Dera Dan
The answer is that the normal closure of a subnormal nilpotent subgroup is
nilpotent. Essentially is is Fitting's Theorem which says that the product
of two nilpotent normal subgroups is nilpotent. Consider the chain
K=K_1<K_2<\cdots <K_n=G for some K nilpotent and subnormal in G, where
each K-_{i+1} is normal in K_{i+1}.
Note that K_2 contains the normal closure of K in K_3 but the normal
closure of K in K_3 is the product of finite many K_3 conjugates of K all
of which are normal in K_2. So the normal closure of K in K_3 is nilpotent
by Fitting and note that we could replace K_2 by this normal closure and
work our way up the chain, or use induction on the defect.
if we don't insist that the groups are finite you get locally nilpotence
instead.
Alan
School of Mathematics
University of East Anglia
Norwich
UK NR4 7TJ
Mobile Phone 07876024281
Home Telephone (Norwich) +44 (0)1603 612177
For information on Maths Association go to www.m-a.org.uk.
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To read my blog go to http://www.alansthunks.blogspot.com/
On Thu, 26 Apr 2007, Dan Lanke wrote:
Dear GAP Forum,
I am trying to find an example of a finite group G with a subnormal nilpotent
subgroup H such that the normal closure of H in G is not nilpotent. I am not
sure if such an example exists. I used GAP to see that no such example
exists for |G| < 190. I run into memory problems with the computer for
larger groups. Any suggestions please?
Thanks,
DL
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