Let $G$ be a finite group and $x\in G$ be an element of order $p$ ($p$ prime). Suppose that $x\in P$, where $P$ is some $p$-Sylow subgroup of $G$.

I could not prove the following:

$x$ is not conjugate in $G$ to any other element of $P$ if and only if $x$ commutes with none of its conjugates in $G$ other than itself.

Is it easy? Does it depend on some group theory result?

  • $\begingroup$ Oh, dear! Those negations in a double-implication claim make things pretty hard to understand. Is this the original wording or you can write down the original one? Is this from some book, perhaps? $\endgroup$ – Timbuc Feb 5 '15 at 8:42
  • $\begingroup$ And the intended meaning must be "$x$ commutes with none of its conjugates in $G$ other than $x$ itself"! $\endgroup$ – Derek Holt Feb 5 '15 at 9:52

Here are some hints - more details on request.

Suppose first that $x \not\in Z(P)$. Then $x$ is conjugate to another element of $P$ (that's clear) and also $x$ must commute with a conjugate in $P$ other than $x$ - to see that, conjugate $x$ by an element of $H \setminus C_P(x)$, where $H$ is a subgroup with $C_P(x) \lhd H \le P$. So neither of your conditions holds in this case.

So we can assume that $x \in Z(P)$. If $x$ commutes with some conjugate $y \ne x$, then $\langle x,y \rangle$ is contained in some Sylow $p$-subgroup of $C_G(x)$, which we can then conjugate in $C_G(x)$ to $P$ to fins such a $y \in P$. Conversely, if $x$ is conjugate to some other element $y$ of $P$, then $y$ is a conjugate that it commutes wih.


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