Infinite groups with elements of finite order

[eddyski3]

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

 

[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.

 

[Bacle]

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

 

[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?

 

[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.

 

[Ben Niehoff]

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.

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.

 

[eddyski3]

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

 

[Ben Niehoff]

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?

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.

 

[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.

 

[Ben Niehoff]

Sounds like you're missing some basic definitions:

http://en.wikipedia.org/wiki/Direct_product_of_groups

 

[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.