Exercise 12.1.14

Proof.

(a)

Write $\rho =(12\dots n)\in S_n$ and $H=⟨\rho ⟩$. Let $\mathit{\tau H}$ be any coset relative to $H$, with $\tau \in S_n$. We must prove that there exists a unique $\sigma \in \mathit{\tau H}$ such that $\sigma (1)=1$.

$\text{∙}$

Existence. Let $k={\tau }^{-1}(1)$ and $\sigma =\tau {\rho }^{k-1}$. Then $\sigma \in \mathit{\tau H}$, and $$\sigma (1)=(\tau {\rho }^{k-1})(1)=\tau (k)=1.$$

$\text{∙}$

Unicity. If $\mathit{\sigma H}={\sigma }^{\prime }H$, with $\sigma (1)={\sigma }^{\prime }(1)=1$, then ${\sigma }^{\prime }\in \mathit{\sigma H}$, so $${\sigma }^{\prime }=\sigma {\rho }^l,l\in \mathbb{Z}.$$

Since ${\sigma }^{\prime }(1)=1$, we have $\sigma ({\rho }^l(1))=1=\sigma (1)$ and $\sigma$ is one-to-one, so ${\rho }^l(1)=1$, therefore $l\equiv 0(\mathit{mod}n)$, so ${\rho }^l=e$ and $\sigma ={\sigma }^{\prime }$.

(b)

As $H=⟨\rho ⟩$ is the stabilizer of ${𝜃}_i={\alpha }_i^n$, the value of $\tau \cdot {𝜃}_i$ are the all the same when $\tau$ is in $\mathit{\sigma H}$, where $\sigma$ is the unique representative of the coset $\mathit{\tau H}$ such that $\sigma (1)=1$. We obtain the elements of the orbit ${\mathcal{}O}_{{𝜃}_i}$ under the action of $S_n$, by taking the value of $\sigma \cdot {𝜃}_i$ with $\sigma (1)=1$. $${\mathcal{}O}_{{𝜃}_i}=\{\sigma \cdot {𝜃}_i|\sigma \in S_n,\sigma (1)=1\}.$$

Moreover these values are distinct. Indeed, if $\sigma \cdot {𝜃}_i={\sigma }^{\prime }\cdot {𝜃}_i$, where $\sigma (1)={\sigma }^{\prime }(1)=1$, then ${\sigma {}^{\prime }}^{-1}\sigma \in H$, so $\mathit{\sigma H}={\sigma }^{\prime }H$. By part (a) (unicity), we obtain $\sigma ={\sigma }^{\prime }$. (Thus $|{\mathcal{}O}_{{𝜃}_i}|=(n-1)!$ is the degree of the Lagrange resolvent.)

So the resolvent is the product

$$R(x)=\underset{\sigma \in S_n,\sigma (1)=1}{\prod }(x-\sigma \cdot {\alpha }_i^n).$$

As Lagrange says, the roots of the resolvent of ${𝜃}_i={\alpha }_i^n$ come from the permutations of the $n-1$ roots $x_2,\dots ,x_n$ that ignore the root $x_1$.