Inducting

As a corollary, this can be inducted finitely many times. Given Galois extensions ${\displaystyle K_{1},\ldots ,K_{n}/k}$ where ${\displaystyle K_{i+1}\cap (K_{1}\cdots K_{i})=k,}$ then there is an isomorphism of the corresponding Galois groups:

${\displaystyle \operatorname {Gal} (K_{1}\cdots K_{n}/k)\cong \operatorname {Gal} (K_{1}/k)\times \cdots \times \operatorname {Gal} (K_{n}/k).}$

Cardinality of the Galois group and the degree of the field extension

One of the basic propositions required for completely determining the Galois groups^[3] of a finite field extension is the following: Given a polynomial ${\displaystyle f(x)\in F[x]}$, let ${\displaystyle E/F}$ be its splitting field extension. Then the order of the Galois group is equal to the degree of the field extension; that is,

${\displaystyle \left|\operatorname {Gal} (E/F)\right|=[E:F]}$

Eisenstein's criterion

A useful tool for determining the Galois group of a polynomial comes from Eisenstein's criterion. If a polynomial ${\displaystyle f\in F[x]}$ factors into irreducible polynomials ${\displaystyle f=f_{1}\cdots f_{k}}$ the Galois group of ${\displaystyle f}$ can be determined using the Galois groups of each ${\displaystyle f_{i}}$ since the Galois group of ${\displaystyle f}$ contains each of the Galois groups of the ${\displaystyle f_{i}.}$

Quadratic extensions

The degree two field extension ${\displaystyle \mathbb {Q} ({\sqrt {2}})/\mathbb {Q} }$ has the Galois group ${\displaystyle \operatorname {Gal} (\mathbb {Q} ({\sqrt {2}})/\mathbb {Q} )}$ with two elements, the identity automorphism and the automorphism ${\displaystyle \sigma }$ which exchanges ${\displaystyle {\sqrt {2}}}$ and ${\displaystyle -{\sqrt {2}}}$. This example generalizes for a prime number ${\displaystyle p\in \mathbb {N} .}$

Product of quadratic extensions

Using the lattice structure of Galois groups, for non-equal prime numbers ${\displaystyle p_{1},\ldots ,p_{k}}$ the Galois group of ${\displaystyle \mathbb {Q} \left({\sqrt {p_{1}}},\ldots ,{\sqrt {p_{k}}}\right)/\mathbb {Q} }$ is

${\displaystyle \operatorname {Gal} \left(\mathbb {Q} ({\sqrt {p_{1}}},\ldots ,{\sqrt {p_{k}}})/\mathbb {Q} \right)\cong \operatorname {Gal} \left(\mathbb {Q} ({\sqrt {p_{1}}})/\mathbb {Q} \right)\times \cdots \times \operatorname {Gal} \left(\mathbb {Q} ({\sqrt {p_{k}}})/\mathbb {Q} \right)\cong (\mathbb {Z} /2\mathbb {Z} )^{k}}$

Cyclotomic extensions

Another useful class of examples comes from the splitting fields of cyclotomic polynomials. These are polynomials ${\displaystyle \Phi _{n}}$ defined as

${\displaystyle \Phi _{n}(x)=\prod _{\begin{matrix}1\leq k\leq n\\\gcd(k,n)=1\end{matrix}}\left(x-e^{\frac {2ik\pi }{n}}\right)}$

whose degree is ${\displaystyle \phi (n)}$, Euler's totient function at ${\displaystyle n}$. Then, the splitting field over ${\displaystyle \mathbb {Q} }$ is ${\displaystyle \mathbb {Q} (\zeta _{n})}$ and has automorphisms ${\displaystyle \sigma _{a}}$ sending ${\displaystyle \zeta _{n}\mapsto \zeta _{n}^{a}}$ for ${\displaystyle 1\leq a<n}$ relatively prime to ${\displaystyle n}$. Since the degree of the field is equal to the degree of the polynomial, these automorphisms generate the Galois group.^[5] If ${\displaystyle n=p_{1}^{a_{1}}\cdots p_{k}^{a_{k}},}$ then

${\displaystyle \operatorname {Gal} (\mathbb {Q} (\zeta _{n})/\mathbb {Q} )\cong \prod _{a_{i}}\operatorname {Gal} \left(\mathbb {Q} (\zeta _{p_{i}^{a_{i}}})/\mathbb {Q} \right)}$

If ${\displaystyle n}$ is a prime ${\displaystyle p}$, then a corollary of this is

${\displaystyle \operatorname {Gal} (\mathbb {Q} (\zeta _{p})/\mathbb {Q} )\cong \mathbb {Z} /(p-1)\mathbb {Z} }$

In fact, any finite abelian group can be found as the Galois group of some subfield of a cyclotomic field extension by the Kronecker–Weber theorem.

Finite fields

Another useful class of examples of Galois groups with finite abelian groups comes from finite fields. If q is a prime power, and if ${\displaystyle F=\mathbb {F} _{q}}$ and ${\displaystyle E=\mathbb {F} _{q^{n}}}$ denote the Galois fields of order ${\displaystyle q}$ and ${\displaystyle q^{n}}$ respectively, then ${\displaystyle \operatorname {Gal} (E/F)}$ is cyclic of order n and generated by the Frobenius homomorphism.

Degree 4 examples

The field extension ${\displaystyle \mathbb {Q} ({\sqrt {2}},{\sqrt {3}})/\mathbb {Q} }$ is an example of a degree ${\displaystyle 4}$ field extension.^[6] This has two automorphisms ${\displaystyle \sigma ,\tau }$ where ${\displaystyle \sigma ({\sqrt {2}})=-{\sqrt {2}}}$ and ${\displaystyle \tau ({\sqrt {3}})=-{\sqrt {3}}.}$ Since these two generators define a group of order ${\displaystyle 4}$, the Klein four-group, they determine the entire Galois group.^[3]

Another example is given from the splitting field ${\displaystyle E/\mathbb {Q} }$ of the polynomial

${\displaystyle f(x)=x^{4}+x^{3}+x^{2}+x+1}$

Note because ${\displaystyle (x-1)f(x)=x^{5}-1,}$ the roots of ${\displaystyle f(x)}$ are ${\displaystyle \exp \left({\tfrac {2k\pi i}{5}}\right).}$ There are automorphisms

${\displaystyle {\begin{cases}\sigma _{l}:E\to E\\\exp \left({\frac {2\pi i}{5}}\right)\mapsto \left(\exp \left({\frac {2\pi i}{5}}\right)\right)^{l}\end{cases}}}$

generating a group of order ${\displaystyle 4}$. Since ${\displaystyle \sigma _{2}}$ generates this group, the Galois group is isomorphic to ${\displaystyle \mathbb {Z} /4\mathbb {Z} }$.

Quaternion group

The Quaternion group can be found as the Galois group of a field extension of ${\displaystyle \mathbb {Q} }$. For example, the field extension

${\displaystyle \mathbb {Q} \left({\sqrt {2}},{\sqrt {3}},{\sqrt {(2+{\sqrt {2}})(3+{\sqrt {3}})}}\right)}$

has the prescribed Galois group.^[7]

Symmetric group of prime order

If ${\displaystyle f}$ is an irreducible polynomial of prime degree ${\displaystyle p}$ with rational coefficients and exactly two non-real roots, then the Galois group of ${\displaystyle f}$ is the full symmetric group ${\displaystyle S_{p}.}$^[2]

For example, ${\displaystyle f(x)=x^{5}-4x+2\in \mathbb {Q} [x]}$ is irreducible from Eisenstein's criterion. Plotting the graph of ${\displaystyle f}$ with graphing software or paper shows it has three real roots, hence two complex roots, showing its Galois group is ${\displaystyle S_{5}}$.