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.

Please give a hint. Please do not give solution. Thanks!

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 $n$.

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

1 Answer 1


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?


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