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Dear Gap Forum members,

again two remarks:

First, please do not abuse the GAP Forum for questions that can

clearly be answered by a look into the maual, in this case even by

just reading and understanding my Forum answer of last time. The Forum

mail goes to several hundred GAP users worldwide and they do not want

their mail folders filled with such correspondence. We do try to

answer all Forum questions in the Forum, but if then there are return

questions of the kind 'I do not understand' then please send these

either directly to the person who had answered the first question or

to gap-trouble, which is read only by the GAP support team.

Then to your letter:

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.

I had explained in my last letter that you get the number of subgroups

of order n of a group g by

f := Sum(Filtered(c, cl -> Size(cl[1]) = n), x -> Size(x));

where

c := ConjugacyClassesSubgroups(g);

If you understand how such a nested function works - and that is

clearly explained in the GAP Tutorial - it is obvious that you get the

list of conjugacy classes of subgroups of oder n by

cn := Filtered(c, cl -> Size(cl[1]) = n);

Finally, I am afraid you 'did your math by hand' wrong:

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)>.

All subgroups that you list are generated by a pair of elements (x,y)

of which x is in the first cyclic factor and y is in the second. But

there are subgroups of order 12 for which this is not the case. Let me

for clarity call a generator of the first cyclic factor A and a

generator of the second cyclic factor B. Then e.g. the subgroup

generated by

A^2*B^2 and B^6

is not of this type.

Hope this settles the case, if you have further questions, please write

directly to me (or ask your supervisor).

Joachim Neubueser

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