# Relation between order of the group, its center, and its conjugacy class

I am not sure how to relate the conjugacy class of a group to the order of the group and its centralizer.

Say I have a group $$G$$ of order $$16$$ and a center $$Z(G)$$ of size $$2$$. Would there be a conjugacy class of size $$2$$ as well?

The first theorem that comes into my mind is that a group of order $$p^3$$ is either abelian or its center has size $$p$$. But this is quite different from what I want to prove.

• The notation $Z(G)$ refers to the center of the group itself which is not the same thing as a centralizer of a particular element. Rather, the centralizer of an element $g$ is denoted $C_G(g)$ (although admittedly some sources might use $Z_G(g)$). Do you understand the difference between center and centralizer? | A group acts on itself by conjugation. The conjugacy classes are the orbits, and the centralizers are the stabilizers. So by the orbit stabilizer theorem, the conjugacy class of an element $g$ has size equal to the index $[G:C_G(g)]$ where $C_G(g)$ is the centralizer of $g$.
– anon
Apr 23, 2021 at 19:44
• The only relation the center has to all of this is that $Z(G)\le C_G(g)$ for all $g$, so we can say any conjugacy class size $[G:C_G(g)]$ must be a divisor of $[G:Z(G)]$.
– anon
Apr 23, 2021 at 19:45
• Shoot. My bad. I should be claiming a center of size 2. Apr 23, 2021 at 19:46

Of course! Let $$G$$ act on itself by conjugation; that is, $$g\cdot x=gxg^{-1}$$. Then $$G$$ divides itself into orbits; each orbit is a conjugacy class. The center is the set of size-$$1$$ conjugacy classes; that is, the fixed points.
By the orbit-stabilizer relation, each orbit has size dividing the order of $$G$$. In this case, that there are (possibly) orbits of size $$1$$, $$2$$, $$4$$, $$8$$, and $$16$$. Let there be $$n_1$$-, $$n_2$$-, $$n_4$$-, $$n_8$$-, and $$n_{16}$$-many such orbits (respectively). Then we know that $$n_1=2$$, and seek to show $$n_2>0$$. Since $$G$$ is not one giant orbit, we must have $$n_{16}=0$$. Counting elements, $$16=|G|=n_1+2n_2+4n_4+8n_8$$ Substituting in $$n_1$$, we obtain
\begin{align} 2\cdot 7=14 &=2n_2+4n_4+8n_8\\ &=2(n_2+2n_4+4n_8). \end{align}
Now divide by two and take the result mod $$2$$. Then $$1\equiv n_2\pmod{2}$$ In particular, $$n_2$$ cannot be $$0$$, as desired.
Obviously, this argument can be generalized: if $$G$$ has order $$p^n$$ and a center of size $$p^k$$, then $$G$$ must have a conjugacy class of size $$p^k$$ as well.