= Solution
Take
$$
G=C_p\times C_{p-1}\cong C_{p(p-1)}
$$
and let $N=C_p\times\{1\}$. Take
$$
G'=\operatorname{AGL}_1(\mathbb F_p)
=\mathbb F_p\rtimes\mathbb F_p^\times,
$$
the <affine group over a finite field>, and let $N'$ be its translation subgroup. Both normal subgroups have order $p$, and
$$
G/N\cong C_{p-1}\cong G'/N'.
$$
The groups are not isomorphic: $G$ is abelian, while $G'$ is nonabelian.
For $p=3$, let
$$
L=\mathbb Q(\zeta_7).
$$
Then $\operatorname{Gal}(L/\mathbb Q)\cong C_6=G$. The fixed field of its subgroup $N$ of order three is its unique quadratic subfield,
$$
\boxed{L^N=\mathbb Q(\sqrt{-7})}.
$$
Let
$$
L'=\mathbb Q(\sqrt[3]{2},\zeta_3),
$$
the <splitting field> of $X^3-2$. Its Galois group is $S_3\cong\operatorname{AGL}_1(\mathbb F_3)=G'$. The normal subgroup $N'=A_3$ fixes the quadratic discriminant field, so
$$
\boxed{(L')^{N'}=\mathbb Q(\sqrt{-3})=\mathbb Q(\zeta_3)}.
$$
Now let $p>3$. Part (a), with $r=p(p-1)$, supplies a cyclic Galois extension with group $G$. For $G'$, take the splitting field
$$
L'=\mathbb Q(\zeta_p,2^{1/p})
$$
of $X^p-2$. 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. Hence
$$
[L':\mathbb Q(\zeta_p)]=p,
\qquad
[L':\mathbb Q]=p(p-1).
$$
The automorphisms
$$
\sigma(2^{1/p})=\zeta_p2^{1/p},
\quad \sigma(\zeta_p)=\zeta_p,
$$
and, for $a\in\mathbb F_p^\times$,
$$
\tau_a(2^{1/p})=2^{1/p},
\quad \tau_a(\zeta_p)=\zeta_p^a
$$
satisfy
$$
\tau_a\sigma\tau_a^{-1}=\sigma^a.
$$
They generate all $p(p-1)$ automorphisms, proving
$$
\boxed{
\operatorname{Gal}(L'/\mathbb Q)
\cong C_p\rtimes\mathbb F_p^\times
\cong\operatorname{AGL}_1(\mathbb F_p)=G'
}.
$$
This is the <affine Galois group of the splitting field of x to the p minus two>.
Back to article page
