Geometrically, $\mu_n$ are the vertices of a regular polygon inscribed in a unit circle. Let $\zeta_n=\exp(2\pi\ci/n)$ ($n$th root of unity), then we can write $$\begin{align*} \mu_n(\C)&=\left\{\exp\left(\frac{2\pi\ci a}{n}\right)\text{ such that }1\leq a\leq n\right\}\\ &=\{\zeta_n^a\text{ such that }1\leq a\leq n\} \end{align*}$$ Thus, $\mu_n$ is a cyclic subgroup of order $n$, that is $\mu_n\simeq\Z/n\Z$ and thus for every divisor $d|n$, there is exactly one subgroup of order $d$. Another avatar of these subgroups are the cyclotomic polynomials $\Phi_d(X)$. The generators of the cyclic group $\mu_n$ are called the primitive $n$th roots of unity. These are precisely $0\leq m<n$ such that $\gcd(m,n)=1$ and are denoted as $\tio{(\Z/n\Z)}$.
  All the fractions $a/n$ can be written in a reduced form $m/d$ where $d|n$ and $\gcd(m,d)=1$. Let $$\begin{align*} \mu_d^0(\C)&=\left\{\exp\left(\frac{2\pi\ci m}{d}\right) \text{ such that }1\leq m< d\right\}\\ &=\left\{\zeta\in \mu_d\text{ such that }\zeta^j\neq 1 \text{ for }1\leq j<d\right\} \end{align*}$$ Thus, writing $\mu_n$ by reducing $a/n$ we have $$\begin{align*} \mu_n(\C)&=\bigcup_{d|n}\left\{ \exp\left(\frac{2\pi\ci m}{d}\right)\text{ such that }\gcd({m,d})=1, 1\leq m< d \right\}\\ &=\bigcup_{d|n}\mu^0_d(\C) \end{align*}$$

For $m=1$, we see that $n$th root of unity is a root of $\Phi_n(X)$. Some examples, $$\begin{align*} \Phi_1(X)&=X-1\\ \Phi_2(X)&=X+1\\ \Phi_3(X)&=X^2+X+1\\ \Phi_4(X)&=X^2+1\\ \Phi_5(X)&=X^4+X^3+X^2+X+1\\ \Phi_6(X)&=X^2-X+1\\ \Phi_7(X)&=X^6+X^5+X^4+X^3+X^2+X+1\\ \Phi_8(X)&=X^4+1\\ \Phi_9(X)&=X^6+X^3+1\\ \end{align*}$$

Proof All $n$th roots of unity have order $d|n$, that is a primitive $d$th root of $1$. Inversely, if $\zeta$ is a primitive $d$th root of $1$. We deduce that the set of $n$th roots of $1$ is the disjoint union (parametrised by the divisors $d$ of $n$) of $d$th primitive roots. Since $X^n-1=\prod(X-\zeta)$, the product being understood over $n$th roots of unity, we deduce the formula $$X^n-1=\prod_{d|n}\Phi_d(X)$$ We will use the fact that if $F(X),G(X)\in\Z[X]$ with $G(X)$ monic then the quotient of $F(X)$ divided by $G(X)$, say $Q(X)\in\Z[X]$. This follows from euclidean division.
Starting from $\Phi_1(X)=X-1$ and noticing the LHS is $X^n-1$ we can divide by $\Phi_1(X)$ to get the remaining terms in $\Z[X]$ by the above fact. Thus, we obtain by induction on $\Z$ that $\Phi_d$ has integer coefficients for all $d|n$, thus also for $d=n$.
Some examples, $$\begin{align*} X-1&=\Phi_1\\ X^2 − 1 &= (X − 1)(X + 1) = \Phi_1\Phi_2\\ X^3 − 1 &= (X − 1)(X^2 + X + 1) = \Phi_1\Phi_3\\ X^4 − 1& = (X − 1)(X + 1)(X^2 + 1) = \Phi_1 \Phi_2 \Phi_4\\ X^5 − 1 &= (X − 1)(X^4 + X^3 + X^2 + X + 1) = \Phi_1 \Phi_5\\ X^6 − 1 &= (X − 1)(X + 1)(X^2 + X + 1)(X^2 − X + 1) = \Phi_1 \Phi_2 \Phi_3 \Phi_6\\ X^7 − 1 &= (X − 1)(X^6 + X^5 + X^4 + X^3 + X^2 + X + 1) = \Phi_1 \Phi_7\\ X^8 − 1 &= (X − 1)(X + 1)(X^2 + 1)(X^4 + 1) = \Phi_1 \Phi_2 \Phi_4 \Phi_8\\ X^9 − 1 &= (X − 1)(X^2 + X + 1)(X^6 + X^3 + 1) = \Phi_1 \Phi_3 \Phi_9\\ \end{align*}$$

Proof : Let $\zeta_n$ be the $n$th root of unity (and thus is a root of $\Phi_n$). It suffices to prove that $\Phi_n|P$ if $P$ is the minimal polynomial of $\zeta_n$, or all primitive roots are roots of $P$. Let $p$ be a prime number not dividing $n$, and $\zeta$ a root of $P$ (minimal among the $\zeta_n$), then $\zeta$ is necessarily primitive because $P|\Phi_n$.
Lemma 3. $\zeta^p$ is a root of $P$.


Let $\zeta'$ be any root of $\Phi_n$, thus it is primitive, then $\zeta'=\zeta^m$ with $\gcd(m,n)=1$ where $\zeta$ is root of $P$. Now, $m$ can be decomposed into prime factors and repeated application of the lemma gives $\zeta'$ as a root of $P$ and hence $\Phi_n|P$. For example if $m=p_1^2p_2$, then $\zeta'=(((\zeta)^{p_1})^{p_1})^{p_2}$ and inductive application of lemma shows $\zeta'$ is a root of $P$.

Let $n$ be an integer greater than $1$, and prime to the $\Char k$ where $k$ is a field. This assures that $X^n-1$ and its derivative $nX^{n-1}$ do not have a common zero in $\Omega$ (algebraic closure of $k$) and thus $X^n-1$ is a separable polynomial. The set $\mu_n(\Omega)$ of the roots in $\Omega$ is thus of size $n$ and is a finite subgroup of $\Omega^\times$, hence is cyclic and isomorphic to $\Z/n\Z$.
  We recall that, by definition, a primitive $n$th root of unity of $1$ is generator of the cyclic group $\mu_n(\Omega)$. We can make a choice of $\zeta_n$ as a generator. Let $\tio{(\Z/n\Z)}$ denote the multiplicative group of invertible elements of $\Z/n\Z$, then the other primitive $n$ the roots are $\zeta_n^m$ where $m$ is prime to $n$, or $m\in\tio{(\Z/n\Z)}$. $$\tio{(\Z/n\Z)}=\text{Set of positive integers less than $n$ and prime to $n$}$$
  Since $\mu_n(\Omega)$ is generated by $\zeta_n^m$ (hence, choice of primitive root does not matter), the field of decomposition of $X^n-1$ is simply $k[\zeta_n]$ (or $k[\zeta_n^m]$ ) and thus is Galois on $k$. We note that $|\gal(k[\zeta_n]/k)|\leq n$ since the polynomial $X^n-1$ has degree $n$. We denote $G_n:=\gal(k[\zeta_n]/k)$. As noted before $\zeta_n$ can be replaced with any $\zeta_n^m$, since any $\zeta_n^m$ generates $\tio{(\Z/n\Z)}$.

Proof : As noted before the splitting field of $X^n-1$ is $k[\zeta_n]$. Since, the extension $k[\zeta_n]/k$ is algebraic and contains all conjugates of $\zeta_n$, the extension is Galois and the degree of extension is $\leq n$ where $n$ is degree of $X^n-1$.

Proof: We can write the image of $\zeta_n$ under the action of an element $g\in G_n$ uniquely as $$g(\zeta_n)=\zeta_n^{\chi(g)} \text{ where }\chi(g)\in\tio{(\Z/n\Z)}.$$ $\zeta\in\mu_n$ can be written as $\zeta=\zeta_n^m$ with $m\in \tio{(\Z/n\Z)}$. Existence can be shown by the following $$g(\zeta)=g(\zeta_n^m)=(g(\zeta_n))^m=(\zeta_n^{\chi(g)})^m=(\zeta_n^m)^{\chi(g)}=\zeta^{\chi(g)}$$ For uniqueness, let there be a second morphism $$\chi':G_n\ra\tio{(\Z/n\Z)}$$ such that for all $g\in G_n$ and $\zeta\in\mu_n$ we have $$g(\zeta)=\zeta^{\chi'(g)}.$$ The above gives for all $g\in G_n$ and $\zeta\in\mu_n$ $$1=\zeta^{\chi(g)-\chi'(g)}.$$ Setting $\zeta=\zeta_n$ we deduce that ${\chi(g)-\chi'(g)}$ is divisible by order of $\zeta_n\in\mu_n$ which is $n$. But, $\chi$ maps into $\tio{(\Z/n\Z)}$, hence giving $\chi(g)-\chi'(g)=0.$

Proof Let $g,g'\in G_n$ and $\zeta\in\mu_n$, we have $$(gg')(\zeta)=g(\zeta^{\chi(g')})=\zeta^{\chi(g)\chi(g')}$$ thus giving $\chi(gg')=\chi(g)\chi(g')$. We also have $(gg^{-1})(\zeta)=\zeta=\zeta^{\chi(g)\chi(g^{-1})}$ hence $\chi(g^{-1})=(\chi(g))^{-1}$. Since $\zeta_n$ generates $k[\zeta_n]$, and $\zeta_n^m$ with $m\in\tio{(\Z/n\Z)}$ is not $1$, we deduce that $\chi$ is injective and and we can identify $G_n$ and its image $\chi(G_n)$.

Proof $\Q[\zeta_n]/\Q$ is a cyclotomic extension, and thus Galois (see lemma above). Since, the cyclotomic polynomial $\Phi_n(X)$ is irreducible over $\Q$ of degree equal to size $|\tio{(\Z/n\Z)}|$. We get $$|\gal(\Q[\zeta_n]/\Q)|=[\Q[\zeta_n]:\Q]= |\tio{(\Z/n\Z)}|$$ But we have already shown $\chi$ maps $\gal(\Q[\zeta_n]/\Q)$ injectively into $\tio{(\Z/n\Z)}$. Hence, we get the isomorphism.

Let $d|n$ so that $\Q[\zeta_n]\supset\Q[\zeta_d]$. The Galois Correspondence predicts that $\Q[\zeta_d]$ corresponds to a subgroup of $\gal(\Q[\zeta_n]/\Q)$ with cardinality $\varphi(n)/\varphi(d)$ (where $\varphi$ is the Euler totient function and $\varphi(n)=\card\tio{(\Z/n\Z)}$). Subgroup of $\gal(\Q[\zeta_n]/\Q)$ must be the kernel of the surjection $$\gal(\Q[\zeta_n]/\Q)\ra\gal(\Q[\zeta_d]/\Q)$$ Since, $\gal(\Q[\zeta_n]/\Q)\simeq \tio{(\Z/n\Z)}$ by Theorem 5, the above surjection can be written as $$\tio{(\Z/n\Z)}\ra \tio{(\Z/d\Z)}$$

Proof Denote $\lcm(n,m)=\alpha,\gcd(n,m)=\delta,K=\Q[\zeta_\alpha]$ and $$\Gamma_d=\Ker\left(\tio{(\Z/\alpha\Z)}\ra \tio{(\Z/d\Z)}\right)\text{ where }d|\alpha$$ There are two sub-fields $K_i=\Q[\zeta_i], i=n,m$ of $K$, defined (via Galois Correspondence) after the preceding, by the subgroups $\Gamma_i,i=n,m$.
Lemma Let $K/k$ be a Galois extension with Galois group $G$ and $K_i/k, i=1,2$ be two sub extensions defined by subgroups $G_1,G_2$ of of $G$. Then $K_1K_2$ corresponds to $G_1\cap G_2$ while $K_1\cap K_2$ corresponds to subgroup of $G$ generated by $G_1$ and $G_2$. Furthermore, the stabilizer $G_x$ of $x\in K$ in $G$ corresponds to field $k[x]$ generated by $x$.


Using the Lemma above we have to simply show $\Gamma_n\cap\Gamma_m=\{1\}$ (which will prove that $K_nK_m=\Q[\alpha]$) and that the group generated by $\Gamma_n$ and $\Gamma_m$ is $\Gamma_\delta$ (proving $K_n\cap K_m=\Q[\zeta_\delta]$).
  The first part is clear: Let $g\mod\alpha\in\tio{(\Z/n\Z)}$ is in the intersection of $\Gamma_n\cap \Gamma_n$, , that is to say that $n$ and $m$ divide $g-1$, in other words $\alpha|(g-1)$. Since $g-1$ is always prime to $g$, it has to be the case that $g=1$.
  For the second statement, using chinese lemma, we can suppose that $n,m$ are powers of $p$, of the form $p^\nu,p^\mu$ with (say) $0\leq\nu\leq\mu$ such that $\delta=p^\nu$. The case where $\nu$ or $\mu$ is zero is trivial, hence suppose $\nu,\mu>0$. We have then $$\Gamma_i=1+p^i\Z/p^\mu\Z,\qquad i=\nu,\mu$$ and thus the group generated is $\Gamma_\nu=\Gamma_\delta.\qquad\qquad\blacksquare$

Remark Another proof that is less galois. Denote $\lcm(n,m)=\varpi$ and $\gcd(n,m)=\delta$. The first point is elementary. Denote $\varpi=n\nu, \varpi=m\mu$ where $\nu$ and $\mu$ are relatively prime. This then gives $$\zeta_n=\zeta^\nu_{\varpi}\qquad\text{and}\qquad \zeta_m=\zeta_{\varpi}^\nu$$ proving $\Q[\zeta_n,\zeta_m]\subset\Q[\zeta_\varpi]$.
  The second equality is more subtle. Denote now $n=\delta\nu,m=\delta\mu$ where $\mu$ and $\nu$ are relatively prime. We have $\zeta_\delta=\zeta_n^\nu=\zeta_m^\mu$ giving the inclusion $$\Q[\zeta_\delta]\subset\Q[\zeta_n]\cap\Q[\zeta_m]$$ Inversely, since the cycltomic extensions $\Q[\zeta_n]\Q$ is galois, we know the degrees of the various extensions thanks to to theorem on galois extension of composites , which are summarized as follows.