Find the bijection between Kummer's field and Galois subgroup 
Let $\mathbb{k}$ is field and $\exists \xi \in \mathbb{k}^*,\ O(\xi) = n$   I need find the bijection between k-isomorphic finite separable
  normal extension $\mathbb{k}$ (i.e. Kummer's field with an exponent n) with abelian
  Galois group with exponent n and finite subgroup
  $\mathbb{k}^*/\mathbb{k}^{*n}$

So, I've seen this theorem (Robert B. Ash, "Abstract Algebra: The Basic Graduate Year", theorem 6.7.5, chapter 6, p.21): 

Let $E/F$ be a finite extension, and assume that F contains a
  primitive n-th root of unity $\xi$. Then E/F is a Kummer extension
  whose Galois group G has an exponent dividing n iff there are
  nonzero elements $a_1, ..., a_r \in F$ such that E is a splitting
  field of $(X^n - a_1)\cdot \cdot \cdot (X^n - a_r)$ over F.

I think it's very similar to my problem. If it is, may I use this theorem for my problem? I just have no other ideas how I may prove my problem.
 A: Kummer's theory gives a complete and explicit description of the finite abelian extensions of exponent n (i.e. Galois extensions whose groups are finite, abelian, killed by the exponent n). You can find an account o in any book on Galois theory. But the way you enunciate the main result is somewhat imprecise. It should read as follows:
Let $k$ be a field of characteristic not dividing $n$, containing the group $\mu_n$ of $n$-th roots of unity. Then every finite abelian extension $K/k$ of exponent $n$ is obtained by adding to $k$ the $n$-th roots of the representatives in $k^*$ of a uniquely defined finite subgroup $R_{K/k}$ of $k^*/(k^*)^{n}$ (called the Kummer radical of $K/k$). In more precise mathematical terms, the bijection between $K/k$ and $R_{K/k}$ can be expressed as a  duality $Gal(K/k)\cong Hom(R_{K/k},\mu_n)$, which is described explicitly by : $s\in Gal(K/k) \to f_s\in Hom(R_{K/k},\mu_n)$, where $f_s$ is defined by $f_s([a])=s(\sqrt [n]a)/\sqrt [n]a \in \mu_n$, where $[a]\in R_{K/k}$ is represented by $a\in k^*$.
A: Notations:

Let k be a field containing m-th roots of unity, where m is coprime to the characteristic of k (if not $0$). Denote by $a^{1/m}$ any m-th root of unity, and the group of m-th roots of unity by $\mu_m$. Let $B$ be a subgroup of $k^*$ containing $(k^*)^m$. Denote by $k(B^{1/m})$ the compositum of all $k(a^{1/m})$ where $a\in B$.

Theorem:

Let $K_B=k(B^{1/m})$, then $K_B$ is abelian of exponent $m$ over $k$. The map $\text{Gal}(K_B/k)\times B\rightarrow \mu_m$, defined as $(\sigma, a)\mapsto \sigma\alpha/\alpha$ where $\alpha^m=a$, is a bi-group homomorphism with trivial left kernel and right kernel $(k^*)^m.$ $K_B/k$ is finite iff $B/(k^*)^m$ is finite and $B/(k^*)^m\cong \text{Hom}(\text{Gal}(K_B/k),\mu_m), [K_B:k]=(B:(k^*)^m)$

Proof.  By definition, $K_B$ contains all roots of $x^m-a$ where $a\in B$, which has no multiple root. Hence $K_B$ is the splitting field of a family of separable polynomials and is thus Galois.
For any $\sigma \in\text{Gal}(K_B/k)$ and $\alpha$ an m-th root of some $a\in B, \sigma\alpha=\alpha\mu_{\sigma}$ where $\mu_{\sigma}$ is an m-th root of unity determined by $\sigma$. Therefore, $\text{Gal}(K_B/k)$ is an abelian group and thus $K_B/k$ is abelian. In addition we have $\sigma^m\alpha=\mu_{\sigma}^m\alpha=\alpha$ for each $\alpha\in K_B$, so $K_B/k$ is of exponent $m$.
The map defined above is well-defined because m-th roots of unity are contained in $k$ and thus are fixed by any element of $\text{Gal}(K_B/k)$. It is a homomorphism in $\sigma$ since for $\sigma,\tau\in \text{Gal}(K_B/k),(\sigma\circ\tau,a)\mapsto\sigma(\tau\alpha)/\alpha=\sigma(\alpha\mu_{\tau})/\alpha=\mu_{\sigma}\mu_{\tau}$ and $(\sigma,a)\mapsto\mu_{\sigma}$, similarly for $\tau$. It is a homomorphism in $a$ since $\sigma$ is a homomorphism.
If $\sigma\alpha/\alpha=1$ for any $\alpha^m=a\in B, \sigma$ fixes all elements of $K_B$ and is thus the identity homomorphism. On the other hand, if $\sigma\alpha/\alpha=1$ for any $\sigma\in \text{Gal}(K_B/k)$, then $\alpha$ must lie in $k^*$, so its m-th power $a$ lies in $(k^*)^m$.
The above bi-group homomorphism induces an injective homomorphism from $\text{Gal}(K_B/k)$ to $\text{Hom}(B/(k^*)^m,\mu_m)$, and similarly an injective homomorphism from $B/(k^*)^m$ to $\text{Hom}(\text{Gal}(K_B/k),\mu_m)$. Therefore $K_B/k$ is finite iff $B/(k^*)^m$ is. Since $\text{Gal}(K_B/k)$ is finite abelian, it is isomorphic to its dual $\text{Hom}(\text{Gal}(K_B/k),\mu_m)$. Suppose $k^*$ has exponent $r$ strictly less than its order as a group, then any element in $k^*$ is a root of the polynomial $x^r-1$, so the number of roots in $k^*$ alone exceeds the degree of the polynomial, contradiction. Hence $r=|k^*|$ and thus $k^*$ is cyclic and a fortiori abelian, and so is $B$ since it's a subgroup of $k^*$. Then $|\text{Hom}(\text{Gal}(K_B/k),\mu_m)|=[K_B:k]\ge (B:(k^*)^m)$, and the other injection gives the reverse inequality. Hence the injection is in fact an isomorphism, that is, $\text{Hom}(\text{Gal}(K_B/k)\cong B/(k^*)^m$, and in particular, $[K_B:k]= (B:(k^*)^m)$.
Theorem:

The map $B\mapsto K_B=k(B^{1/m})$ defines a bijection between the set of subgroups of $k^*$ containing $(k^*)^m$ and the set of abelian extensions of $k$ of exponent $m$.

Proof.  Let $B_1, B_2$ be two subgroups of $k^*$ containing $(k^*)^m.$ If $B_1\subset B_2$, clearly $K_{B_1}\subset K_{B_2}$. Conversely, suppose $K_{B_1}\subset K_{B_2}$.  Given any element $b\in B_1, k(b^{1/m})$ is contained in a finitely generated subextension of $K_{B_2}$, so we may assume without loss of generality that let $K_{B_2}$ is finitely generated, hence finite. Now, let $B_b$ be the smallest subgroup of $k^*$ generated by $B_2$ and $b$. Then $K_{B_b}=K_{B_2}$ and hence by the previous theorem, $(B_2:(k^*)^m)=[K_{B_2}:k]=[K_{B_b}:k]=(B_b:(k^*)^m)$. Hence $B_2=B_b$ and in particular $b\in B_2$.  Since we may pick $b$ to be any element in $B_1$, this implies that $B_1\subset B_2$.
Given any abelian extension $K/k$ of exponent $m$ and any $b\in K, k(b^{1/m})$ is contained in a finitely generated subextension of $K$, which we denote by $K_b$. Then $\text{Gal}(K_b/k)$ is finite abelian and thus by the fundamental theorem of finite abelian groups, $\text{Gal}(K_b/k)\cong \Bbb Z/d_1\Bbb Z\times\cdots\times \Bbb Z/d_r\Bbb Z$ where $d_i\mid d_{i+1}$. Let $K_i$ be the fixed field of the group obtained from replacing the i-th position in the direct product by $\{1\}$, then by the fundamental theorem of Galois theory, $K_i/k$ is Galois because its corresponding automorphism group is a normal subgroup of $\text{Gal}(K_b/k)$. Then the restriction of $\text{Gal}(K_b/k)$ on $\text{Gal}(K_i/k)$, which is a surjective homomorphism with kernel $\text{Gal}(K_b/K_i)$, induces an isomorphism $\text{Gal}(K_i/k)\cong(\text{Gal}(K_b/k))/(\text{Gal}(K_b/K_i))$ by the first isomorphism theorem. But $K_i$ is the fixed field of $\Bbb Z/d_1\Bbb Z\times\cdots\times\Bbb \{1\}\times\Bbb Z/d_{i+1}\Bbb Z\times \cdots\times \Bbb Z/d_r\Bbb Z$ and so $\text{Gal}(K_b/K_i)\cong \Bbb Z/d_i\Bbb Z$, which implies that $K_i/k$ is a cyclic extension.
Since the primitive m-th root of unity $\mu$ is in $k$, it is fixed by any automorphism in $\text{Gal}(K_i/k)$ and hence has norm $1$. Therefore by Hilbert's theorem $90,$ the generator of $\text{Gal}(K_i/k)$ sends any $\beta\in K_i\setminus k$ to $\mu\beta$. Then for $j=1,\cdots,m/d_i, \mu^j\beta$ must be the roots of the minimal polynomial of $\beta$ in $k$, and they are all distinct, so $K_i=k(\beta)$ for some $\beta\in K_i\setminus k$ such that $\beta^{m/d_i}\in k$. The compositum of all $K_i$ is the fixed field of the intersection of all of their corresponding Galois group, which is clearly the identity. Hence $K_b=K_1\cdots K_r$ where each $K_i$ is generated by an element $\beta_i$ such that $\beta_i^{m/d_i}\in k$, and thus $K_b=k(\gamma)$ where $\gamma$ has minimal polynomial $x^m-c$ over $k$. In other words, $K_b=k(c^{1/m})$ and then $K$ is the compositum of all $K_b$, so $K=K_B$ where $B=\langle c^{1/m}: c\in k^*\rangle(k^*)^m$, which concludes our proof.
