Exercise 14.3.13

Proof. (a) Take any ordered $n$-tuple $(a_1,\dots ,a_n)$ of distinct elements of $\{1,\dots n\}$. Then consider the map

Then $\{a_1,\dots ,a_n\}\subset \{1,\dots ,n\}$ and, since the $a_i$ are distinct, $|\{a_1,\dots ,a_n\}|=n$, so that $\{a_1,\dots ,a_n\}=\{1,\dots ,n\}$. This shows that $\sigma$ is surjective, and since the $a_i$ are distinct, $\sigma$ is injective, thus $\sigma$ is a permutation: $\sigma \in S_n$

( $\sigma =\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 2\hfill & \hfill \cdots \hfill & \hfill n\hfill \\ \hfill a_1\hfill & \hfill a_2\hfill & \hfill \cdots \hfill & \hfill a_n\hfill \end{array}\right)$ is the only permutation such that $\sigma (i)=a_i,i=1,\dots ,n$).

Since

the orbit of $(1,\dots ,n)$ is the whole set $P_n$ of ordered $n$-tuples of distinct elements of $\{1,\dots ,n\}$. This proves that there is only one orbit, and that $G$ acts transtitively on $P_n$, i.e. $S_n$ is $n$-transitive. (b) Take any ordered $(n-2)$-tuple $(a_1,\dots ,a_{n-2})$ of distinct elements of $\{1,\dots ,n\}$. Name $a,b$ the two remaining elements:

Consider the two permutations

Then $\sigma \cdot (1,\dots ,n-2)=(a_1,\dots ,a_{n-2})=\tau \cdot (1,\dots ,n-2)$.

But $\tau =(ab)\sigma$, thus one of the two permutations $\sigma ,\tau$ is in $A_n$, therefore $(a_1,\dots ,a_{n-2})$ is in the orbit $A_n\cdot (1,\dots ,n-2)$ of $(1,\dots ,n-2)\in P_{n-2}$. Therefore $A_n$ acts transitively on $P_{n-2}$. $A_n$ is $(n-2)$-transitive. □