Bug with subgroup method?

[Altario]

SageMath version 9.0, Release Date: 2020-01-01

Using Python 3.8.5.

Ubuntu 20.04 LTS

64bit

Hi,

I want to create the multiplicative group (Z/7Z)*={1,2,3,4,5,6}

I did these steps:

sage: n=7                                                                                            

sage: Zn=Zmod(n)                                                                                     

sage: G=Zn.unit_group() 

sage: list(G)

[1, f, f^2, f^3, f^4, f^5]                                                                   

sage: G.inject_variables()                                                                           

Defining f

Then I want to create the subgroups H generated by f^2=2 mod 7 which is {1,2,4}.

I did the following steps:

sage: H = G.subgroup([f^2])                                                                          

sage: list(H)                                                                                        

[1, f, f^2]

sage: Zn(f)                                                                                          

3

"There seems to be a bug in the subgroup method: H should consist of [1, f^2, f^4]."

 cf:

https://ask.sagemath.org/question/57703/how-can-i-manipulate-a-multiplicative-group-of-zmodn/

Is there a solution for this ? Or I miss something?

[David Roe]

I agree that this is a bug.  There are several issues, and I don't know if there's an easy fix.

1. The elements of the enumeration are produced by passing in exponent vectors in terms of the generators of H:

sage: H([1])

f

sage: H([1])^3

1

It's unfortunate that we choose the same letter to represent the generator of H as we do for the generator of G; this is just the default variable name for multiplicative groups.

2. The generators of H are elements of G, not of H, but they print in a reasonable way:

sage: H.0

f^2

sage: (H.0).parent()

Multiplicative Abelian group isomorphic to C6
sage: H.0.parent() is G                                                                                                                
True

I think that they should be elements of H, but that will mean they print incorrectly unless we update the element class.

3. There is no coercion from H to G:

sage: G.has_coerce_map_from(H)                                                                                                          
False

There should be.

I'm not going to be able to work on this soon, but am happy to help advise anyone who wants to.

David

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[Travis Scrimshaw]

+1 for having the ambient group having a coercion from a subgroup. One other bug is that

sage: H.gen(0).parent() is G

True

sage: H([2])
f^2
sage: _.parent()
Multiplicative Abelian subgroup isomorphic to C3 generated by {f^2}

Best,

Travis

[Samuel Lelievre]

Also discussed at Ask Sage:

- Ask Sage question 57703
  Multiplicative group of Zmod(n)
  https://ask.sagemath.org/question/57703

[Mickaël Hamdad]

And Is there a way to list H directly with numerical values ?

Le jeu. 24 juin 2021 à 21:48, Mickaël Hamdad <hamdad....@gmail.com> a écrit :

Thank you very much!!!!

Just what H.0 mean ?

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[Mickaël Hamdad]

Thank you very much!!!!

Just what H.0 mean ?

Le jeu. 24 juin 2021 à 10:21, Samuel Lelievre <samuel....@gmail.com> a écrit :

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[David Roe]

H.0 is Sage notation for the 1st generator of H.

sage: preparse("H.0")                                                                                                                  
'H.gen(0)'

As for getting the value, it depends on what kind of element you have.  

sage: H.0._exponents                                                                                                                    
(2,)
sage: H.0.value()                                                                                                                      
2
sage: H([1])._exponents                                                                                                                
(1,)
sage: H([1]).value()

Traceback (most recent call last):

...

AttributeError... 

Also remember that the exponent vectors depend on the context: H.0 is an element of G (inappropriately in my opinion) and is thus expressed in terms of a generator of G, while H([1]) is in terms of a generator H.

David

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[Mickaël Hamdad]

Thank you

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