For a transitive permutation group $G,$ show that there is some $\sigma \in G$ such that $\sigma(a) \neq a$ for all $a \in A.$

Question

Here is a problem that I have been working on. I was able to prove part A, but am having problems with part B. Thanks!

Let $G$ be a permutation group acting on a finite set $A.$ If $g\in G,$ let $A^g$ be the set of fixed points of $G$ in $A,$ so $A^g = \{a \in A : ga = a\}.$ If $a\in A,$ let $G_a$ be the stabilizer of $a$ in $G;$ so $G_a = \{g\in G : ga=a\}.$

A) Show that $\sum_{g\in G}|A^g| = \sum_{a\in A} |G_a|.$

B) If $G$ be a transitive permutation group acting on a finite set $A$ and $|A| > 1,$ then show that there is some $\sigma \in G$ such that $\sigma(a) \neq a$ for all $a \in A.$

Answer 1

For $a \in A$ let $Ga:=\{ g \cdot x \mid g \in G \}$ denote the $G$-orbit of $a$ and $A / G$ denote the $G$-orbit space of $A$, i.e., the set of $G$-orbits of $A$. From the first part of the problem and the orbit-stabalizer theorem we can deduce Burnside's lemma: $$ \lvert A / G \rvert = \sum_{a \in A} \frac{1}{|Ga|} = \sum_{a \in A} \frac{|G_a|}{|G|} = \frac{1}{|G|} \sum_{g \in G} |A^g|. $$ If $G$ acts transitively on $A$, then the left-hand side of the displayed equation is $1$. Since the identity element $e \in G$ satisfies $|A^e| = |A| \geq 2$, there must some $g \in G$ such that $A^g = \varnothing$, as otherwise the right-most sum is strictly greater than $|G|$.

Answer 2

Start: If $G$ is transitive then you can compute $|G_a|$ in terms of $G$ and $A$ using orbit-stabilizer; now simplify the right-hand side of the formula in the first part and proceed.