# Infinite groups with elements of finite order

1. Oct 4, 2011

### eddyski3

Can anyone think of an example of an infinite group that has elements with a finite order?

2. Oct 4, 2011

### Ben Niehoff

"Infinite", as in having an infinite number of elements? The circle group U(1) comes to mind, having all of the cyclic groups Z_n as subgroups.

3. Oct 4, 2011

### Bacle

How about (0,1) , in Z(+)Z/2 ?

4. Oct 5, 2011

### eddyski3

Yes, as in having an infinite number of elements. What if we wanted a group with order infinity that had a relatively small number of elements of finite order?

5. Oct 5, 2011

### Bacle

Ben:
I imagine eddy wants the elements to have finite order under the operation of the ambient group, not under the operation of the subgroups.

6. Oct 5, 2011

### Ben Niehoff

But in my example they are the same. For example, $e^{i \pi /4}$ has finite order in U(1), while $e^{i \pi / \sqrt{2}}$ has infinite order.

7. Oct 5, 2011

### eddyski3

How does the element 1 have finite order in the group you described Bacle?

8. Oct 5, 2011

### Ben Niehoff

Yet another construction is to take any finite group G and construct the group $H = G \times \mathbb{Z}$. Then H has elements of finite order given by (g, 0), and all elements of the form (g, z) for $z \neq 0$ have infinite order.

9. Oct 5, 2011

### eddyski3

Can you explain this further Ben? The way I see it, since G is finite, when you add two elements of H the result might not be in H.

10. Oct 5, 2011

### Ben Niehoff

11. Oct 5, 2011

### Bacle

Eddyski3:

The group Z(+)Z/2 is the direct sum of groups, which carries the operation of the first group in the first component, and the operation of the second group in the second component, so that:

(a,b)+(a',b'):= (a+'Za', b+'Z/2b') , so you add the first

components as if you were in Z, and the other components as if you were in Z/2,

so (0,1)+(0,1)=(0+0,1+1)=(0,0).

Ben:
You're right, I missed your point; any rational multiple of pi will have finite order.

Last edited: Oct 5, 2011