# Show ZXZ/<1,1> is an infinite cyclic group.

1. Jan 11, 2009

### Daveyboy

1. The problem statement, all variables and given/known data

Show ZXZ/<1,1> is an infinite cyclic group.

2. Relevant equations

3. The attempt at a solution
<1,1> = {...(-1,-1), (0,0), (1,1),...}

implies ZXZ/<1,1> = {<1,0>+<1,1>, <0,1>+<1,1>} which is isomorphic to ZXZ.

But ZXZ is not cyclic, is my description of the factor group wrong , is it not isomorphic to ZXZ, or am I missing other points?

2. Jan 11, 2009

### tim_lou

what characterize an infinite cyclic group?

1. generated by a single element
2. infinite order

can you put an element canonically into a certain form?
so say you have (m, n), what can I write it as?

3. Jan 11, 2009

### Daveyboy

if (m,n) is an element of one of the left cosets of the factor group then,
if m=n (m,n) is in <1,1>. if m/=n then (m,n) = (m,n) + <1,1>. (because (0,0) is
generated by <1,1>). I'm not sure what this implies though.

4. Jan 11, 2009

### Daveyboy

so actually <1,0> + <1,1> generated all elements of ZXZ. Is it acceptable to say ZXZ/<1,1> is isomorphic to Z+0Z which is isomorphic to Z? (Which is generated by <1> or <-1>)

5. Jan 11, 2009

### tim_lou

<1,1> is just the identity. So, yeah.

6. Jan 11, 2009

### Daveyboy

... this is the additive group, (0,0) is the identity, and I'm using the notation <1,1> to imply that this is the cyclic generator. I hope that clears up any confusion.

I'm not sure if the conclusion I've come to is correct.

ZXZ/<1,1> = {Z+<1,1>}
Clearly Z is infinite and cyclic generated by <1> or <-1>.
So, ZXZ/<1,1> is a cyclic infinite group.

7. Jan 11, 2009

### tim_lou

well, (0,0)=(1,1) since you are modding out the diagonal set.
(1,0) is the generator (or using coset notations (1,0)+<(1,1)>). Your conclusion is definitely correct (as in post 4)

8. Jan 11, 2009

### Daveyboy

awesome thanks.