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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.

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FNH
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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 $\,H\,$ ...!)

DonAntonio
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  • Why don't post your answer here? –  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 ?!

     – FNH Jan 19 '13 at 12:27
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    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 ? – FNH 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
  • About my answer: can you give the exact place where this exercise is in Dummit & Foote? – DonAntonio Jan 20 '13 at 22:45
  • page 106 , the exercise number 11 , it's in section 4 in chapter 3 which talks about holder program and composition series . – FNH Jan 20 '13 at 23:11
  • Ok: the commutator subgroups (or derived subgroup) was defined in exercises 40-41 in page 89 (section 3.1), but characteristic subgroup indeed is presented later (page 135 or so). Anyway, you can prove directly that the commutator subgroup of a normal subgroup is again normal in the big group... – DonAntonio Jan 21 '13 at 03:37
  • Read "Added 2" in the answer by Arturo Magidin in the link provided by YACP above. I think it works out an answer using only the available info in D&F book at page 89. It is pretty messy and long since after some pages, with some new stuff, the claim follows pretty easily, as in my answer. – DonAntonio Jan 21 '13 at 03:54