Ask Your Question
1

does as_permutation_group() respect generators?

asked 2018-02-01 20:22:04 +0200

amontero90 gravatar image

updated 2020-06-01 10:47:39 +0200

FrédéricC gravatar image

From sage documentation I can construct a group as follows:

sage: m1 = matrix(GF(11), [[1,2],[3,4]])
sage: m2 = matrix(GF(11), [[1,3],[10,0]])
sage: G = MatrixGroup(m1, m2);  G
Matrix group over Finite Field of size 11 with 2 generators (
[1 2]  [ 1  3]
[3 4], [10  0]
)

Now I use the following code:

sage: G_p=G.as_permutation_group(); G_p
Permutation Group with generators [(1,2,3,4,5,6,7,8,9,10,11)(12,13,15,14,16)(17,18), (1,3,5,7,9,11,2,4,6,8,10)(12,14,13,16,15)(17,18)]

In this case it is true that the generators of G are in correspondence with the generators of G_p. Is this always the case? I mean, if I compute a very complicated group G with lots of generators and its corresponding permutation group G_p, then is it true that G.gens()[i] corresponds to G_p.gens[i]?

edit retag flag offensive close merge delete

Comments

I am not good on Algebra, exactly on representation theory, but In my understanding, that is always. It means all finite groups can be represented by matrix group. Perhaps, link below may be able to help you. Sorry if it does not help you. https://math.stackexchange.com/questions/85308/can-every-group-be-represented-by-a-group-of-matrices (link text)

dimahphone gravatar imagedimahphone ( 2018-02-05 05:13:32 +0200 )edit

1 Answer

Sort by » oldest newest most voted
0

answered 2018-02-06 23:56:22 +0200

dan_fulea gravatar image

The code under the method can be accessed, with the notations / objects from the posted question, as follows:

G.as_permutation_group??

After the doc string we get the following active lines of code:

    from sage.groups.perm_gps.permgroup import PermutationGroup
    if not self.is_finite():
        raise NotImplementedError("Group must be finite.")
    n = self.degree()
    MS = MatrixSpace(self.base_ring(), n, n)
    mats = [] # initializing list of mats by which the gens act on self
    for g in self.gens():
        p = MS(g.matrix())
        m = p.rows()
        mats.append(m)
    mats_str = str(gap([[list(r) for r in m] for m in mats]))
    gap.eval("iso:=IsomorphismPermGroup(Group("+mats_str+"))")
    if algorithm == "smaller":
        gap.eval("small:= SmallerDegreePermutationRepresentation( Image( iso ) );")
        C = gap("Image( small )")
    else:
        C = gap("Image( iso )")
    return PermutationGroup(gap_group=C, canonicalize=False)

(There is one more indent...)

Here, the group G from the example takes the role of self.

So let us see how the things work in this special case, we adapt the above code...

from sage.groups.perm_gps.permgroup import PermutationGroup

F  = GF(11)
m1 = matrix( F, [[1,2],[ 3,4]] )
m2 = matrix( F, [[1,3],[10,0]] )
G  = MatrixGroup( m1, m2 )

print "Is G finite? %s" % G.is_finite()
n  = G.degree()
print "Which is the degree of G? %s"    % n
print "Is G.base_ring() equal to F? %s" % bool( F == G.base_ring() )
MS = MatrixSpace(F, n, n)
mats = []    # initializing list of mats by which the gens act on self
print ( "Are the generators of G exactly m1, m2 (in this order)? %s"
        % bool( G.gens() == (m1, m2) ) )

for g in G.gens():
    p = MS( g.matrix() )
    m = p.rows()
    mats.append( m )

mats_str = str( gap( [ [ list(r) for r in m ]
                       for m in mats ] ) )

gap.eval( "iso:=IsomorphismPermGroup(Group("+mats_str+"))" )
#if algorithm == "smaller":
#    gap.eval("small:= SmallerDegreePermutationRepresentation( Image( iso ) );")
#    C = gap("Image( small )")
#else:
#    C = gap("Image( iso )")

C = gap( "Image( iso )" )
PG = PermutationGroup( gap_group=C, canonicalize=False )

print "mats_str is the following string:"
print mats_str

The results are as follows:

Is G finite? True
Which is the degree of G? 2
Is G.base_ring() equal to F? True
Are the generators of G exactly m1, m2 (in this order)? True
''
mats_str is the following string:
[ [ [ Z(11)^0, Z(11) ], [ Z(11)^8, Z(11)^2 ] ], 
  [ [ Z(11)^0, Z(11)^8 ], [ Z(11)^5, 0*Z(11) ] ] ]

So our group is finite. (Since it is a part of a finite matrix group, but sage can also detect / compute this.)

It is realizea as a group of $2\times 2$ matrices. (Degree is two.)

The base ring is $F$ the field with eleven elements.

Now a first point which is relevant for the question. The two given generators are taken exactly in this form, without refinament or sorting, so the matrices m1, m2 used explicitly in the construction of $G$ are exactly the generators of $G$.

Then the code translates the entries $1,2,3,4$ and $1,3,10,0$ into elements of the gap field representation for $F$, and still preserves accurately the order. The matrix string used in the final gap constructor is thus:

mats_str is the following string:
[ [ [ Z(11)^0, Z(11) ], [ Z(11)^8, Z(11)^2 ] ], 
  [ [ Z(11)^0, Z(11)^8 ], [ Z(11)^5, 0*Z(11) ] ] ]

Now we go through the gap tunnel, construct the object C,

sage: C
Group( [ ( 1, 2, 3, 4, 5, 6, 7, 8, 9,10,11)(12,13,15,14,16)(17,18), 
  ( 1, 3, 5, 7, 9,11, 2, 4, 6, 8,10)(12,14,13,16,15)(17,18) ] )

which collects two permutations with the corresponding same orders. (Here, we cannot distinguish between them, in the given example).

For instance, the order of m1 is $11\cdot 5\cdot 2$, the product of the (relatively prime) orders of the cycles involved in C.1.

sage: m1^( 11*5*2 )
[1 0]
[0 1]
sage: m1^( 5*2 )
[6 8]
[1 7]
sage: m1^( 11*2 )
[9 0]
[0 9]
sage: m1^( 11*5 )
[10  0]
[ 0 10]

So the constructions seems to respect at the two points of passage the generators and their order.

edit flag offensive delete link more

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

Question Tools

1 follower

Stats

Asked: 2018-02-01 20:20:47 +0200

Seen: 341 times

Last updated: Feb 06 '18