# how can I manipulate a multiplicative group of Zmod(n)

Hi,

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

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]


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

I did the following steps:

sage: f = G.gen()
sage: H = G.subgroup([f^2])
sage: list(H)
[1, f, f^2]


My problem is when I did

sage: Zn(f)
3


But here f needs to be 2. How can I solve this? Replace f by value?

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For those interested, there is a parallel discussion on that topic on sage-devel : https://groups.google.com/g/sage-deve...

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### Going through the group of units

This implementation suffers from the defect that

• elements of H have G as their parent
• but they display in terms of a generator of H

Here is a slightly convoluted way to work around this.

Define G and H:

sage: n = 7
sage: Zn = Zmod(n)
sage: G = Zn.unit_group()
sage: f = G.gen()
sage: H = G.subgroup([f^2])


List the elements of H (this displays using a different f):

sage: Hlist = list(H)
sage: Hlist
[1, f, f^2]


List the elements of H expressed in G:

sage: HlistG = list(prod((a^b for a, b in zip(H.gens(), h.list())), G.one()) for h in H)
sage: HlistG
[1, f^2, f^4]


Get their values in Zn:

sage: HlistZn = [h.value() for h in HlistG]
sage: HlistZn
[1, 2, 4]


I opened a ticket to make this happen more naturally:

The cyclic ring also has a method multiplicative_subgroups.

That method lists generating tuples for its multiplicative subgroups:

sage: n = 7
sage: Zn = Zmod(n)
sage: Sub = Zn.multiplicative_subgroups()
sage: Sub
((3,), (2,), (6,), ())


Sadly they do not give a hold on the subgroups as such.

sage: H = Sub
sage: H
(2,)
sage: parent(H)
<class 'tuple'>


Neither do the generators for these subgroups have these subgroups as parents.

Instead, they are simply elements of the initial cyclic ring.

sage: h = H
sage: parent(h)
Ring of integers modulo 7

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