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For distinct or equal prime numbers $\ell ,\lambda$, the polynomial $X^{\ell }-\lambda$ is Eisenstein at $\lambda$. Its splitting field over $\mathbb{Q}({\zeta }_{\ell })$ has Galois group a subgroup of $C_{\ell }$, which is nontrivial because a degree-$\ell$ radical cannot belong to a degree-$(\ell -1)$ cyclotomic field. Thus the full Galois group has order $\ell (\ell -1)$. On roots indexed by $j\in {\mathbb{F}}_{\ell }$, all automorphisms act as $j\mapsto uj+b$. Equality of orders identifies the group with the full affine group over a finite field.

// Target: algebra.bigb

A nonidentity involution in the affine group over a finite field ${\mathbb{F}}_{\ell }$, for odd prime $\ell$, has linear coefficient $-1$: a translation has order $\ell$, while $u^2=1$ forces $u=\pm 1$. Its unique fixed point is $b/2$, and its remaining orbits have size two. This orbit count computes primes in an intermediate field as double cosets for an order-two decomposition group acting on the radical roots.

// Target: number-theory.bigb

Write $\alpha ={\lambda }^{1/5}$, $M=\mathbb{Q}(\alpha )$ and $E=\mathbb{Q}({\zeta }_5)$. The Eisenstein criterion at $\lambda$ gives $[M:\mathbb{Q}]=5$, whereas the prime cyclotomic field degree gives $[E:\mathbb{Q}]=4$. The splitting field $F=E(\alpha )$ is a Galois extension of $E$, and an automorphism fixing $E$ sends $\alpha$ to ${\zeta }_5^b\alpha$. Thus its Galois group embeds in $C_5$. It is nontrivial, since $M$ cannot be contained in the degree-four field $E$. Therefore $[F:E]=5$ and $[F:\mathbb{Q}]=20$.

Index the five roots by $j\in {\mathbb{F}}_5$, with root ${\zeta }_5^j\alpha$. Every automorphism has the form ${\zeta }_5\mapsto {\zeta }_5^u$, $\alpha \mapsto {\zeta }_5^b\alpha$ and acts on indices as $j\mapsto uj+b$, with $u\in {\mathbb{F}}_5^{\times }$. This is a faithful inclusion of $G=\mathrm{Gal}(F/\mathbb{Q})$ into the affine group over a finite field, and both groups have order $20$, so it is an isomorphism. This is the affine Galois group of a prime-radical splitting field.

Let $D$ be the specified decomposition group. Its nonidentity element is an involution. If its linear coefficient were $1$, it would be a nontrivial translation, which has order $5$. Hence it acts as $j\mapsto -j+b$. There is exactly one fixed root, at $j=b/2$, and the other four roots form two pairs. By involution orbits in an affine group of odd prime degree, $D$ has three orbits on the five roots. Those roots identify with $G/\mathrm{Gal}(F/M)$, so the double-coset formula gives three primes in $M$.

For $E$, the embeddings identify with $G/\mathrm{Gal}(F/E)\simeq {\mathbb{F}}_5^{\times }$. The image of $D$ is $\{1,-1\}$ and acts by multiplication, giving the two orbits $\{1,-1\}$ and $\{2,-2\}$. ThereforeThe orbit argument counts primes whether or not $p$ is ramified; cyclicity of the given order-two decomposition group does not require that it be generated by an unramified Frobenius element.

Let $\zeta$ be a primitive fifth root of unity. Over ${\mathbb{F}}_p$, the Frobenius automorphism acts by $\zeta \mapsto {\zeta }^p$, so the splitting field of $X^5-1$ has degree equal to the multiplicative order of $p$ modulo $5$. Its Galois group is generated by Frobenius and is cyclic. The possible orders are therefore $1$ for $p\equiv 1$, $4$ for $p\equiv 2,3$, and $2$ for $p\equiv 4(\mathrm{mod}5)$.

Put $\alpha =5^{1/5}$. The polynomial $X^5-5$ satisfies the Eisenstein criterion at $5$, so $[\mathbb{Q}(\alpha ):\mathbb{Q}]=5$. Also $[\mathbb{Q}(\zeta ):\mathbb{Q}]=\phi (5)=4$. The intersection of these two fields has degree dividing both $5$ and $4$, hence is $\mathbb{Q}$. Since the splitting field is $\mathbb{Q}(\alpha ,\zeta )$,

Label the five roots by $\alpha {\zeta }^i$, $i\in {\mathbb{F}}_5$. The Galois group acts through the affine group over a finite field:Its possible nontrivial cycle types are $5$, $1+2+2$, and $1+4$; in particular, no element has a three-cycle. By the Dedekind factorization cycle test, the reduction of $X^5-5$ modulo any prime $p\ne 5$ therefore has no irreducible cubic factor.

The multiplier of the Frobenius element is $a=p(\mathrm{mod}5)$. If $p\equiv 2$ or $3$, then $a$ has order four; every affine map with this multiplier has one fixed point and one four-cycle. Hence $X^5-5$ is a product of a linear factor and an irreducible quartic. If $p\equiv 1$, Frobenius is a translation $i\mapsto i+b$: it is the identity when $b=0$, giving complete splitting, and is a five-cycle when $b\ne 0$, giving an irreducible polynomial.

Takeand let $N=C_p\times \{1\}$. Takethe affine group over a finite field, and let $N^{\prime }$ be its translation subgroup. Both normal subgroups have order $p$, andThe groups are not isomorphic: $G$ is abelian, while $G^{\prime }$ is nonabelian.

For $p=3$, letThen $\mathrm{Gal}(L/\mathbb{Q})\cong C_6=G$. The fixed field of its subgroup $N$ of order three is its unique quadratic subfield,

Letthe splitting field of $X^3-2$. Its Galois group is $S_3\cong {\mathrm{AGL}}_1({\mathbb{F}}_3)=G^{\prime }$. The normal subgroup $N^{\prime }=A_3$ fixes the quadratic discriminant field, so

Now let $p>3$. Part (a), with $r=p(p-1)$, supplies a cyclic Galois extension with group $G$. For $G^{\prime }$, take the splitting fieldof $X^p-2$. The prime $2$ is unramified in $\mathbb{Q}({\zeta }_p)$, so at every prime ideal above $2$ its valuation is one. It cannot therefore be a $p$th power in the cyclotomic field. HenceThe automorphismsand, for $a\in {\mathbb{F}}_p^{\times }$,satisfyThey generate all $p(p-1)$ automorphisms, provingThis is the affine Galois group of the splitting field of x to the p minus two.

A group action is sharply two-transitive when exactly one element sends each ordered distinct pair to each other ordered distinct pair. The affine group over a finite field provides an example: solve $ax+b=u$ and $ay+b=v$ uniquely, giving $a=(v-u)/(y-x)$.