Example 9

Do the groups G and H given below have Z-equivalent copies which lie in a finite unimodular group?

G is generated by

6
0 0 0 0 0 -1
0 0 0 0 1 -1
1 0 0 0 0  0
0 1 0 0 0  0
0 0 1 0 0  0
0 0 0 1 0  0

6
0 -1 0 0 0  0
1  0 0 0 0  0
0  0 0 0 0 -1
0  0 0 0 1  0
0  0 1 0 0  0
0  0 0 1 0  0

Used Programs

Solution

1. Note first that the answer is positive if and only if G and H are contained in Z-equivalent Bravais groups. Therefore write each of the generating sets into files called 'g' and 'h' and make sure that you add "#g1" as top line to have the files in bravais_TYP format.
2. Call

            Bravais_grp g > gb
            Bravais_grp h > hb
         

   to write the Bravais groups of G and H into files 'gb' and 'hb'. (Note, we have assumed already that G and H are finite. By calling

            Is_finite g
            Is_finite h
         

   this could have been checked beforehand.)
3. Call

            Bravais_inclusions -S gb
            Bravais_inclusions -S hb
         

   to get lists of names for the Z-classes of all Bravais groups containing G resp. H. We get

            Bravais groups which contain a Z-equivalent subgroup
            Symbol: 6-2'  homogeneously d.: 1 zclass: 1
            Symbol: 6-2  homogeneously d.: 1 zclass: 1
            Symbol: 6-2  homogeneously d.: 2 zclass: 1
            Symbol: 6-2  homogeneously d.: 3 zclass: 1
         

   and

            Bravais groups which contain a Z-equivalent subgroup
            Symbol: 4-1';2-1  homogeneously d.: 1 zclass: 1
            Symbol: 4-1;2-1  homogeneously d.: 1 zclass: 1
            Symbol: 4-1;2-1  homogeneously d.: 2 zclass: 1
            Symbol: 6-1  homogeneously d.: 1 zclass: 1
         

   There are no common names. Hence there is no finite subgroup of GL_6(Z) containig G and a GL₆(Z)-conjugate of H.