Galois group of irreducible polynomial $x^q - \theta$ over $K$, where $q$ prime and char$K \neq q$ 
Suppose $x^q - \theta$ is irreducible over $K$, where $q$ prime and char$K \neq q$. Let $\omega$ be a primitive $q$th root of unity, and let $[K(\omega):K] = j$. Show that the Galois group of $x^q - \theta$ is generated by elements $\sigma$ and $\tau$ satisfying
$\sigma^q = \tau^j = 1$, $\sigma^k \tau = \tau \sigma$,
where $\overline{k}$ is a generator of the multiplicative group $\mathbb{Z}_q^*$.

I've worked through this problem, except I'm getting $\overline{k}$ is a generator of $\Gamma(L:K(\alpha))$, which is isomorphic to a subgroup of the multiplicative group $\mathbb{Z}_q^*$, where $\alpha$ is a root of $x^q - \theta$ and $L = K(\alpha, \omega)$.
Could somebody tell me if the problem is correct as stated?
Right now I'm thinking it is correct, and I've made a mistake in obtaining $\tau$.
What I'm doing is starting with $i: K(\alpha) \to K(\alpha)$ and extending to $\phi: L \to L$ with $\phi(\omega) = \omega^k$, where $\overline{k}$ is a generator of $\mathbb{Z}_q^*$. Then $o(\phi) = q - 1$, and take $\tau = \phi^{(q - 1)/j}$.
Is this a valid approach or have I made a mistake already?
 A: Let us keep your notations $K(\omega)$ (of degree $j$ over $K$), $K(\alpha)$ (of degree $q$ over $K$), adding only $L=K(\omega,\alpha)$, which is the splitting field of $X^q - \theta$  over $K$, say with Galois group $G$. Because $q$ is a prime and $j <{q-1}$, obviously the degree over $K$ of $K(\alpha)\cap K(\omega)$ must be $1$ and the two extensions $K(\alpha),K(\omega)$are linearly disjoint. It follows (draw a parallelogram of extensions) that $G$ admits two visible cyclic subgroups: on the one hand, $G(L/K(\alpha))\cong G(K(\omega)/K)$, say generated by $\tau$, of order $j$; on the other hand, $G(L/K(\alpha))$, of prime degree $q=[K(\alpha):K]$, hence also cyclic, say generated by $\sigma$. Note that $G$ is not a direct product because $K(\alpha)/K$ is not Galois.
It remains only to check directly the relation $\sigma^{k} \tau=\tau \sigma$ by  computing the galois action on $\alpha$ and $\omega$. By definition: 1) $\sigma(\omega)=\omega$ and $\tau(\omega)=\omega^{k_\tau}$, where the map $\tau  \in G(L/K(\alpha))\to k_{\tau}\in(\mathbf Z/q)^* $ is a group homomorphism ; 2) analogously,  $\tau (\alpha)=\alpha$ and $\sigma (\alpha)$ is a conjugate of $\alpha$ over $K(\omega)$, e.g. $\omega^{h(\sigma)}\alpha$. The rest of the calculation is obvious.
