**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.1 Exercise 4.1.3**

Suppose $G \leq S_A$ is an abelian and transitive subgroup. Show that $\sigma(a) \neq a$ for all $\sigma \neq 1$ in $G$ and all $a \in A$. Deduce that $|G| = |A|$.

Solution: For each $a \in A$, we have $$1 = \bigcap_{\sigma \in G} \sigma \mathsf{stab}(a) \sigma^{-1} = \bigcap_{\sigma \in G} \sigma \sigma^{-1} \mathsf{stab}(a) = \bigcap_{\sigma \in G} \mathsf{stab}(a) = \mathsf{stab}(a)$$ since $G$ is abelian. So for all $a \in A$ and $\sigma \neq 1$, we have $\sigma(a) \neq a$.

Because the action of $G$ is transitive, if we fix $a \in A$ then for every $b \in A$ there exists $\sigma \in G$ such that $b = \sigma \cdot a$. Suppose now that $\sigma \cdot a = \tau \cdot a$; then $\tau^{-1} \sigma \cdot a = a$, so that $\tau^{-1}\sigma \in \mathsf{stab}(a) = 1$. Hence $\sigma = \tau$. Thus we can say, for a fixed $a \in A$, that for every $b \in A$ there exists a unique $\sigma \in G$ such that $\sigma \cdot a = b$. In this way we define an injective mapping $\varphi_a : A \rightarrow G$. Hence $|A| \leq |G|$.

Now with $a \in A$ fixed, we define $\psi_a : G \rightarrow A$ by $\psi_a(\sigma) = \sigma \cdot a$. This mapping is injective since if $\sigma \cdot a = \tau \cdot a$, we have $\tau^{-1} \sigma \in \mathsf{stab}(a) = 1$. Thus $|G| \leq |A|$.

Hence $|G| = |A|$.