Galois Group is Group

Theorem
Let $L/K$ be a normal extension and


 * $\operatorname{Gal}(L/K) = \{\sigma:L\mapsto L \mid \sigma$ is an automorphism which fixes $K$ point-wise$\}$.

Then $\operatorname{Gal}(L/K)$ forms a group under the operation of composition.

Proof
First, we show that $\operatorname{Gal}(L/K)$ is closed under composition.

Suppose $\sigma, \tau\in\operatorname{Gal}(L/K)$.

Let $\rho = \sigma\circ\tau$. If $k\in K$, then


 * $\rho(k) = \sigma(\tau(k)) = \sigma(k) = k$,

so $\rho$ fixes $K$.

By Composition of Isomorphisms, $\rho$ is an automorphism.

Thus, $\operatorname{Gal}(L/K)$ is closed under composition.

Second, if $\sigma\in \operatorname{Gal}(L/K)$, then by Inverse Isomorphism, $\sigma^{-1}$ is an automorphism and for each $k\in K$, $\sigma(k) = k$ implies that $\sigma^{-1}(k) = k$.

Thus, $\sigma^{-1}\in \operatorname{Gal}(L/K)$.

Finally, the identity map is always an automorphism and it obviously fixes $K$.

Therefore, $\operatorname{Gal}(L/K)$ is a group.