On conjugacy class size of an element and its order. Let $G$ be a finite group and $x\in G$. Also we denote the conjugacy class of $x$ in $G$ by $x^G$. I want to know if there is any relation between $|x|$ and $|x^G|$? Suggestions would be appreciated.
 A: Consider any dihedral group $G=D_{2n}$ of order $2n$ with $n$ even.  Then $Z(D_{2n})\cong\mathbb{Z}_2$, but the reflections are also elements of order two with conjugacy class of size $n/2$.  So on one hand there exists $x$ with
$$|x|=2, |x^G|=1,$$
and also a $y$ with
$$|y|=2, |y^G|=n/2.$$
So $|y^G|$ could be even or odd, and of arbitrarily large size, if we pick the right $n$.  So, no, there's not really any fully general relation you can assert, though in certain groups (such as these dihedral groups) you can enumerate various possibilities and then just assert whatever relations you want that are satisfied.
A: The following shows a nice relation if both order and cardinality of the conjugacy class are a $p$-power, $p$ a prime.
Proposition Let $G$ be a finite group, $x \in G$ and assume  $|x^G|$ is a $p$-power. Then $x \in P$, for all $P \in Syl_p(G)$ if and only if $|x|$ is a $p$-power.
Proof If $x \in ⋂_{P \in Syl_p(G)}P$, then $|x|$ is a $p$-power. So assume conversely that $|x|$ is a $p$-power. Then there must exist a $P \in Syl_p(G)$, such that $x \in P$. (Why? $\langle x \rangle$ is a $p$-subgroup and by Sylow Theory contained in some Sylow $p$-subgroup). Now $|x^G|=[G:C_G(x)]$, the index of the centralizer of $x$, is a $p$-power. This means that the subgroups $P$ and $C_G(x)$ have relatively prime indices. But, as is well-known, this means $G=PC_G(x)$. So if $g \in G$, then $g=s.c$, with $s \in P$ and $c \in C_G(x)$. And this yields $P^g=P^{sc}=P^c$. So the conjugates of $P$ are in fact the conjugates of $C_G(x)$ acting on $P$. But we had $x \in P$ and then $c^{−1}xc=x \in P^c$. This shows that $x \in P$ for all $P \in Syl_p(G)$.□
