# A Isomorphism concepts,( example periods elliptic functions )

1. Feb 4, 2017

### binbagsss

Hi,
I have the following:

Let $\Omega$ be a discrete subgroup of $C$, the complex plane.
If:
i) $\Omega = \{nw_1 | n \in Z\}$, then $\Omega$ is isomorphic to $Z$.
ii) $\Omega = \{nw_1 + mw_2 | m,n \in Z\}$ where $w_1/w_2 \notin R$ , then $\Omega$ is isomorphic to $Z$ x $Z$

So from what I understand isomorphic is a map that is one to one between two sets that preserves the binary relatione exisising between elements, that is $f(x*y)=f(x)*f(y)$ (1), where $*$ is the operation the map is isomorphic to. So to define a isomorphism you need to define:

- two sets
- the map between them
- the relevant operation which is preserved, defined by (1)

QUESTION 1)
So, my book doens't say which operation, is it addition, it also doesn't say which map - is the map to take the integer with the map $f = n$ in case i) and $f=n+m$ in case 2, under the operation addition it is then easy to show that (1) is obeyed in both cases?

QUESTION 2)
By the wording it seems to imply the fact that $w_1/w_2 \notin R$ is significant for there to be an isomorphism to $Z$ x$Z$, I don't at all understand why, can someone explain?

2. Feb 4, 2017

### Orodruin

Staff Emeritus
Your second case is not a map from $\Omega$ to $Z\times Z$ and as such cannot be an isomorphism.

What happens if you take, say, $w_2 = 2 w_1$?

3. Feb 4, 2017

### binbagsss

So the first case is wrong to?

What would a map to $Z$ x $Z$ look like? two integers multiplied together? do you take $mn$ instead? so the relevant operation is multiplication, not addition?

I dont think I am in a position to answer your second question until I can answer the first..

4. Feb 4, 2017

### Orodruin

Staff Emeritus
$Z \times Z$ is the set of 2-tuples of integers, not the set of integers.

5. Feb 4, 2017

### binbagsss

edit: if you take $w1=2w2$ there is ambiguity about what $n$ and $m$ are i.e- the map would no longer be one-to-one?

the map is not $nm$ ? no?

6. Feb 4, 2017

### Orodruin

Staff Emeritus
Right. Although, if I am not wrong (I may be, I have been up for quite some time), it would be sufficient to have $w_1/w_2 \notin \mathbb Q$. Of course, since $\mathbb Q \subset \mathbb R$, $w_1/w_2 \notin \mathbb R$ implies $w_1/w_2 \notin \mathbb Q$.

8. Feb 4, 2017

### Orodruin

Staff Emeritus
Yes.

9. Feb 4, 2017