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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!

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    $\begingroup$ 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 $g$. $\endgroup$
    – anon
    Apr 23, 2021 at 19:44
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    $\begingroup$ 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 $[G:Z(G)]$. $\endgroup$
    – anon
    Apr 23, 2021 at 19:45
  • $\begingroup$ Shoot. My bad. I should be claiming a center of size 2. $\endgroup$
    – 1642920877
    Apr 23, 2021 at 19:46

1 Answer 1

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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.

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