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How to find (describe) all groups which have 3 conjugacy classes?

Thanks in advance!

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  • $\begingroup$ $|Z(G)| \leq 3$. So there is not any p-Group with $p \not= 3$. I'll try to obtain more. $\endgroup$ – Ivan Di Liberti Dec 26 '12 at 14:15
  • $\begingroup$ It's good to post the work you've done up until now, whenever you post a question. $\endgroup$ – rschwieb Dec 26 '12 at 14:33
  • $\begingroup$ I would guess that this is very difficult (probably impossible) for infinite groups. There are many infinite groups with just two conjugacy classes. Perhaps the exercise was just about finite groups? $\endgroup$ – Derek Holt Dec 26 '12 at 17:34
  • $\begingroup$ math.stackexchange.com/questions/52350/… $\endgroup$ – Ivan Di Liberti Dec 26 '12 at 21:03
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Well, the unit element is always a conjugacy class by itself, just as any element is in any abelian group, so you need other two classes...for example, the (cyclic) abelian group of order $\,3\,$, but also the permutation group $\,S_3\,$, which only has transpositions and $\,3-$cycles.

It could be now a nice exercise to show the above are the only finite groups, up to isomorphism, with three conjugacy classes.

About infinite groups I don't know: constructions like HNN show there can be very funny groups all the non-unit elements of which are conjugated, so this may require lots of care

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  • $\begingroup$ Are you sure that there is not any other? $\endgroup$ – Ivan Di Liberti Dec 26 '12 at 14:38
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    $\begingroup$ In fact I remembered once having done this, yet I wasn't completely sure. The following paper though supports this: link.springer.com/article/10.1007%2FBF02764723 $\endgroup$ – DonAntonio Dec 26 '12 at 18:30
  • $\begingroup$ I'm trying to understand why. I just discovered that if we call $C_1, C_2, C_3$ the conjugacy classes than $|G|= \frac{|C_2||C_3|}{|C_2||C_3|-(|C_2|+|C_3|)}$, calling $C_1$ the class of $e$. This seems to be possible just for 3 and 6 (Wolfram computing). $\endgroup$ – Ivan Di Liberti Dec 26 '12 at 19:37
  • $\begingroup$ I'd rather write the above as $$|G|=1+\frac{|G|}{|C_2|}+\frac{|G|}{|C_3|}\Longleftrightarrow 1=\frac{1}{|G|}+\frac{1}{|C_2|}+\frac{1}{|C_3|}$$ Observe that we can assume $\,|G|>3\,$ and non abelian, so it can't be that $\,|C_2|\geq 4\,$ and etc. Try to finish this. $\endgroup$ – DonAntonio Dec 27 '12 at 0:40

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