create my own solution.

Thank you all.

share|cite|improve this question

marked as duplicate by YACP, Stefan Hansen, Davide Giraudo, rschwieb, TMM Jan 19 '13 at 14:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Hint: Isomorphism theorems – Geoff Robinson Aug 23 '11 at 17:39
Yes, right! Because using third theorem isomorphism $[H\cap G^{(m-1)}\cap G^{(m)}]=1$. And for the second relation $(H\cap G^{(m-1)})G^{(m)}<G^{(m-1)}$ since $G^{(m)}<G^{(m-1)}$. Is it right? – stacy Aug 24 '11 at 9:25
See the answer by Arturo Magidin at…. – Rick Jan 20 '12 at 15:15
@stacy: Did you get your answer? :-) – Babak S. Sep 4 '12 at 18:38
See this page for a good idea. It doesn't have any answers explicitly posted, but it will get you started. – Clayton Jan 19 '13 at 5:48

Hints (for you to prove):

1) It is true that

$$H\geq H'\geq\ldots\geq H^{(n)}=1\,\, ,\,\, \text{for some}\,\,\,n\in\Bbb N$$

2) Show that $\,H^{(n-1)}\triangleleft G\,\;\;$ (Yes, not only in $\,H\,$ ...!)

share|cite|improve this answer
Why don't post your answer here? – user26857 Jan 19 '13 at 10:31
Thank you for this hints ! i think I can slove it as follows , H is normal subgroup of G , so H is solvabe also . let 1=Hn is normal of H(n-1) is normal of ... is normal of H where H(i-1)\Hi is abelian let Hi = 1 then H(i-1) = H(n-1) and H(n-1) is abelian from this , then H(n-1) is normal of G done ! Is this right ?! – Maths Lover Jan 19 '13 at 12:27
Well: $\,H^{(n-1)}\,$ is abelian because its derived group is trivial, but he's normal in $\,G\,$ not for being abelian (there're lots of groups with non-normal abelian subgroups!), but because it is a characteristic (even fully invariant) subgroup of a normal subgroup of $\,G\,$... – DonAntonio Jan 19 '13 at 13:47
but I didn't study derived Group or characteristic ! it's not mentioned untill this moment in my text ! i use dummit and foote ! is there another way to do it ? can you give me example to a group whose a abelian subgroup , and this subgroup is not normal ? – Maths Lover Jan 20 '13 at 19:16
The subgroup $\,\{(1)\,,\,(12)\}\leq S_3\,$ is abelian and very not normal, to give perhaps the minimal and easiest example. – DonAntonio Jan 20 '13 at 22:44

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