1
$\begingroup$

I am preparing for the exam, and one of the questions on the study guide asks:

Let $G$ be a group and let $H$ be a subgroup of $G$. Define group operation $H \times G \rightarrow G$ by $(h,g)=hgh^{-1}$. This makes $G$ into $H-set$. This is called conjugation action. Find all orbits.

I have attempted solving the problem by figuring out if there are repetitive orbits so we can narrow down the list of all possible orbits, but since we don’t know anything about H or G and their specific structure, my attempt was not successful.

I am just starting in abstract algebra. We have just started talking about group acting on sets and have not proved much yet. Any help would be greatly appreciated.

$\endgroup$
  • $\begingroup$ I'm not quite sure what they want as a characterization of the orbits. Perhaps I'm missing something, but without more information, I don't see much concrete that can be said. $\endgroup$ – jgon Nov 25 '18 at 22:47
  • $\begingroup$ That was my issue too. The previous exercise asked to find all orbits for operation defined by $(h,g)=hg$. I have figured out that all orbits are just the set of right cosets of $H$ in $G$. But this time, I could not see anything much. $\endgroup$ – Renat Sergazinov Nov 25 '18 at 22:48
  • $\begingroup$ There is something similar to this on page 96 of this book: marinazahara22.files.wordpress.com/2013/10/… $\endgroup$ – Seth Nov 25 '18 at 22:52
  • $\begingroup$ @Seth While that, or more accurately an adaptation of Thm 2.11.1 on page 97 of the linked pdf (which is a specialization of orbit-stabilizer) would find the size of the orbits, I'm not sure anything in there actually characterizes the orbits in a way I would say answers the question: "Find all orbits." $\endgroup$ – jgon Nov 25 '18 at 22:56
1
$\begingroup$

All the orbits are the conjugacy classes of an element in $G$. You can divide elements in $G$ into two parts, those that are in the center $Z(G)$ of $G$, and those that are not in the center.

Let $x \in Z(G)$, then $x$ is fixed by conjugation by any elements in $G$. Hence when you look at the orbit-stabilizer formular, you have that $|G|=|Z(G)|+\sum_{x \in conjugacyclasses}{\frac{|G|}{|stab(x)|}}$. you can use this formula to show for example group of order $p^2$ is abelian as $p||G|$ and $p|\frac{|G|}{|stab(x)|}$ for all $x$ in the sum. so $p||Z(G)|$. Hence $G$ has no trivial center. And you see that $G \cong Z/p^2Z$ or $G \cong (Z/pZ)^2$ which are both abelian.

$\endgroup$
  • $\begingroup$ Not super helpful, you're hardly addressing the question. There's nothing in your answer about a subgroup $H$, nor about an actual characterization of the orbits. $\endgroup$ – jgon Nov 25 '18 at 22:52

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.