I'm doing a course on elliptic curves. We're working with a field $K$ with $\mu_n \subset K$ ($K$ contains all $n$th roots of unity) and $\mbox{char}(K)\not|\;n$.
I'm trying to understand the proof of
There is a bijection $$\begin{Bmatrix}\mbox{finite subgroups of }\Delta \subset \frac{K^\times}{(K^\times)^n}\end{Bmatrix}$$ $$\leftrightarrow$$ $$ \begin{Bmatrix}\mbox{finite abelian extensions }L/K\mbox{ of exponent dividing }n\end{Bmatrix}$$ given by $$\begin{array}{rll} \Delta & \mapsto & L=K(\sqrt[n]{\Delta} \\ \\ \frac{K^\times \cap (L^\times)^n}{(K^\times)^n} & \leftarrow & L. \end{array}$$
Given $\Delta \subset \frac{K^\times}{(K^\times)^n}$, we define $L=K(\sqrt[n]{\Delta})$ and $$\Delta'=\frac{K^\times \cap (L^\times)^n}{(K^\times)^n}$$ We want to show that $\Delta=\Delta'$. Clearly $\Delta \subset \Delta'$. Then we have $$L=K(\sqrt[n]{\Delta})\subset K(\sqrt[n]{\Delta'})\subset L,$$ so these inclusions are equalities. Then the lecturer writes
by Lemma 10.1, $|\Delta|=|\Delta'|$.
Where Lemma 10.1 says
Let $\Delta \subset \frac{K^\times}{(K^\times)^n}$ be a finite subgroup. Let $L=K(\sqrt[n]{\Delta})$. Then $L/K$ is Galois with $$\mbox{Gal}(L/K) \cong \mbox{Hom}(\Delta, \mu_n).$$
But I don't see why we immediately know that $\Delta'$ is finite, which I think we need to know to apply the lemma.
Why are there only finitely many
$$\left(\sum_{i=1}^m a_i\sqrt[n]{x_i}\right)^n\in K \hspace{10mm} a_i \in K, x_i \in \Delta$$
up to being in the same $(K^\times)^n$ coset?
Even if $\Delta \subseteq \frac{K^\times}{(K^\times)^n}$ is not finite, we still get injective homomorphisms $$\operatorname{Gal}(L|K) \hookrightarrow \operatorname{Hom}(\Delta, \mu_n),\quad \Delta \hookrightarrow \operatorname{Hom}(\operatorname{Gal}(L|K),\mu_n)$$ from the Kummer pairing (look at the proof of Lemma 10.1 again; we didn't use the finiteness until the very end when comparing the orders of the groups). In particular, from the second injective map we get that $\Delta$ is finite if $K(\sqrt[n]{\Delta})|K$ is finite.