Galois Group is Group

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Theorem

Let $L / K$ be a normal extension.

Let $\Gal {L / K}$ be the Galois group of $L / K$.


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


Proof

The Galois group $\Gal {L / K}$ of $L / K$ is defined as:

$\Gal {L / K} = \left\{ {\sigma: L \to L: \sigma}\right.$ is an automorphism of $L$ such that $\sigma$ fixes $K$ point-wise$\left.\right\}$


We have that $\Gal {L / K}$ is a subset of the automorphism group on $L / K$.


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

Thus $\Gal {L / K}$ is not the empty set.


First, to show that $\Gal {L / K}$ is closed under composition of mappings.

Let $\sigma, \tau \in \Gal {L / K}$.

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

If $k \in K$, then:

$\map \rho k = \map \sigma {\map \tau k} = \map \sigma k = k$

and so $\rho$ fixes $K$.

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

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


Next, by Inverse of Algebraic Structure Isomorphism is Isomorphism, $\sigma^{-1}$ is an automorphism, and:

$\forall k \in K: \map \sigma k = k \implies \map {\sigma^{-1} } k = k$

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


The result follows from the Two-Step Subgroup Test.

That is, $\Gal {L / K}$ is a group.

$\blacksquare$