2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.