# Sizes of Conjugacy Classes

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.

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Are you asking for general notions, or are you asking about a group of order 30 specifically? (The "say 30" seems to imply that this is a number picked more or less at random). The order of a conjugacy class must divide the order of the group, and except for the trivial group, it cannot be equal to the whole group. –  Arturo Magidin Dec 2 '11 at 17:30
I'm asking about the order 30 specifically. In my notes, we were told that the order of the conjugacy class doesn't have to divide the order of the group, but the order of centralizers has to divide the order of the group. Was I mistold that? –  user711 Dec 2 '11 at 17:35
For groups of size 30, the possible class sizes are 1, 2, 3, 5, 15, but of course one cannot have all of these sizes inside a single group. You can prove this using the class equation. Possible sizes within a single group of size 30 are {1}, {1,2,3}, {1,2,5}, and {1,2,15}. This seems to require a little more, but only sylow theorems really. –  Jack Schmidt Dec 2 '11 at 17:35
Thanks, Jack. Can you show how you obtained those sizes? –  user711 Dec 2 '11 at 17:42
@user711: If the class equation tells you that the order of a centralizer times the order of a conjugacy class gives the order of the group, doesn't that tell you that both factors on the left divide the order of the group? –  Marc van Leeuwen Dec 2 '11 at 17:53

The order of a conjugacy class must always divide the order of the group. This follows from a theorem sometimes called the "orbit-stabilizer" theorem: The size of a group = the size or an orbit $\times$ the size of a corresponding stabilizer.

Consider $G$ acting on itself by conjugation: $g \circ x = gxg^{-1}$

Pick some $x \in G$. Then $\mathrm{orbit}(x) = \{ g \circ x \;|\; g \in G\} = \{ gxg^{-1} \;|\; g \in G\}$ (this is the conjugacy class of $x$) and $\mathrm{stabilizer}(x) = \{ g \in G \;|\; g \circ x=x \} = \{ g \in G \;|\; gxg^{-1}=x \} = \{ g \in G \;|\; gx=xg \}$ (this is the centralizer of $x$).

So the size of a conjugacy class times the size of a corresponding centralizer is equal to the size of the group.

So for a group of order 30. The size of the conjugacy classes are limited to divisors of 30: 1, 2, 3, 5, 6, 10, 15, and 30. However, these classes also partition $G$, so their sizes must sum to the order of $G$. The identity is in a conjugacy class by itself, so you have to have a bunch of divisors which add up to 30 which include at least one 1 [this means we can't have a conjugacy class of which exhausts the whole group -- i.e. 30 is not an option]. Another limitation [on any class equation] is that the conjugacy classes of order 1 correspond to elements which commute with everything (i.e. elements of the center). So the 1's in the class equation need to add to a divisor of 30 as well (since the order of the center must divide the order of the group). So for example, $1+1+1+1+2+2+2+10+10=30$ is no good. Since the order of the center (a subgroup) would be $1+1+1+1=4$ which does not divide 30. That's about all the low hanging fruit.

Of course, more can be said. For example, the class equation cannot be $1+1+2+2+2+2+5+15=30$. If it was, the center of the group would have order 2 so that $G$ mod its center would have order 15. However, every group of order 15 is isomorphic to the cyclic group of order 15. This is a problem since if $G$ quotient its center is cyclic, then $G$ must be abelian so that the center is the whole group. Thus this particular equation is ruled out. Other than piecemeal tricks like this, the Sylow theorems tend to help you find lots of restrictions on the class equation.

[A note about orbit-stablizer theorem: If you let $G$ act on the left cosets of some subgroup $H$, then the orbit of $H$ is the set of all left cosets and the stabilizer is $H$ itself. In this case the orbit-stablizer theorem says the size of a subgropu times the number of cosets is the size of the group. That's Lagrange's thoerem. As you might imagine, since Lagrange's thorem is so important, a generalization of his theorem should show up quite often.]

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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.

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Thanks!${}{}{}$ –  Gerry Myerson Dec 3 '11 at 21:42