List members of each subgroup of integers modulo n

asked 2014-02-10 14:31:48 +0100

Martin
1 ●1 ●2 ●2

Hi - I am new to Sage so apologies if this is a dumb question.

For a given n, I would like to print out the list of subgroups of the multiplicative group modulo n as a list of the members of the subgroup. Here's one of the things I have tried so far:

R = Zmod(20) grouplist = R.multiplicative_subgroups() for subgroup in grouplist: print subgroup (11, 17) (9, 11) (3,) (11,) (19,) (17,) (9,) ()

As you can see, this gives me the generators of the sub-groups, but what I'd like is the complete list of members of the subgroup for each subgroup. Something along these lines - where members() is the method I would like, but don't know if it exists!:

R = Zmod(20) grouplist = R.multiplicative_subgroups() for subgroup in grouplist: print subgroup.members() (1, 3, 7, 9, 11, 13, 17, 19) (1, 9, 11, 19) (1, 3, 7, 9) (1, 11) (1, 19) (1, 9, 13, 17) (1, 9) (1)

Thanks for any help, Martin

answered 2014-02-10 15:48:53 +0100

Luca
890 ●6 ●21

multiplicative_subgroups returns only lists of generators, rather than proper group objects. This is, in my opinion, unfortunate. You can work it around with a bit of code. This function returns the list of elements spanned by a list G of generators

def list_elts(G):
    exps = [xrange(g.multiplicative_order()) for g in G]
    return [prod(g^e for g, e in zip(G, exp)) for exp in CartesianProduct(*exps)]

So, you may call

sage: R = Zmod(20)
sage: map(list_elts, R.multiplicative_subgroups())
[[1, 17, 9, 13, 11, 7, 19, 3],
 [1, 11, 9, 19],
 [1, 3, 9, 7],
 [1, 11],
 [1, 19],
 [1, 17, 9, 13],
 [1, 9],
 [1]]

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Thank you - that did the trick!

Martin ( 2014-02-10 17:11:12 +0100 )edit

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