Dear Gap Forum,
First, I want to apologize for the last message. I had used absolute value
symbols and they appeared to you as =E2=94=82. Had I known this would have
happened, I would have written out "the order of Bn."
Second, I thank Joachim Neubueser for their response, which seemed to work
in the cases I described below. Unfortunately the need was for cases that
are not yet solved. Again, I tried the below mentioned commands on the 3
cases mentioned earlier:
Zp X Zp X ... XZp where p is prime
Zp^n X Zp^n where p is prime and n is a natural number
Zp1p2 X Zp1p2 where p1 and p2 are distinct primes
and it seemed to work for those tested.
I also tried it on Z12 X Z12 and received 28 subgroups of order 12. Doing this by hand only yields 24 subgroups of order 12. <(0, 1)>, <(1, 0)>, <(1, 1)>, <(1, 2)>, <(1, 3)>, <(1, 4)>, <(1, 5)>, <(1, 6)>, <(1, 7)>, <(1, 8)>, <(1, 9)>, <(1, 10)>, <(1, 11)>, <(2, 1)>, <(2, 3)>, <(2, 5)>, <(3, 1)>, <(3, 2)>, <(3, 4)>, <(3, 7)>, <(4, 1)>, <(4, 3)>, <(4, 5)>, and <(6, 1)>.
I redid my math by hand and get the same answer. I'm not sure how to get
GAP to show the 28 subgroups it thinks are of order 12 so that I can compare
results. Anyone having any idea on either question are more than welcome to
respond.
Again, my original question should have read:
Bn is the collection of all subgroups of B with order n.
Our goal is to find all groups so that the order of Bn will divide the order
of the group.
We are looking at finite groups only.
Or as restated by Joachim Neubueser:
I want to study groups in which the number of subgroups of a certain order
divides the group order.
How can I use GAP to find the number of subgroups of a given order in a
given finite group?
Thank you again for your help.
ChadT ---------------------------- There are three aspects of this letter to be answered:
First: Please do not use exotic font encoding in letters to the
Forum. I, likely for one, have not been able to decode it and also
some younger colleagues who are pretty good at these things, have not
been able to do it in reasonable time. Please rather use plain text or
at best some TEX.
If I understand right what you say without using the encoded formulae,
then you want to study groups in which the number of subgroups of a
certain order divides the group order and for this you ask how you can
use GAP to find the number of subgroups of a given order in a given
finite group.
So secondly let me answer this last question.
The function
ConjugacyClassesSubgroups(g)
will give you a list of all conjugacy classes of subgroups of the
group g.
You can the use the various functions for lists to get all information
about subgroups. E.g. to get the number of subgroups of order 4 in the
symmetric group S_4 type:
g := SymmetricGroup(4); c := ConjugacyClassesSubgroups(g); f := Sum(Filtered(c, cl -> Size(cl[1]) = 4), x -> Size(x));
and you will get the result that there are 7 subgrups of order 4 in S_4.
For the many possibilities of using list functions in GAP read the
manual chapter on lists. Note in the above that Size(cl) will give you
the size of the class, i.e. the number of conjugates, but Size(cl[1])
will give you the size, i.e. order of the first group in the class!
Finally, I am afraid that the problem you want to investigate does not
look very promising to me. The condition that the number of subgroups
of given order should divide the group order looks rather unnatural to
me.
There are on the other hand some papers on the number of subgroups of
given order, mainly using the idea of the Eulerian functions of a
group, that might interest you. See for instance:
G.E. Wall
Some applications of the Eulerian Functions of a Finite Group.
Journal Austral. Math. Soc. 2 (1961) 35 - 59.
as well as papers quoted therein and also possibly papers that quote
this one and that you may find via the Science Quotation Index.
With kind regards Joachim Neubueser
Chad T. Lower
a.k.a. Chopper
http://chadtlower.tripod.com/
Fight Spam! Join CAUCE! == http://www.cauce.org/
Greet one another with the kiss of love. Peace to all of you that are in Christ. I Peter 5:14
The suespammers.org mail server is located in California; do not send
unsolicited bulk e-mail or unsolicited commercial e-mail to my
suespammers.org address.
_________________________________________________________________ Get your FREE download of MSN Explorer at http://explorer.msn.com/intl.asp
Miles-Receive-Header: reply