Galois Group is Group

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

Let $\operatorname{Gal} \left({L / K}\right)$ be the Galois group of $L / K$.

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

Proof
By definition, the Galois group $\operatorname{Gal} \left({L / K}\right)$ of $L / K$ is the mathematical object defined as:
 * $\operatorname{Gal} \left({L / K}\right) = \left\{{\sigma: L \to L: \sigma}\right.$ is an automorphism of $L$ such that $\sigma$ fixes $K$ point-wise$\left.\right\}$

We have that $\operatorname{Gal} \left({L / K}\right)$ is a subset of the group of automorphisms on $L / K$.

We initially note that the Identity Mapping is Automorphism which trivially fixes $K$.

Thus $\operatorname{Gal} \left({L / K}\right)$ is not the empty set.


 * Step 1: We show that $\operatorname{Gal} \left({L / K}\right)$ is closed under composition of mappings.

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

Let $\rho = \sigma \circ \tau$.

If $k \in K$, then:
 * $\rho(k) = \sigma(\tau(k)) = \sigma(k) = k$

and so $\rho$ fixes $K$.

By Composite of Isomorphisms: Algebraic Structure, $\rho$ is an automorphism.

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


 * Step 2: Let $\sigma \in \operatorname{Gal} \left({L / K}\right)$.

Then by Inverse Isomorphism, $\sigma^{-1}$ is an automorphism, and:
 * $\forall k \in K: \sigma(k) = k \implies \sigma^{-1}(k) = k$

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

The result follows from the Two-Step Subgroup Test.

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