Galois group of $x^{15}-1$ Let $\zeta$, $\eta$, $\omega$ denote the primitive fifteenth, fifth, and cube roots of unity.
a) Describe all the automorphisms in $G=G(\mathbb Q (\zeta)/ \mathbb Q)$.
b) Show that $G$ is isomorphic to a direct product of two cyclic groups. Construct this isomorphism.
The group $|G|=|G(\mathbb Q (\zeta)/ \mathbb Q)|= [\mathbb Q (\zeta):\mathbb Q)]=U_{15}$
Now what is the generator of $\mathbb F_{15}^*$? We have to find the primitive root modulo 15. I am confused.
Please help. Thanks in advance.

When constructing an automorphism of $\mathbb{Q}(\zeta)$, the question is where $\zeta$ goes to. But because roots from $x^{15}-1$ should be interchanged, there are only $15$ possibilities. These are the possibilities:
$$f(\zeta)\in\{\zeta,\zeta^2,\zeta^4,\zeta^7,\zeta^8,\zeta^{11},\zeta^{13},\zeta^{14}\}$$
There are no other possibilities, because if for example $f(\zeta)=\zeta^3$, then $f(\zeta^5)=\zeta^{15}=1=f(1)$, but elements from $\mathbb{Q}$ should be fixed and the automorphism $f$ is invertible.
Now we want to construct an ismorphism between this group and $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/4\mathbb{Z}$. This is a natural choice, because there are no other possibilities to write a group with eight elements as a product of two cyclic groups.
A possible isomorphism from $G$ to $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/4\mathbb{Z}$ is the following:
\begin{align}
G&\rightarrow\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/4\mathbb{Z}\\
\zeta&\mapsto(0,0)\\
\zeta^2&\mapsto(0,1)\\
\zeta^4&\mapsto(0,2)\\
\zeta^7&\mapsto(1,3)\\
\zeta^8&\mapsto(0,3)\\
\zeta^{11}&\mapsto(1,2)\\
\zeta^{13}&\mapsto(1,1)\\
\zeta^{14}&\mapsto(1,0)
\end{align}
Because this group with eight elements is not cyclic (there does not exist an element with order eight), $(\mathbb{Z}/15\mathbb{Z})^\star$ does not have a primitive root.
