  
  [1X2 [33X[0;0YThe main functions[133X[101X
  
  
  [1X2.1 [33X[0;0YZassenhaus Conjecture[133X[101X
  
  [33X[0;0YThis  function checks whether the Zassenhaus Conjecture ((ZC) for short, cf.
  Section  [14X5.1[114X) can be proved using the HeLP method with the data available in
  GAP.[133X
  
  [1X2.1-1 HeLP_ZC[101X
  
  [29X[2XHeLP_ZC[102X( [3XOrdinaryCharacterTable|Group[103X ) [32X function
  [6XReturns:[106X  [33X[0;10Y[9Xtrue[109X if (ZC) can be solved using the given data, [9Xfalse[109X otherwise[133X
  
  [33X[0;0Y[9XHeLP_ZC[109X checks whether the Zassenhaus Conjecture can be solved for the given
  group  using  the  HeLP  method,  the  Wagner  test  and  all character data
  available.  The argument of the function can be either an ordinary character
  table  or  a  group.  In  the  second  case  it  will  first  calculate  the
  corresponding  ordinary  character  table.  If  the  group  in  question  is
  nilpotent,  the  Zassenhaus Conjecture holds by a result of A. Weiss and the
  function will return [9Xtrue[109X without performing any calculations.[133X
  
  [33X[0;0YIf  the  group is not solvable, the function will check all orders which are
  divisors  of  the  exponent  of the group. If the group is solvable, it will
  only check the orders of group elements, as there can't be any torsion units
  of  another  order.  The  function  will use the ordinary table and, for the
  primes  [23Xp[123X  for  which the group is not [23Xp[123X-solvable, all [23Xp[123X-Brauer tables which
  are  available in GAP to produce as many constraints on the torsion units as
  possible.  Additionally,  the  Wagner  test  is  applied to the results, cf.
  Section  [14X5.4[114X.  In  case  the  information suffices to obtain a proof for the
  Zassenhaus Conjecture for this group the function will return [9Xtrue[109X and [9Xfalse[109X
  otherwise.  The  possible  partial augmentations for elements of order [22Xk[122X and
  all its powers will also be stored in the list entry [9XHeLP_sol[k][109X.[133X
  
  [33X[0;0YThe  prior  computed  partial augmentations in [9XHeLP_sol[109X will not be used and
  will  be  overwritten.  If  you  do  not  like  the  last  fact,  please use
  [2XHeLP_AllOrders[102X ([14X3.3-1[114X).[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XG := AlternatingGroup(5);[127X[104X
    [4X[28XAlt( [ 1 .. 5 ] )[128X[104X
    [4X[25Xgap>[125X [27XHeLP_ZC(G);[127X[104X
    [4X[28Xtrue[128X[104X
    [4X[25Xgap>[125X [27XC := CharacterTable("A5");[127X[104X
    [4X[28XCharacterTable( "A5" )[128X[104X
    [4X[25Xgap>[125X [27XHeLP_ZC(C);[127X[104X
    [4X[28Xtrue[128X[104X
    [4X[25Xgap>[125X [27XHeLP_sol;[127X[104X
    [4X[28X[ [ [ [ 1 ] ] ], [ [ [ 1 ] ] ], [ [ [ 1 ] ] ],, [128X[104X
    [4X[28X  [ [ [ 0, 1 ] ], [ [ 1, 0 ] ] ], [  ],,,, [  ],,,,, [  ],,,,,,,,,,,,,,, [  ] [128X[104X
    [4X[28X ][128X[104X
    [4X[25Xgap>[125X [27XHeLP_PrintSolution(); [127X[104X
    [4X[28XSolutions for elements of order 2:[128X[104X
    [4X[28X[ [         u ],[128X[104X
    [4X[28X  [  [ "2a" ] ],[128X[104X
    [4X[28X  [       --- ],[128X[104X
    [4X[28X  [     [ 1 ] ] ][128X[104X
    [4X[28XSolutions for elements of order 3:[128X[104X
    [4X[28X[ [         u ],[128X[104X
    [4X[28X  [  [ "3a" ] ],[128X[104X
    [4X[28X  [       --- ],[128X[104X
    [4X[28X  [     [ 1 ] ] ][128X[104X
    [4X[28XSolutions for elements of order 5:[128X[104X
    [4X[28X[ [               u ],[128X[104X
    [4X[28X  [  [ "5a", "5b" ] ],[128X[104X
    [4X[28X  [             --- ],[128X[104X
    [4X[28X  [        [ 0, 1 ] ],[128X[104X
    [4X[28X  [        [ 1, 0 ] ] ][128X[104X
    [4X[28XThere are no admissible partial augmentations for elements of order 6.[128X[104X
    [4X[28XThere are no admissible partial augmentations for elements of order 10.[128X[104X
    [4X[28XThere are no admissible partial augmentations for elements of order 15.[128X[104X
    [4X[28XThere are no admissible partial augmentations for elements of order 30.[128X[104X
  [4X[32X[104X
  
  [33X[0;0YThis  is  the classical example of Luthar and Passi to verify the Zassenhaus
  Conjecture  for  the alternating group of degree 5, cf. [LP89]. In the first
  call  of  [9XHeLP_ZC[109X  this is checked using the character table computed by GAP
  using  the  given  group,  the second call uses the character table from the
  character table library. The entries of [9XHeLP_sol[109X are[133X
  
  [30X    [33X[0;6Ylists  with  entries  0  and  1  (at  the  spots  1, 2, 3 and 5, which
        correspond to torsion units that are conjugate to group elements),[133X
  
  [30X    [33X[0;6Yempty  lists (at the spots 6, 10, 15 and 30, stating that there are no
        admissible partial augmentations for these orders),[133X
  
  [30X    [33X[0;6Yor  are  not bound (these orders were not checked as they don't divide
        the exponent of the group).[133X
  
  [33X[0;0YThe function [2XHeLP_PrintSolution[102X ([14X3.8-1[114X) can be used to display the result in
  a pretty way.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XC := CharacterTable( "A6" );[127X[104X
    [4X[28XCharacterTable( "A6" )[128X[104X
    [4X[25Xgap>[125X [27XSetInfoLevel(HeLP_Info, 2);[127X[104X
    [4X[25Xgap>[125X [27XHeLP_ZC(C);[127X[104X
    [4X[28X#I  Checking order 2.[128X[104X
    [4X[28X#I  Checking order 3.[128X[104X
    [4X[28X#I  Checking order 4.[128X[104X
    [4X[28X#I  Checking order 5.[128X[104X
    [4X[28X#I  Checking order 6.[128X[104X
    [4X[28X#I  Checking order 10.[128X[104X
    [4X[28X#I  Checking order 12.[128X[104X
    [4X[28X#I  Checking order 15.[128X[104X
    [4X[28X#I  Checking order 20.[128X[104X
    [4X[28X#I  Checking order 30.[128X[104X
    [4X[28X#I  Checking order 60.[128X[104X
    [4X[28X#I  ZC can't be solved, using the given data, for the orders: [ 6 ].[128X[104X
    [4X[28Xfalse[128X[104X
    [4X[25Xgap>[125X [27XHeLP_sol[6];                                          [127X[104X
    [4X[28X[ [ [ 1 ], [ 0, 1 ], [ -2, 2, 1 ] ], [ [ 1 ], [ 1, 0 ], [ -2, 1, 2 ] ] ][128X[104X
    [4X[25Xgap>[125X [27XHeLP_PrintSolution(6);                                [127X[104X
    [4X[28XSolutions for elements of order 6:[128X[104X
    [4X[28X[ [                   u^3,                   u^2,                     u ],[128X[104X
    [4X[28X  [              [ "2a" ],        [ "3a", "3b" ],  [ "2a", "3a", "3b" ] ],[128X[104X
    [4X[28X  [                   ---,                   ---,                   --- ],[128X[104X
    [4X[28X  [                 [ 1 ],              [ 0, 1 ],          [ -2, 2, 1 ] ],[128X[104X
    [4X[28X  [                 [ 1 ],              [ 1, 0 ],          [ -2, 1, 2 ] ] ][128X[104X
    [4X[25Xgap>[125X [27XSetInfoLevel(HeLP_Info, 1);[127X[104X
  [4X[32X[104X
  
  [33X[0;0YThis  is  the  example  M.  Hertweck deals with in his article [Her08c]. The
  HeLP-method  is  not sufficient to verify the Zassenhaus Conjecture for this
  group.  There  are  two tuples of possible partial augmentations for torsion
  units of order 6 which are admissible by the HeLP method. M. Hertweck used a
  different argument to eliminate these possibilities.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XG := SmallGroup(48,30);;[127X[104X
    [4X[25Xgap>[125X [27XStructureDescription(G);[127X[104X
    [4X[28X"A4 : C4"[128X[104X
    [4X[25Xgap>[125X [27XHeLP_ZC(G);[127X[104X
    [4X[28X#I  ZC can't be solved, using the given data, for the orders: [ 4 ].[128X[104X
    [4X[28Xfalse[128X[104X
    [4X[25Xgap>[125X [27XSize(HeLP_sol[4]);[127X[104X
    [4X[28X10[128X[104X
  [4X[32X[104X
  
  [33X[0;0YThe  group SmallGroup(48,30) is the smallest group for which the HeLP method
  does not suffice to prove the Zassenhaus Conjecture. However (ZC) was proved
  for this group in [HK06], Proposition 4.2.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XC1 := CharacterTable(SymmetricGroup(5));[127X[104X
    [4X[28XCharacterTable( Sym( [ 1 .. 5 ] ) )[128X[104X
    [4X[25Xgap>[125X [27XHeLP_ZC(C1);[127X[104X
    [4X[28X#I  The Brauer tables for the following primes are not available: [ 2, 3, 5 ].[128X[104X
    [4X[28X#I  ZC can't be solved, using the given data, for the orders: [ 4, 6 ].[128X[104X
    [4X[28Xfalse[128X[104X
    [4X[25Xgap>[125X [27XC2 := CharacterTable("S5");[127X[104X
    [4X[28XCharacterTable( "A5.2" )[128X[104X
    [4X[25Xgap>[125X [27XHeLP_ZC(C2); [127X[104X
    [4X[28Xtrue[128X[104X
  [4X[32X[104X
  
  [33X[0;0YThis  example  demonstrates  the  advantage of using the GAP character table
  library:  Since  GAP can't compute the Brauer tables from the ordinary table
  of  [23XS_5[123X  in  the  current  implementation,  they  are  not used in the first
  calculation.  But  in  the  second  calculation  [9XHeLP_ZC[109X accesses the Brauer
  tables  from  the  library  and can prove the Zassenhaus Conjecture for this
  group,  see  [Her07], Section 5. This example might of course change as soon
  as GAP will be able to compute the needed Brauer tables.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XC := CharacterTable("M11");[127X[104X
    [4X[28XCharacterTable( "M11" )[128X[104X
    [4X[25Xgap>[125X [27XHeLP_ZC(C);[127X[104X
    [4X[28X#I  ZC can't be solved, using the given data, for the orders: [ 4, 6, 8 ].[128X[104X
    [4X[28Xfalse[128X[104X
    [4X[25Xgap>[125X [27XHeLP_sol[12];[127X[104X
    [4X[28X[  ][128X[104X
    [4X[25Xgap>[125X [27XHeLP_PrintSolution(8);[127X[104X
    [4X[28XSolutions for elements of order 8:[128X[104X
    [4X[28X[ [      u^4,                         u^2,                           u ],[128X[104X
    [4X[28X  [ [ "2a" ],              [ "2a", "4a" ],  [ "2a", "4a", "8a", "8b" ] ],[128X[104X
    [4X[28X  [      ---,                         ---,                         --- ],[128X[104X
    [4X[28X  [    [ 1 ],                    [ 0, 1 ],              [ 0, 0, 0, 1 ] ],[128X[104X
    [4X[28X  [    [ 1 ],                    [ 0, 1 ],              [ 0, 0, 1, 0 ] ],[128X[104X
    [4X[28X  [    [ 1 ],                    [ 0, 1 ],             [ 0, 2, -1, 0 ] ],[128X[104X
    [4X[28X  [    [ 1 ],                    [ 0, 1 ],             [ 0, 2, 0, -1 ] ],[128X[104X
    [4X[28X  [    [ 1 ],                   [ 2, -1 ],              [ 0, 0, 0, 1 ] ],[128X[104X
    [4X[28X  [    [ 1 ],                   [ 2, -1 ],              [ 0, 0, 1, 0 ] ],[128X[104X
    [4X[28X  [    [ 1 ],                   [ 2, -1 ],             [ 0, 2, -1, 0 ] ],[128X[104X
    [4X[28X  [    [ 1 ],                   [ 2, -1 ],             [ 0, 2, 0, -1 ] ] ][128X[104X
  [4X[32X[104X
  
  [33X[0;0YComparing this example to the result in [BK07a] one sees, that the existence
  of   elements  of  order  12  in  [23X\mathrm{V}(\mathbb{Z}M_{11})[123X  may  not  be
  eliminated  using only the HeLP method. This may be done however by applying
  also  the  Wagner  test,  cf.  Section  [14X5.4[114X and the example for the function
  [2XHeLP_WagnerTest[102X ([14X3.7-1[114X).[133X
  
  [33X[0;0YThis  example  also  demonstrates, why also the partial augmentations of the
  powers of [23Xu[123X must be stored (and not only the partial augmentations of [23Xu[123X). To
  prove  that  all  elements  of  order  [23X8[123X in [23X\mathrm{V}(\mathbb{Z}M_{11})[123X are
  rationally  conjugate  to group elements, it is not enough to prove that all
  elements  [23Xu[123X  of  order  [23X8[123X  in  [23X\mathrm{V}(\mathbb{Z}M_{11})[123X have all partial
  augmentations  [23X1[123X  and [23X0[123X, as the fifth and sixth possibility from above still
  could  exist  in [23X\mathrm{V}(\mathbb{Z}M_{11})[123X, which would not be rationally
  conjugate to group elements.[133X
  
  
  [1X2.2 [33X[0;0YPrime Graph Question[133X[101X
  
  [33X[0;0YThis  function  checks whether the Prime Graph Question ((PQ) for short, cf.
  Section  [14X5.1[114X)  can be verified using the HeLP method with the data available
  in GAP.[133X
  
  [1X2.2-1 HeLP_PQ[101X
  
  [29X[2XHeLP_PQ[102X( [3XOrdinaryCharacterTable|Group[103X ) [32X function
  [6XReturns:[106X  [33X[0;10Y[9Xtrue[109X if (PQ) can be solved using the given data, [9Xfalse[109X otherwise[133X
  
  [33X[0;0Y[9XHeLP_PQ[109X  checks  whether  an affirmative answer for the Prime Graph Question
  for  the  given  group  can  be  obtained  using the HeLP method, the Wagner
  restrictions  and  the  data  available. The argument of the function can be
  either  an  ordinary  character table or a group. In the second case it will
  first  calculate the corresponding ordinary character table. If the group in
  question  is solvable, the Prime Graph Question has an affirmative answer by
  a result of W. Kimmerle and the function will return [9Xtrue[109X without performing
  any calculations.[133X
  
  [33X[0;0YIf  the group is non-solvable, the ordinary character table and all [23Xp[123X-Brauer
  tables  for  primes  [23Xp[123X  for  which the group is not [23Xp[123X-solvable and which are
  available  in GAP will be used to produce as many constraints on the torsion
  units  as possible. Additionally, the Wagner test is applied to the results,
  cf.  Section  [14X5.4[114X. In case the information suffices to obtain an affirmative
  answer  for  the  Prime Graph Question, the function will return [9Xtrue[109X and it
  will  return [9Xfalse[109X otherwise. Let [23Xp[123X and [23Xq[123X be distinct primes such that there
  are  elements  of order [23Xp[123X and [23Xq[123X in [23XG[123X but no elements of order [23Xpq[123X. Then for [23Xk[123X
  being  [23Xp[123X,  [23Xq[123X or [23Xpq[123X the function will save the possible partial augmentations
  for  elements  of  order  [23Xk[123X and its (non-trivial) powers in [9XHeLP_sol[k][109X. The
  function also does not use the previously computed partial augmentations for
  elements  of these orders but will overwrite the content of [9XHeLP_sol[109X. If you
  do not like the last fact, please use [2XHeLP_AllOrdersPQ[102X ([14X3.3-2[114X).[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XC := CharacterTable("A7");[127X[104X
    [4X[28XCharacterTable( "A7" )[128X[104X
    [4X[25Xgap>[125X [27XHeLP_PQ(C);[127X[104X
    [4X[28Xtrue[128X[104X
  [4X[32X[104X
  
  [33X[0;0YThe  Prime  Graph  Question  for the alternating group of degree 7 was first
  proved by M. Salim [Sal11].[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XC := CharacterTable("L2(19)");[127X[104X
    [4X[28XCharacterTable( "L2(19)" )[128X[104X
    [4X[25Xgap>[125X [27XHeLP_PQ(C);                   [127X[104X
    [4X[28Xtrue[128X[104X
    [4X[25Xgap>[125X [27XHeLP_ZC(C);[127X[104X
    [4X[28X#I  (ZC) can't be solved, using the given data, for the orders: [ 10 ].[128X[104X
    [4X[28Xfalse[128X[104X
    [4X[25Xgap>[125X [27XHeLP_sol[10];[127X[104X
    [4X[28X[ [ [ 1 ], [ 0, 1 ], [ 0, -1, 1, 0, 1 ] ], [128X[104X
    [4X[28X  [ [ 1 ], [ 0, 1 ], [ 0, 0, 0, 1, 0 ] ], [128X[104X
    [4X[28X  [ [ 1 ], [ 1, 0 ], [ 0, 0, 0, 0, 1 ] ], [128X[104X
    [4X[28X  [ [ 1 ], [ 1, 0 ], [ 0, 1, -1, 1, 0 ] ] ][128X[104X
  [4X[32X[104X
  
  [33X[0;0YThe  HeLP  method provides an affirmative answer to the Prime Graph Question
  for  the  group  L2(19),  although  the  method doesn't solve the Zassenhaus
  Conjecture  for  that  group,  as  there  are  two  sets of possible partial
  augmentations  for  units  of  order  10  left,  which  do not correspond to
  elements  which  are  rationally conjugate to group elements. The Zassenhaus
  Conjecture for this group is proved in [BM14].[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XC1 := CharacterTable(PSL(2,7));                                  [127X[104X
    [4X[28XCharacterTable( Group([ (3,7,5)(4,8,6), (1,2,6)(3,4,8) ]) )[128X[104X
    [4X[25Xgap>[125X [27XHeLP_PQ(C1);[127X[104X
    [4X[28X#I  The Brauer tables for the following primes are not available: [ 2, 3, 7 ].[128X[104X
    [4X[28X#I  PQ can't be solved, using the given data, for the orders: [ 6 ].[128X[104X
    [4X[28Xfalse[128X[104X
    [4X[25Xgap>[125X [27XC2 := CharacterTable("L2(7)");  [127X[104X
    [4X[28XCharacterTable( "L3(2)" )[128X[104X
    [4X[25Xgap>[125X [27XHeLP_PQ(C2);                  [127X[104X
    [4X[28Xtrue[128X[104X
  [4X[32X[104X
  
  [33X[0;0YThis  example  demonstrates  the  advantage  of  using  tables  from the GAP
  character  table  library:  Since  GAP  can  not  compute  the Brauer tables
  corresponding  to  [9XC1[109X they are not used in the first calculation. But in the
  second  calculation  [9XHeLP_PQ[109X accesses the Brauer tables from the library and
  can  prove  the Prime Graph Question for this group, see [Her07], Section 6.
  This example might change, as soon as GAP will be able to compute the Brauer
  tables needed.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XSetInfoLevel(HeLP_Info,2);[127X[104X
    [4X[25Xgap>[125X [27XC := CharacterTable("A6");[127X[104X
    [4X[28XCharacterTable( "A6" )[128X[104X
    [4X[25Xgap>[125X [27XHeLP_PQ(C);[127X[104X
    [4X[28X#I  Checking order 2.[128X[104X
    [4X[28X#I  Checking order 3.[128X[104X
    [4X[28X#I  Checking order 5.[128X[104X
    [4X[28X#I  Checking order 6.[128X[104X
    [4X[28X#I  Checking order 10.[128X[104X
    [4X[28X#I  Checking order 15.[128X[104X
    [4X[28X#I  PQ can't be solved, using the given data, for the orders: [ 6 ].[128X[104X
    [4X[28Xfalse[128X[104X
    [4X[25Xgap>[125X [27XSetInfoLevel(HeLP_Info,1);[127X[104X
  [4X[32X[104X
  
  [33X[0;0YThe  Prime  Graph Question can not be confirmed for the alternating group of
  degree  6  with  the HeLP-method. This group is handled in [Her08c] by other
  means.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XC := CharacterTable("L2(49)");[127X[104X
    [4X[28XCharacterTable( "L2(49)" )[128X[104X
    [4X[25Xgap>[125X [27XHeLP_PQ(C);[127X[104X
    [4X[28X#I  The Brauer tables for the following primes are not available: [ 7 ].[128X[104X
    [4X[28X#I  (PQ) can't be solved, using the given data, for the orders: [ 10, 15 ].[128X[104X
    [4X[28Xfalse[128X[104X
  [4X[32X[104X
  
  [33X[0;0YThis  example  shows  the limitations of the program. Using the Brauer table
  for  the  prime  7  one  can  prove (PQ) for PSL(2,49), but this data is not
  available  in GAP at the moment. The fact that there are no torsion units of
  order 10 and 15 was proved in [Her07], Proposition 6.7. See also the example
  in  Section  [14X4.5[114X.  The  other critical orders were handled in a more general
  context in [BM16].[133X
  
