Exercise 4.3.26 (Fixed point free elements)

Proof. Let $a$ be any element of $A$. Since $|A|>1$, there is some $b\in A$ such that $b\ne a$.

If $G_a=G$, then for all $\sigma \in G$, $\sigma (a)=a$, so there is no $\sigma \in G$ such that $\sigma (a)=b$. This contradicts the hypothesis, which affirms that the action of $G$ on $A$ is transitive. Therefore, for all $a\in G$,

Assume for the sake of contradiction that there is no $\sigma \in G$ such that $\sigma (a)\ne a$ for all $a\in A$. Equivalently, for all $\sigma \in G$, there is some $a\in A$ such that $\sigma (a)=a$. In other words, if $\sigma \in G$, then $\sigma \in G_a$ for some $a\in A$, so

Pick some element $x\in A$. Since the action of $G$ on $A$ is transitive, there is some $\sigma \in G$ such that $\sigma (x)=a$. Then

indeed, for all $\tau \in G$,

Hence

(Indeed, if $\tau \in G$, then $\tau \in G_a$ for some $a\in A$, so $\tau \in \sigma G_x{\sigma }^{-1}$ for some $\sigma \in G$.)

Since $G_x$ is a proper subgroup of $G$, Exercise 24 shows that (1) is impossible: $G$ is not the union of the conjugates of any proper subgroup. This contradiction shows that there is some fixed point free $\sigma \in G$.

If $G$ is a transitive permutation group on a finite set $A$ with $|A|>1$, then

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