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Old November 4th, 2009, 05:23 PM
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Default conjugacy classes

Sorry for all the recent posts, I don't want to throw all my homework questions into one big post, so instead I'm breaking them down. If this bothers anyone let me know though.

Anyways, another problem was to find the conjugacy classes of the infinite dihedral group. For those who need memory refreshing, the infinite dihedral group is: /{s,t|s^2=e, ts=st^{-1}/} with e being the identity element.

I arrived at these three conjugacy classes:
\{t^a, t^{-a}\}\{e\}\{s, t^{2a}s, t^{-2a}s\}
Is there anything I missed? And if I did miss it, can you give me a hint as to how you found it?
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Old November 4th, 2009, 06:56 PM
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You are right in so much as the elements of your conjugacy classes are certainly conjugates, but there are surely more than three conjugacy classes - you have really exhibited three classes of conjugacy classes. For each a!=0, the elements of t^a and t^-a form a separate class I think. I was wondering only the other day about whether any infinite groups have a fine number of conjugacy classes. It seems it is possible for this to be true for certain finitely generated groups, but I don't think it is possible for a finitely presented group like the infinite dihedral group.
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Old November 4th, 2009, 09:50 PM
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Quote:
Originally Posted by alunw View Post
You are right in so much as the elements of your conjugacy classes are certainly conjugates, but there are surely more than three conjugacy classes - you have really exhibited three classes of conjugacy classes. For each a!=0, the elements of t^a and t^-a form a separate class I think. I was wondering only the other day about whether any infinite groups have a fine number of conjugacy classes. It seems it is possible for this to be true for certain finitely generated groups, but I don't think it is possible for a finitely presented group like the infinite dihedral group.

You may want to read (in Rotman's or Robinson's books) about HNN extensions: there are infinite groups with only two conjugacy classes! Of course, this means that ALL the non-trivial elements of group are conjugated, and the group is then trivially simple.

Tonio
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Old November 5th, 2009, 03:13 AM
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I'm aware of that result - I was reading about it only the other day in this now rather old archive http://www.math.niu.edu/~rusin/known...95/finite.conj , though I don't fully understand it. From that I gathered that the infinite groups with finitely many conjugacy classes known so far are not finitely presented, though finitely generated examples are known. I would be very interested to learn of any known example of an infinite and finitely presented group with only finitely many conjugacy classes.
I have Rotman's book but I haven't yet studied the later chapters properly.
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