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If $G$ acts transitively by permutations on a finite set $A$ with more than one element (i.e. $G$ is a transitive permutation subgroup of the symmetric group $S_A$). Why does $G$ necessarily contain an element which has no fixed points (i.e. $g$ such that $g \cdot a \neq a$ for any $a \in A$)?


The hint I have is to think about, given $a \in A$, what fraction of elements of $G$ fixes $a$. I'm not sure how to go about this hint...

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2 Answers 2

up vote 5 down vote accepted

Here's another solution, in case you haven't heard about Burnside's lemma:

Given $a \in A$, the subset $G_a \subseteq G$ which fixes $a$ (called the stabilizer of $a$ in $G$) forms a subgroup of $G$. Since $G$ acts transitively, each of these point stabilizers is conjugate, i.e. for each $b \in A$ we have $G_b = g^{-1}G_ag$ for some $g \in G$. If we suppose that every element of $G$ has a fixed point, then each $\sigma \in G$ is contained in some stabilizer $G_b$. Then $\sigma \in \bigcup_{g\in G} g^{-1}G_ag$ for all $\sigma \in G$. But this can't happen since $G$ cannot be the union of the conjugates of any proper subgroup. (See if you can prove this.)

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3  
Actually, Bruno's argument is a very nice proof of the fact that the conjugates of a proper subgroup do not cover the group :) –  Mariano Suárez-Alvarez Feb 6 '12 at 2:12
    
Thank you! I got it all! And I learned so many additional things with this exercise (such as that a finite $G$ is never the union of the conjugates of a proper subgroup -- that's interesting!). –  Rick Feb 6 '12 at 15:49

By Burnside's lemma, you have

$$\frac{1}{|G|}\sum_{g \in G}|\text{fix }g| = |A/G|=1$$

Since $1 \in G$ has $|A|>1$ fixed points, at least one of the terms in the sum must be $0$.

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