# cardinality of centralizer of an element

Suppose $$G$$ is finite group of order $$p^dn$$ where $$d$$ and $$n$$ are positive integers and $$p$$ is a prime that does not divide $$n$$. Show that $$G$$ contains an element of order $$p$$ such that the cardinality of its conjugacy class divides $$n$$

Cardinality of conjugacy class of an element divides the order of group. So it is sufficient to prove that for some element $$g$$ of order $$p$$, $$p^d\mid C(g)$$ where $$C(g)$$ is the centralizer.

Edit: I realized that the above attempt does not make use of (i thought it does make use of) the fact that $$p$$ does not divide $$n$$.

• You have to show, $|C(g)|\ge p^d$ Aug 6, 2020 at 16:23
• @Subhajit Yes. please tell how to proceed. Aug 6, 2020 at 16:28

Hints: sylow p-subgroup always has an element of order $$p$$.

Center of p-group is always non trivial, there is an element of order p which commutes all the elements of sylow p-subgroup.

Can you continue?