Considerations on cyclotomic extensions I stopped in front of some issues regarding certain passages of the theorem 9.4 of  Algebra of Serge Lang, in which we suppose to have  k a field such that $[k (\mu_n):k]=\phi (n)$ where $\mu_n$ is the set of n-th units of k and $\phi $ is the Euler function and n and char (k) are coprime and n is odd. Why have we immediately that  $[k (\mu_m):k]=\phi (m)$ for any integer m that divides n ? Further,  how can I  deduce again    for any integer m that divides n  that if $[k (\beta,\mu_m):k (\mu_m)   ]=m$ for some $\beta $ root of $ X^m-a $ for some a in k then $[k (\beta,\mu_n):k (\mu_n)   ]=m$? I think the relations between different cyclotomic extensions of a same field k aren't clear to me ... thank in advance
 A: Lemma: Assume $m \mid n$.  The homomorphism $$(\mathbb{Z}/n\mathbb{Z})^{\ast} \rightarrow (\mathbb{Z}/m\mathbb{Z})^{\ast}$$ given by $t + n\mathbb{Z} \mapsto t + m\mathbb{Z}$ is surjective.
Proof: Write $n = p_1^{e_1} \cdots p_j^{e_j}$, and $m = p_1^{d_1} \cdots p_j^{d_j}$ with $d_i \leq e_i$.  We have a commutative diagram 
$$\begin{matrix} \mathbb{Z}/n\mathbb{Z} & \rightarrow & \mathbb{Z}/p_1^{e_1}\mathbb{Z} \times \cdots \times \mathbb{Z}/p_j^{e_j} \mathbb{Z} \\ \downarrow & & \downarrow \\
\mathbb{Z}/m\mathbb{Z} & \rightarrow & \mathbb{Z}/p_1^{d_1}\mathbb{Z} \times \cdots \times \mathbb{Z}/p_j^{d_j} \mathbb{Z} \end{matrix}$$
with horizontal isomorphisms, from which you can see it suffices to prove the lemma when $n$ is a power of a prime number.  But in this case it's obvious. $\blacksquare$
Let $\zeta$ be a primitive $n$th root of unity, and $L = K(\zeta)$.  Then $L$ is a Galois extension of $K$, since it is a splitting field and the polynomial $X^n -1$ is separable over $K$.  Let $G = \textrm{Gal}(L/K)$.  There is an injection $$G \rightarrow (\mathbb{Z}/n\mathbb{Z})^{\ast}$$which by your hypothesis is an isomorphism.  This is equivalent to saying that for every $1 \leq t \leq n$ which is relatively prime to $n$, there is a $\sigma \in G$ such that $\sigma \zeta = \zeta^t$.  
Now if $m \mid n$, then $\xi := \zeta^{n/m}$ is a primitive $m$th root of unity.  Again there is an injection of groups $$\textrm{Gal}(K(\xi)/K) \rightarrow (\mathbb{Z}/m\mathbb{Z})^{\ast}$$  To say this map is surjective is to say that for $1 \leq s \leq m$, relatively prime to $m$, there is a $\tau \in \textrm{Gal}(K(\xi)/K)$ such that $\tau(\xi) = \xi^s$.  
And this is the case: since $s$ is relatively prime to $m$, by the lemma there is a $1 \leq t \leq n$, relatively prime to $n$, with $t \equiv s \pmod{m}$.  Let $\sigma$ be the element of $G$ which sends $\zeta$ to $\zeta^t$.  It sends $\zeta^{n/m} = \xi$ to $\zeta^{t(n/m)} = \xi^t$.  Let $\tau$ be the restriction of $\sigma$ to $K(\xi)$.  Then $\tau$ does what you want:
$$\tau \xi = \xi^t = \xi^s$$ 
