Let's prove a general result:
Let $\Phi_n\in\mathbb{Z}[t]$ the nth-cyclotmic polynimoial, which is irreducible in rationals, and the Galois group $G$ of this polynomial over rationals. We claim that $G\cong\mathbb{Z}_{n}^*$.
Since the roots of $\Phi_n$ are the primitive nth-roots of the unity, lets say $\zeta^{m_1},\cdots,\zeta^{m_t}$, where $t=\phi(n)$ because there are by definition of phi-function $\phi(n)$ coprime numbers with $n$ less than it. So the Galois group $G$ is formed by the automorphisms $\tau_k$ with $1\leq k\leq t$ such as $\tau_k(\zeta)=\zeta^{m_k}$. So the homomorphism
$$\gamma\;:\; G\to \mathbb{Z}_n^*\;, \tau_k\to [m_k]_n$$
is an isomorphism. Since the splitting field of cyclotomical polynomial is $\mathbb{Q}(\zeta)$ for a $\zeta$ nth-primitve root of unity, and $ord(\mathbb{Z}_n^*)=\phi(n)$, we get the result desired.
To see that is an isomorphism:
-Well defined since $gcd(m_k,n)=1\implies [m_k]_n\in\mathbb{Z}_n^*$.
-Is an homomorphism:
$$(\tau_k\cdot\tau_l)(\zeta)=\tau_l(\tau_k(\zeta))=\zeta^{m_l\cdot m_k}$$
-The injectivity : Lets $\tau_k,\tau_l\in G$ with $1\leq k,l\leq t$ and $k\not =l$. So $\gamma(\tau_k)=[m_k]$ and $\gamma(\tau_l)=[m_l]$, but $[m_k]=[m_l]\iff m_k\equiv m_l\;mod\;n$ which is impossible for the choosen $k$ and $l$ . The surjectivity follows from that the degree of the cyclotomical polynomila of nth-order is $\phi(n)$.
To see the proof of the irreducibility of cyclotomical polynomial of any order just check this great answer: showing that $n$th cyclotomic polynomial $\Phi_n(x)$ is irreducible over $\mathbb{Q}$
