1
$\begingroup$

Given group $G$, $|G| = p^m$, $p$ is prime. Group $G$ acts on itself by conjugation. Prove that all orbits contain either one element or $p^k$ elements.

I don't really understand what it means "acts on itself by conjugation". From the definition it is like we have a group action and instead of $S$ we pick $G$, forgetting about its group operation, then we map $c_g : G \rightarrow G$, so $c_g(x) = gxg^{-1}$. And in this case, i think i need to rewrite the definition of orbit and stabilizer, something like that : $Orb(x) = \{gxg^{-1}|g \in G \}$.

$\endgroup$
  • $\begingroup$ yes except that $G.g = \{c_h(g) \mid h \in G\}$ $\endgroup$ – mercio Oct 23 '16 at 23:45
  • $\begingroup$ This much is true for any finite group acting on any set in any way: the orbits' size is always a divisor of the group's order, as this size is some subgroup's (the stabilizer's) index in the group. $\endgroup$ – DonAntonio Oct 23 '16 at 23:46
2
$\begingroup$

Hint: Orbit stabilizer theorem.

For G to act in itself by conjugation means that $g\cdot h=ghg^{-1}$, so the orbit has the form $G\cdot g=\{hgh^{-1}:h\in G\}$.

$\endgroup$
1
$\begingroup$

To be precise, given a group $G$ and a set $T$ then a group action is a mapping from $G\times T \rightarrow T$ that satisfies the two group action axioms.

So if we take $T$ to be $G$ and then map the order pair $(g,x)$ to $gxg^{-1}$ this map is a group action. In order to show this you do have to use the fact that $T$, which is $G$ in this case, has a B.O. (so you cannot just forget/not use the fact that $G$ is a group when "using it" as $T$). Also as you study this " conjugation" group action (for any group) you will find you still use the fact that "$T$" has a B.O. when looking at things like the orbits of elements.

$\mbox {Orb}(x) =\{gxg^{-1}: g\in G\}$ (also called the conjugacy class of $x$ in $G$) and $\mbox{ Stab(x) }=\{g\in G: gxg^{-1}=x \}$ (also called the centralizer of $x$ in $G$)

$\endgroup$
  • $\begingroup$ I want to ask you a stupid question. What does B.O. mean? $\endgroup$ – False Promise Oct 24 '16 at 1:01
  • 1
    $\begingroup$ Its not a stupid question. It stands for "binary operation". $\endgroup$ – Kevin Bowman Oct 24 '16 at 1:02
  • $\begingroup$ So, in this particular case Stab$(x)$ can be denoted as $Z(G)$? $\endgroup$ – False Promise Oct 24 '16 at 1:24
  • 1
    $\begingroup$ No. $Z(G)=\{z\in G: zg=gz \mbox { for all } g\in G\}$. The elements in the stabilizer of $x$ are all the elements of $G$ that commute with $x$, they may not necessarily commute this other elements of $G$. Elements in the centre of $G$ commute with every element of $G$ (which includes commuting with $x$). $\endgroup$ – Kevin Bowman Oct 24 '16 at 1:58
  • $\begingroup$ Glad to have helped $\endgroup$ – Kevin Bowman Oct 24 '16 at 2:01

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.