Isomorphism concepts,( example periods elliptic functions )

[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?

Many thanks in advance

 

[Orodruin]

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?

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

By the wording it seems to imply the fact that w1/w2∉Rw1/w2∉Rw_1/w_2 \notin R is significant for there to be an isomorphism to ZZZ xZZ Z, I don't at all understand why, can someone explain?

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

 

[binbagsss]

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, $w_2 = 2 w_1$?

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 don't think I am in a position to answer your second question until I can answer the first..

 

[Orodruin]

So the first case is wrong to?

What would a map to $Z$ x $Z$ look like? two integers multiplied together?

I don't think I am in a position to answer your second question until I can answer the first..

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

 

[binbagsss]

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

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?

 

[Orodruin]

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?

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 $w_1/w_2 \notin \mathbb Q$.

the map is not $nm$ ? no?

No, assuming that by $nm$ you mean to multiply $n$ and ##m#, it is an integer, not a 2-tuple of integers.

 

[binbagsss]

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 $w_1/w_2 \notin \mathbb Q$.No, assuming that by $nm$ you mean to multiply $n$ and ##m#, it is an integer, not a 2-tuple of integers.

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 $w_1/w_2 \notin \mathbb Q$.No, assuming that by $nm$ you mean to multiply $n$ and ##m#, it is an integer, not a 2-tuple of integers.

Okay do you take $(n,m)$, the operation is then addition ?

 

[Orodruin]

Okay do you take $(n,m)$, the operation is then addition ?

Yes.

 

[binbagsss]

Yes.

thank you for your help