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## Main Question or Discussion Point

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

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

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Ben Niehoff

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

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

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Ben Niehoff

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

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

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

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Ben Niehoff

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

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Ben Niehoff

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

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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+'_{Z}a', b+'_{Z/2}b') , 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.

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+'

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.

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