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This question already has an answer here:

why is there only finite number of (finite or infinite)groups with a fixed number of conjugacy classes? I know this is classical ,so plz give me a reference if you have. thank you

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marked as duplicate by Mikko Korhonen, Dietrich Burde, Ali Caglayan, Aditya Hase, UserX Nov 27 '14 at 17:58

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  • $\begingroup$ group is not necessarily finite in my case $\endgroup$ – joda Nov 27 '14 at 16:32
  • $\begingroup$ I am not sure that there are finitely many isomorphism types with a given finite number of conjugacy classes if you allow infinite groups. I think there are infinite groups of prime exponent $p$ with $p$ conjugacy classes, for example, and I think there are infinite groups with two conjugacy classes. $\endgroup$ – Geoff Robinson Nov 27 '14 at 16:43
  • $\begingroup$ are u sure? can u give me a reference plz? $\endgroup$ – joda Nov 27 '14 at 16:47
  • $\begingroup$ You can embed any torsion free group in a group with only two conjugacy classes. This is a standard application of HNN extensions. $\endgroup$ – Derek Holt Nov 27 '14 at 20:43
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In 1903 Edmund Landau proved that, for any positive integer $k$, there are only finitely many finite groups, up to isomorphism, with exactly $k$ conjugacy classes. I think the paper is to be found in the Math. Annalen 56, in German (Über die Klassenzahl der binären quadratischen Formen von negativer Discriminante). See also here.

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  • $\begingroup$ i cant read german but it seems it is in finite case at the end. has it proven for all groups in this? $\endgroup$ – joda Nov 27 '14 at 16:48
  • $\begingroup$ Yes, for finite groups. See the link in my answer: for infinite groups the situation is different. $\endgroup$ – Nicky Hekster Nov 27 '14 at 19:02

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