This section describes the investigation of the following finitely presented group:
G := langlea, b ; ; ; a6 = 1 rangle We will show especially how to deal with subgroups which have a very large index like those occuring in the prime quotient algorithm.
Define the group and open the subgroup lattice window:
gap> f := FreeGroup(2); <free group on the generators [ f1, f2 ]> gap> g := f/[f.1^6]; <fp group on the generators [ f1, f2 ]> gap> s := GraphicSubgroupLattice(g); <graphic subgroup lattice "GraphicSubgroupLattice">
First compute prime quotients by Prime Quotient
in the Subgroups
menu. You are asked for a prime and a class. Enter 2 and 7
respectively. You get lots of output in the GAP command window and
seven new vertices. Some of the corresponding subgroups have huge
indices. Note that these groups are only represented as kernels of
epimorphisms within GAP. So explicit calculation of a coset table
or a presentation could take very long or be absolutely impossible!
Now compute epimorphisms onto the symmetric group on 3 points by
Epimorphisms (GQuotients)
in the Subgroups
menu, but use a polycyclic
presentation as follows (the reason for this will be explained below):
gap> IdGroup(SymmetricGroup(3)); [ 6, 1 ] gap> s3 := SmallGroup(6,1); <pc group with 2 generators> gap> IMAGE_GROUP := s3;;
This first determines the identification number of the symmetric group
on 3 points within the small groups library, and then fetches this
group as a polycyclic group. For groups of size less than 1000 this
is often a good way to get a polycyclic presentation. Note that
SymmetricGroup(3)
leads to a permutation group. The last statement
stores the group in a variable which can be used by XGAP.
Select vertex G, then click on Epimorphisms (GQuotient)
in the
Subgroups
menu and select User defined
in the window that pops up.
This will always use the group stored in the global variable
IMAGE_GROUP
. GAP finds three epimorphisms. Display the three
kernels by selecting display in the epimorphisms window.
Note that the new vertex 9 will be drawn on the line between vertices 2 and 3 because there is not yet a vertex in the level corresponding to index 6. You can move it aside by dragging it with the mouse to some better position within its level.
Now select vertices 8 and 11 and calculate the intersection of the two subgroups of indices 137438953472 and 6 respectively. GAP can calculate this intersection by calculating the subdirect product of the image groups of the epimorphisms (the index of the subgroup belonging to vertex 12 in G is 412316860416 which is three times of the index of the subgroup belonging to vertex 11). Note that this subdirect product can only be calculated because the two image groups are polycyclic groups. This is the reason why we needed S3 as polycyclic group earlier.
This is now the end of our partial investigation of the (partial)
subgroup lattice of G, close the graphic sheet by selecting close
graphic sheet
from the Sheet
menu.
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