Let $F$ be a field. Let $F^\times$ be the multiplicative group of $F$. Let $(a, b) \in F^\times \times F$. We denote by $\lambda(a, b)$ the permutation $x \rightarrow ax + b$ on $F$. Let $(a, b), (c, d) \in F^\times \times F$. Since $a(cx + d) + b = acx + ad + b$, $\lambda(a, b)\circ\lambda(c, d) = \lambda(ac, ad + b)$. Hence $\{\lambda(a, b) | (a, b) \in F^\times \times F\}$ is a permutation group on $F$. We call this group the affine general linear group of degree $1$ on $F$ and denote it by $AGL(1, F)$. Since $\lambda(a, b)(0) = b$ and $\lambda(a, b)(1) - b = a$, $(a, b)$ is uniquely determined by $\lambda(a, b)$. Hence $|AGL(1, F)| = |F^\times \times F|$.
Proposition Let $p, q$ be prime numbers which may or may not be equal. Let $a$ be an integer which is divisible by $q$, but not not divisible by $q^2$. $X^p - a$ is irreducible in $\mathbb{Q}[X]$ by the Eisenstein's criterion. Let $K$ be the splitting field of $X^p - a$ in $\mathbb{C}$. Then the Galois group $G$ of $K/\mathbb{Q}$ is isomorphic to $AGL(1, F)$, where $F = \mathbb{Z}/p\mathbb{Z}$.
Proof: Let $\zeta$ be a primitive $p$-th root of unity. Let $\alpha$ be a root of $X^p - a$ in $\mathbb{C}$. $\alpha,\alpha\zeta,\dots,\alpha\zeta^{p-1}$ are distinct roots of $X^p - a$. Hence $K = \mathbb{Q}(\alpha,\alpha\zeta,\dots,\alpha\zeta^{p-1})$. Since $\zeta = \alpha\zeta/\alpha \in K$, $K = \mathbb{Q}(\alpha, \zeta)$.
Let $\sigma \in G$. Since $\sigma(\zeta)^p = 1$, there exists an integer $b$ such that $\sigma(\zeta) = \zeta^b$. Clearly $b$ (mod $p$) is uniquely determined by $\sigma$ and $b$ (mod $p) \in F^\times$. On the other hand, there exists an integer $c$ such that $\sigma(\alpha) = \alpha\zeta^c$. Clearly $c$ (mod $p$) is uniquely determined by $\sigma$. We denote $\phi(\sigma) = \lambda(b$ mod $pćc$ mod $p) \in AGL(1, F)$. Hence we get a map $\phi\colon G \rightarrow AGL(1, F)$. Let $\sigma, \tau \in G$. Suppose $\phi(\sigma) = \lambda(b$ mod $p, c$ mod $p$) and $\phi(\tau) = \lambda(d$ mod $p, e$ mod $p$). Then $\sigma\tau(\zeta) = \zeta^{bd}$ and $\sigma\tau(\alpha) = \sigma(\alpha\zeta^e) = \alpha\zeta^c\zeta^{be} = \alpha\zeta^{be + c}$. Hence $\phi(\sigma\tau) = \phi(\sigma)\phi(\tau)$. Hence $\phi$ is a homomorphism. It is easy to see that $\phi$ is injective.
It remains to prove that $\phi$ is surjective. Since $[\mathbb{Q}(\zeta) \colon \mathbb{Q}] = p - 1$, $[K \colon \mathbb{Q}(\zeta)] > 1$. Let $H$ be the Galois group of $K/\mathbb{Q}(\zeta)$. Let $\tau \in H$. There exists an integer $c$ such that $\tau(\alpha) = \alpha\zeta^c$. We define a map $\psi\colon H \rightarrow F$ by $\psi(\tau) = c$ mod $p$. It is easy to see that $\psi$ is an injective homomorphism. Hence $|F| = p$ is divisible by $|H|$. Since $|H| > 1$, $|H| = p$, Hence $[K \colon \mathbb{Q}] = p(p-1)$. Hence $\phi$ is surjective. QED