 Solution

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.