### 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:

### Direct access to generating sets for subgroups

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[1]
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[0]
sage: parent(h)
Ring of integers modulo 7
```

Welcome to Ask Sage! Thank you for your question.

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