TITLE: Sizes of Conjugacy Classes
QUESTION [7 upvotes]: If G is a group of a certain order, say 30, what are the possible sizes of conjugacy classes?
I know that I need to use the Class equation, but I'm not quite sure how to do it.  Suggestions would be appreciated.

REPLY [3 votes]: Gerry asked about class equation ruling out the goofy class sizes 6 and 10.
G can only have elements and subgroups of order 1, 2, 3, 5, 6, 10, 15, 30.  Elements of order k have centralizers whose order is a multiple of k, and elements of order p with centralizers of order a multiple of pq have elements of order pq (Lagrange and basic properties of exponents).
Not 10:  If G has a class $g^G$ of size 10, then $C_G(g)$ has size 3, so g has order 3, and does not centralize any elements of order 2 or 5 or 10 and there are no elements of order 6 or 15 or 30.  The identity class has size 1.
If h has class size {1,2,5}, then $C_G(h)$ has order divisible by {15,15,10}, contradicting our statement about centralizers and orders, so the only possible sizes for non-identity elements are {3,6,10,15}.  Only 10 is not a multiple of 3, so we must have 1+10+10 + {3 stuff adding to 9}, that is 1+10+10+10+3+6 or 1+10+10+10+3+3+3.  The first is impossible, since a class size of 3 implies a centralizer of size 10, and so an element of order 10, so classes for elements of order 2, 5, and 10 all of size 3.  However, even the second is impossible, since the normalizer of the cyclic group of order 10 either has order 10 (and thus we have 4 classes of elements of order 10) or 30 (in which case we have at least 2 classes).
In other words, a class of size 10 implies an element of order 10, but the cyclic subgroup of order 10 has 4 elements of order 10, and the elements of order 3 in G will end up centralizing one of them.
Not 6: This is very similar but a little easier.  We have no elements of order 10 or 15, so other than 6s and the identity's 1, all the class sizes are multiples of 5.  Hence we have to have 1+6+6+6+6+{5 stuff adding to 5}, that is, 1+6+6+6+6+5.  However a class of size 5 implies an element of order 6, so at least 3 classes of size 5, a contradiction.
Specifically not 1+2+2+10+15: In this case we have a class of size 2, so an element of order 15, so elements of order 3, 5, and 15 all with centralizers of order 15, so the class equation begins 1+2+2+2, not just 1+2+2.