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could you please help me with this statement. First of all, is this statement true? If an action of a finite group G on a finite set X with the number of elements in X strictly greater than 1, has a unique orbit, then G contains an element with no fixed points. I've tried to apply the Orbit Stabilizer Theorem, which is saying that the number of orbits of a Group acting on X equal is to 1/ (G) multiplied by the sommation of the number of fixed points by every element in the group G. Well if there only exists one orbit, then all the element g of the group are conjugated 'cause orbits are equivalence classes (conjugacy classes as well?). So x, and y are lying in the same orbit if and only if there exists an element in G such that g(x)=y... Now I'm stuck.

Thank you

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  • $\begingroup$ Actually you are citing Burnside's lemma, not orbit-stabilizer. $\endgroup$
    – anon
    Nov 4, 2013 at 1:20
  • $\begingroup$ Yes it's the "Counting Theorem" right? $\endgroup$
    – Heinz1
    Nov 4, 2013 at 1:24
  • $\begingroup$ What do you mean by "all elements g of the group are conjugated"? Do you mean all elements of $G$ are conjugate? This is both false and irrelevant. (What is true is that all of the point-stabilizers are conjugate, but again this is irrelevant.) Additionally, please know that orbits are subsets of $X$, and conjugacy classes are subsets of $G$. $\endgroup$
    – anon
    Nov 4, 2013 at 1:47
  • $\begingroup$ Does this answer your question? Intuition on the Orbit-Stabilizer Theorem $\endgroup$
    – Babu
    Mar 11, 2022 at 16:10

2 Answers 2

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Burnside's lemma states

$$|X/G|=\frac{1}{|G|}\sum_{g\in G}|X^g|.$$

If $X$ is precisely one orbit, then $|X/G|=1$ and this reads $|G|=\sum\limits_{g\in G}|X^g|$.

Observe that if any one of the summands $|X^g|$ is bigger than $1$, then some other summand has to be $0$ (lest the sum be $>|G|$). Consider $|X^e|$ where $e\in G$ is the identity (note $|X|>1$).

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  • $\begingroup$ Could you explain why one of the other summand must be equal to zero? the set of fixed points of an element g in G is a subset of X. I don't understand the relation with the number of elements in G. Thank youl $\endgroup$
    – Heinz1
    Nov 4, 2013 at 1:31
  • $\begingroup$ if g and h of G are in lying in the same equivalence class of G, then/ X^g / = /X^h/ right?. So if there only exists one orbit. All the elements in G have the same number of fixed points... $\endgroup$
    – Heinz1
    Nov 4, 2013 at 1:34
  • $\begingroup$ @Heinz1 What do you mean by "equivalence class of $G$"? If the sum of $n$ nonnegative numbers is $n$ and one of those numbers is bigger than $1$, then at least one of the numbers must be $0$. Apply this reasoning with $n=|G|$. You don't need to think group-theoretically at all about this: just think numerically about the formula given by Burnside. Please think about my answer. $\endgroup$
    – anon
    Nov 4, 2013 at 1:44
  • $\begingroup$ Yes I understand that. I was just confused about the number of fixed points of two different elements lying in the same orbit. Thank you. $\endgroup$
    – Heinz1
    Nov 4, 2013 at 2:17
  • $\begingroup$ @Heinz1 What does "fixed points of two different elements lying in the same orbit" even mean? You can talk about fixed points of group elements, but elements of the group do not lie in orbits at all. You need to make sure you know what all of the words here mean. Know what a group is, know what a group action is, know what an orbit is, know what a fixed point is. And if you really do understand what I've said, then you know the answer to your question. $\endgroup$
    – anon
    Nov 4, 2013 at 4:41
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Let every element $g\in G$ fix some point, namely $\left|\operatorname{Fix}(g)\right|\ge1$ for every $g\in G$. Therefore: \begin{alignat}{1} \sum_{g\in G}\left|\operatorname{Fix}(g)\right| &= \left|\operatorname{Fix}(e)\right|+\sum_{g\in G\setminus\{e\}}\left|\operatorname{Fix}(g)\right| \\ &\ge |X|+|G|-1 \\ \tag1 \end{alignat} and hence: \begin{alignat}{1} 1&=\#\text{ of orbits} \\ &=\frac{1}{|G|}\sum_{g\in G}\left|\operatorname{Fix}(g)\right| \\ &\stackrel{(1)}{\ge} 1+\frac{|X|-1}{|G|} \\ &\stackrel{|X|>1}{>}1 \end{alignat} Contradiction. So, there is some $\tilde g\in G$ such that $\left|\operatorname{Fix}(\tilde g)\right|=0$, i.e. $\tilde g$ has no fixed point.

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