Exercise 4.1.8 (Doubly transitive action)

Proof. Note that $G$ is doubly transitive if and only if, for any $a,a^{\prime },b,b^{\prime }\in A$ such that

there is $\sigma \in G$ such that

Indeed, suppose that $G$ is doubly transitive on $A$ and assume that $a,a^{\prime },b,b^{\prime }\in A$ are such that $a\ne a^{\prime },b\ne b^{\prime }$. Then, $G$ being transitive, there is some $\tau \in G$ such that $\tau (a)=b$. Note that ${\tau }^{-1}(b^{\prime })\in A-\{a\}$, otherwise ${\tau }^{-1}(b^{\prime })=a$, so $b^{\prime }=\tau (a)=b$, which is false. Since $G_a$ acts transitively on $A-\{a\}$, and $a^{\prime },{\tau }^{-1}(b^{\prime })\in A-\{a\}$, there is some $\lambda \in G_a$ such that

Put $\sigma =\mathit{𝜏𝜆}$. Then

Conversely, suppose that for all $a,a^{\prime },b,b^{\prime }\in A$ such that $a\ne a^{\prime },b\ne b^{\prime }$, there is some $\sigma \in G$ such that $\sigma (a)=b$ and $\sigma (a^{\prime })=b^{\prime }$.

Let $a\in A$. We show that $G_a$ is transitive on the set $A-\{a\}$. Let $u,v\in A-\{a\}$. If we apply the assumed property to $(a,a^{\prime },b,b^{\prime })=(a,u,a,v)$, where $u\ne a,v\ne a$, there is some $\lambda \in G$ such that

So $\lambda \in G_a$ satisfies $\lambda (u)=v$. This shows that the subgroup $G_a$ is transitive on the set $A-\{a\}$, for any $a\in A$.

Suppose that $G_a$ is transitive on $A-\{a\}$ for some $a\in A$. Let $b\in A$, $b\ne a$, and let $u,v\in A-\{b\}$. The preceding argument shows that $G_b$ is transitive on $A-\{b\}$ for all $b\in A$. This explains the “(hence all)” of the statement.

(a)

Let $G_n$ be the stabilizer of $n$ in $G=S_n$. By Exercise 2.2.8, $$G_n=S_{[[1,n]]-\{n\}}\simeq S_{n-1}.$$

Since $S_{n-1}$ is a transitive group on $[[1,n-1]]$ (the transposition $(ab)$ maps $a$ on $b$), then $S_n$ is doubly transitive on $[[1,n]]$ by definition.

(Alternatively, if we use the second definition (1), for all $a,b,a^{\prime },b^{\prime }\in [[1,n]]$ such that $a\ne a^{\prime }$ and $b\ne b^{\prime }$, there are $(n-2)!$ permutations $\sigma \in S_n$ such that $\sigma (a)=b$ and $\sigma (a^{\prime })=b^{\prime }$, of the form $\sigma =\left(\begin{array}{ccc}\hfill a\hfill & \hfill a^{\prime }\hfill & \hfill \cdots \hfill \\ \hfill b\hfill & \hfill b^{\prime }\hfill & \hfill \cdots \hfill \end{array}\right).)$

(b)

First proof (my proof, in the spirit of Exercise 7).

Let $G$ be a doubly transitive group on $A$. Let $a$ be any element of $A$. By definition, $G_a$ acts transitively on $A-\{a\}$. We want to prove that $G_a$ is a maximal subgroup of $G$.

First $G_a\ne G$, since $G$ acts transitively on $A$ (we assume $|A|>1$).

Let $H$ a subgroup of $G$ such that

$$G_a<H\le G.$$

We will prove that $H=G$.

Let

$$B=\{\sigma (a)\mid \sigma \in H\}.$$

Since $G_a<H$, there is some $\sigma \in H-G_a$. Then $b=\sigma (a)\ne a$.

If $x\in A-\{a\}$, since $G_a$ acts transitively on $A-\{a\}$, there there exists some $\tau \in G_a$ such that $\tau (b)=x$. So

$$x=(\mathit{𝜎𝜏})(a),\sigma \in H-G_a,\tau \in H.$$

Since $\mathit{𝜎𝜏}\in H$, $x\in B$. Moreover $a\in B$ (for $\sigma =1$), so $A\subseteq B$, where $B\subseteq A$, so $B=A$:

$$A=\{\sigma (a)\mid \sigma \in H\}.$$

For any $\lambda \in G$, $\lambda (a)\in A$, thus $\lambda (a)=\sigma (a)$ for some $\sigma \in H$.

Therefore ${\sigma }^{-1}\lambda \in G_a\subset H$, so $\lambda \in \mathit{𝜎𝐻}=H$, since $\sigma \in H$. This shows $G\le H$, where $H\le G$, so

$$H=G.$$

Hence $G_a$ is a maximal subgroup of $G$. Since this is true for every $a\in A$, $G$ is primitive on $A$ by Exercise 7.

In conclusion, every doubly transitive group on $A$ is primitive on $A$.

Second proof (adapted from D. Cox’s demonstration in “Galois Theory” p. 430).

Suppose that $G$ is doubly transitive on $A$, but not primitive on $A$. Then there is a non trivial block $B$ for the action of $G$ on $A$. By Exercise 7, the distinct images ${\sigma }_1(B),{\sigma }_2(B),\dots ,{\sigma }_n(B)$ of $B$ under the elements of $G=\{{\sigma }_1,{\sigma }_2,\dots ,{\sigma }_n\}$ (where ${\sigma }_1=1$) form a partition of $A$. Let $R_1,R_2,\dots ,R_k$ (where $R_1={\sigma }_1(B)=B$) be the distinct classes of this partition. Since $|{\sigma }_i(B)|=|B|$ for all indices $i$, we obtain

$$1<|R_j|=|B|<|G|\text{for all }j\in [[1,k]].$$

(In particular, $k>1$, otherwise $B=G$ is not a non trivial block.)

Pick $a\ne b$ in $R_1$, and also pick $c$ in $R_2$ (this is possible, since $k>1$). Since $R_1\cap R_2=\varnothing$, $a\ne c$. By definition, $G_a$ acts transitively on $A-\{a\}$. Therefore we can find $\sigma \in G$ such that

$$\sigma (a)=a\text{and}\sigma (b)=c.$$

But $\sigma (R_1)=\sigma (B)$ is a class $R_j$:

$$\sigma (R_1)=R_j(j\in [[1,k]]).$$

Since $a=\sigma (a)\in \sigma (R_1)=R_j$, then $a\in R_1\cap R_j$, therefore $j=1$.

Since $c=\sigma (b)\in \sigma (R_1)=R_j$, then $c\in R_2\cap R_j$, therefore $j=2$.

The contradiction $j=1=2$ shows that there is no group $G$ doubly transitive on $A$, but not primitive on $A$.

In conclusion, every doubly transitive group on $A$ is primitive on $A$.

Since $D_8$ is not primitive as a permutation group on the four vertices of a square, $D_8$ is not doubly transitive on this set.

(As a confirmation, an isometry cannot map two consecutive vertices of the square on two opposite vertices, since the distances are not the same.) □