# Find cardinalities of all orbits

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 : $Orb(x) = \{gxg^{-1}|g \in G \}$.

• yes except that $G.g = \{c_h(g) \mid h \in G\}$ – mercio Oct 23 '16 at 23:45
• 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. – DonAntonio Oct 23 '16 at 23:46

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 $G\cdot g=\{hgh^{-1}:h\in G\}$.

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 $G$)

• I want to ask you a stupid question. What does B.O. mean? – False Promise Oct 24 '16 at 1:01
• Its not a stupid question. It stands for "binary operation". – Kevin Bowman Oct 24 '16 at 1:02
• So, in this particular case Stab$(x)$ can be denoted as $Z(G)$? – False Promise Oct 24 '16 at 1:24
• 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 $x$). – Kevin Bowman Oct 24 '16 at 1:58
• Glad to have helped – Kevin Bowman Oct 24 '16 at 2:01