Exercise 14.3.19

Proof. (a) Suppose that $G$ is doubly transitive. If $j,k\in \{1,\dots ,n\}\setminus \{i\}$, as $i\ne j,i\ne k$, there exists $\sigma \in G$ such that $\sigma (i)=i,\sigma (j)=k$, so that $\sigma \in G_i$ and $\sigma (j)=k$. This proves that $G_i$ acts transitively on $\{1,\dots ,n\}\setminus \{i\}$.

Conversely, suppose that $G_i$ acts transitively on $\{1,\dots ,n\}\setminus \{i\}$ for some $i\in \{1,\dots ,n\}$. We first show that $G_{i^{\prime }}$ acts also transitively on $\{1,\dots ,n\}\setminus \{i^{\prime }\}$ for all $i^{\prime }\in \{1,\dots ,n\}$.

First, since $G$ is transitive, there is some $\tau \in G$ such that $\tau (i)=i^{\prime }$.

Let $j^{\prime },k^{\prime }\in \{1,\dots ,n\}\setminus \{i^{\prime }\}$, and define $j={\tau }^{-1}(j^{\prime }),k={\tau }^{-1}(k^{\prime })$. Then $j\ne i,k\ne i$, otherwise $j^{\prime }=i^{\prime }$ or $k^{\prime }=i^{\prime }$. Then the hypothesis gives $\sigma \in G_i$ such that $\sigma (j)=k$, that is $\sigma ({\tau }^{-1}(j^{\prime }))={\tau }^{-1}(k^{\prime })$.

Therefore

which shows that ${\sigma }^{\prime }=\mathit{\tau \sigma }{\tau }^{-1}\in G_{i^{\prime }}$ satisfies ${\sigma }^{\prime }(j^{\prime })=k^{\prime }$, thus $G_{i^{\prime }}$ acts transitively on $\{1,\dots ,n\}\setminus \{i^{\prime }\}$, and this is true for all $i^{\prime }$.

Now we prove that $G$ is doubly transitive. Let $(r,s),(r^{\prime },s^{\prime }),r\ne s,r^{\prime }\ne s^{\prime }$ be pairs of distinct elements of $\{1,\dots ,n\}$. Since $G_r$ acts transitively on $\{1,\dots ,n\}\setminus \{r\}$, and $r\ne s$, there exists ${\sigma }_1\in G_r$ such that ${\sigma }_1(r)=r,{\sigma }_1(s)=s^{\prime }$. Note that $r\ne s^{\prime }$, otherwise ${\sigma }_1(r)={\sigma }_1(s),r\ne s$, which is impossible since ${\sigma }_1$ is a permutation. Thus there exists ${\sigma }_2\in G_{s^{\prime }}$ such that ${\sigma }_2(r)=r^{\prime },{\sigma }_2(s^{\prime })=s^{\prime }$, where $r^{\prime }\ne s^{\prime }$. Then $\sigma ={\sigma }_2{\sigma }_1$ satisfies $\sigma (r)=r^{\prime },\sigma (s)=s^{\prime }$. This proves that $G$ is doubly transitive. (b) Suppose first that $G$ is $k$ transitive, and let $(i_1,\dots ,i_{k-1}),(j_1,\dots ,j_{k-1})$ be any $(k-1)$-tuples of distinct elements of $\{1,\dots ,n\}\setminus \{i\}$. Then $(i,i_1,\dots ,i_{k-1}),(i,j_1,\dots ,j_{k-1})$ are $k$-tuples of distinct elements of $\{1,\dots ,n\}$, thus there is some $\sigma \in G$ such that $\sigma \cdot (i,i_1,\dots ,i_{k-1})=(i,j_1,\dots ,j_{k-1})$, so that $\sigma \in G_i$ and $\sigma (i_1)=j_1,\dots ,\sigma (i_{k-1})=j_{k-1}$. This proves that $G_i$ acts $(k-1)$-transitively on $\{1,\dots ,n\}\setminus \{i\}$.

Conversely, suppose that $G_i$ acts $(k-1)$-transitively on $\{1,\dots ,n\}\setminus \{i\}$ for some $i\in \{1,\dots ,n\}$. As in part (a), we first show that $G_{i^{\prime }}$ acts $(k-1)$-transitively on $\{1,\dots ,n\}\setminus \{i^{\prime }\}$ for all $i^{\prime }\in \{1,\dots ,n\}$, using a permutation $\tau \in G$ satisfying $\tau (i)=i^{\prime }$. Take $(i_1^{\prime },\dots ,i_{k-1}^{\prime }),(j_1^{\prime },\dots ,j_{k-1}^{\prime })$ two $(k-1)$-tuples of distinct elements in $\{1,\dots ,n\}\setminus \{i^{\prime }\}$. We define $i_l={\tau }^{-1}(i_l^{\prime }),j_l={\tau }^{-1}(j_l^{\prime }),l=1,\dots ,k-1$. Since ${\tau }^{-1}$ is a permutation, $(i,i_1,\dots ,i_{k-1})$ and $(i,j_1,\dots ,j_{k-1})$ are $k$-tuples of distinct elements. The hypothesis gives $\sigma \in G_i$ such that $\sigma (i_l)=j_l,l=1,\dots ,k-1$. Then ${\sigma }^{\prime }=\mathit{\tau \sigma }{\tau }^{-1}$ satisfies

Therefore $G_{i^{\prime }}$ acts $(k-1)$-transitively on $\{1,\dots ,n\}\setminus \{i^{\prime }\}$, for all $i^{\prime }\in \{1,\dots ,n\}$.

Now let $(r_1,\dots ,r_k),(s_1,\dots ,s_k)$ be any $k$-tuples of distinct elements. There exists ${\sigma }_1\in G_{r_k}$ such that

where $s_1,\dots ,s_{k-1},r_k$ are distinct, since $r_k=s_i$ for some $i\ne k$ implies ${\sigma }_1^{-1}(r_k)={\sigma }_1^{-1}(s_i)$, that is $r_k=r_i$, which is false. Thus there exists ${\sigma }_2\in G_{s_1}$ such that

Then $\sigma ={\sigma }_2{\sigma }_1$ satisfies $\sigma \cdot (r_1,\dots ,r_k)=(s_1,\dots ,s_k)$. This proves that $G$ is $k$-transitive. □