Date: Thu, 25 May 95 18:41:00 -0400
From: Mark Longridge <mark.longridge@canrem.com >
~~~ ~~~ Subject: GAP notes
```On 05/22/95 at 11:13:00 Martin Schoenert said:
```

GAP's 'NumberConjugacyClasses' follows the general usage in
group theory.
The conjugacy class of an element <g> of <G> is the set of elements <h>
that are G-conjugated to <g> (i.e., there exists an element <x> in <G>,
such that <x>^-1 * <g> * <x> = <h>).

```On  05-24-95 (18:16) Jerry Bryan said:
```

Just to give an example that I am familiar with, suppose the group
in question were M itself. Then, NumberConjugacyClasses should yield
10, because the 48 elements in M yield 10 conjugacy classes under
M-conjugation. If anybody who has GAP also has defined M, you
might give it a try.

Ok... let's define C in the context of GAP:

```c := Group(
( 1, 3, 8, 6)( 2, 5, 7, 4)( 9,33,25,17)(10,34,26,18)(11,35,27,19)
(20,12,36,28)(21,13,37,29)
(46,48,43,41)(44,47,45,42)(38,30,22,14)(39,31,23,15)(40,32,24,16),
( 9,11,16,14)(10,13,15,12)( 1,17,41,40)( 4,20,44,37)( 6,22,46,35)
(2,18,42,39)(7,23,47,34)
(30,32,27,25)(28,31,29,26)(19,43,38,3) (21,45,36,5) (24,48,33, 8)
);;
```

M is the same as C but with the central reflection:

```m := Group(
( 1, 3, 8, 6)( 2, 5, 7, 4)( 9,33,25,17)(10,34,26,18)(11,35,27,19)
(20,12,36,28)(21,13,37,29)
(46,48,43,41)(44,47,45,42)(38,30,22,14)(39,31,23,15)(40,32,24,16),
( 9,11,16,14)(10,13,15,12)( 1,17,41,40)( 4,20,44,37)( 6,22,46,35)
(2,18,42,39)(7,23,47,34)
(30,32,27,25)(28,31,29,26)(19,43,38,3) (21,45,36,5) (24,48,33, 8),
(1,8)(3,6)(2,7)(4,5)
(17,24)(19,22)(18,23)(20,21)
(9,16)(11,14)(10,15)(12,13)
(25,32)(27,30)(26,31)(28,29)
(33,40)(35,38)(34,39)(36,37)
(41,48)(43,46)(42,47)(44,45)
);;
```

Then we have

```Size (c) = 24
NumberConjugacyClasses (c) = 5

Size (m) = 48
NumberConjugacyClasses (m) = 10
```

These results concur with Dan's message from Tue, 28 Dec 93 18:40:52 EST
from the archives.

We can also use GAP to calculate the size of the M-conjugacy class of
a given element:

Size (ConjugacyClass (m, cross4)) = 3

Here we see there are three possible 4 Cross order 2 patterns.

I've tried dabbling in some GAP programming. Say we are looking for an
element in the slice group with 4 variants under M-conjugacy....

```a := 0;
x := 0;
z := Elements (slice);

repeat
a := a+1
x := Size (ConjugacyClass (m, Random (slice)));
until a = 768 or x = 4
```

This short program found no elements of size 4 in the slice group.

```-> Mark <-
```