Exercise 14.2.5

Proof. (a) The map ${\phi }_i$ defined in (14.9) is

Let $\lambda =(\tau ;{\mu }_1,\dots ,{\mu }_p),{\lambda }^{\prime }=({\tau }^{\prime };{\mu }_1,\dots ,{\mu }_p)$ be elements of $G_i$. The definition of $G_i$ shows that $\lambda (R_i)={\lambda }^{\prime }(R_i)=R_i$, so that $\tau (i)={\tau }^{\prime }(i)=i$.

By (14.7) (see Exercise 1),

therefore, using ${\tau }^{\prime }(i)=i$,

thus ${\phi }_i$ is a group homomorphism.

Write $G_i^{\prime }={\phi }_i(G_i)\subset S_p$. We prove first that $G_i^{\prime }$ is transitive.

Take any $k$ and $l$ in $\{1,\dots ,p\}$. Since $G$ is transitive, there exists some $\lambda =(\tau ;{\mu }_1,...,{\mu }_k)\in G$ which sends $(i,j)$ on $(i,k)$:

Then $\tau (i)=i$, so that $\lambda \in G_i$ and ${\mu }_i={\phi }_i(\lambda )\in G_i^{\prime }$. Moreover ${\mu }_i(j)=k$. This proves that $G_i^{\prime }$ is a transitive subgroup of $S_p$.

Moreover, $G_i$ is a subgroup of the solvable group $G$, thus $G_i$ is solvable. Then $G_i^{\prime }={\phi }_i(G_i)$ is isomorphic to $G_i∕\ker <!--\; FUNCTION\; APPLICATION\; -->({\phi }_i)$, which is a quotient of a solvable group, thus $G_i^{\prime }$ is solvable. (b) As in the proof of Theorem 14.2.15, let $\sigma =(\tau ;{\mu }_1,\dots ,{\mu }_p)\in G$ be arbitrary, and fix $j$ between $1$ and $p$. By (14.10) with $i=\tau (j)$, $𝜃\in G_i^{\prime }={\phi }_i(G_i)$, thus there exists $\lambda =(\rho ;{\nu }_1,\dots ,{\nu }_p)\in G_i$ such that $𝜃={\phi }_i(\lambda )$, thus $𝜃={\nu }_i$ and $\rho (i)=i$.

Now consider the element $\gamma ={\sigma }^{-1}\mathit{\lambda \sigma }\in G$. Using (14.6) and (14.7), we obtain

If we write $\gamma =({\tau }^{-1}\mathit{\rho \tau };{\lambda }_1,\dots ,{\lambda }_p)$, we obtain

and at the index $j$, using $𝜃={\nu }_i={\nu }_{\tau (j)}$,

Since $i=\tau (j)$ and $\rho (i)=i$,

thus

□