Exercise 7.2.2

Proof. $F\subset K\subset L$.

Let $\tau \in \mathrm{Gal}(L∕\mathit{\sigma K})$. Then $\tau :L\to L$ is an automorphism of $L$, and $\tau (\gamma )=\gamma$ for all $\gamma \in \mathit{\sigma K}$, thus $\tau (\sigma (\alpha ))=\sigma (\alpha )$ for all $\alpha \in K$.

Let $\lambda ={\sigma }^{-1}\mathit{\tau \sigma }\in \mathrm{Gal}(L∕F)$. For all $\alpha \in K$,

Thus $\lambda ={\sigma }^{-1}\mathit{\tau \sigma }\in \mathrm{Gal}(L∕K)$, so $\tau =\mathit{\sigma \lambda }{\sigma }^{-1}\in \sigma \mathrm{Gal}(L∕K){\sigma }^{-1}$:

As the converse inclusion is proved in section 7.2.A,

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