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.
Can someone please give me some advices how to proceed? Thanks!
 A: 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.
