correspondence between totally, tamely ramified extension and value groups

I want to show that if $L/K$ is totally and tamely ramified, then the intermediate fields of $L/K$ correspond 1-1 to subgroups of $w(L^{*})/v(K^{*})$, where $v$, $w$ are the valuations on $L$, $K$.

Since the extension is totally ramified, there is no unramified subextension $T$ (or we can say the maximal unramified extension is $K$ itself). Hence from tamely ramified condition, we have $gcd([L:K],p)=1$, where $p=char(k)$, $k$ is the residue class field of $K$. Then I got $[L:K]=[w(L^{*}):v(K^{*})]$.

To prove the relation is 1-1, I defined a map: $H \mapsto w(H^{*})$. Then how to continue from here to prove it is bijective?

Thank you！

Let $L/K$ be totally tamely ramified of degree $e$. We know that $L=K({\pi}_L)$, an Eisenstein extension with $w({\pi}_L)=1$. Considering the successive powers of ${\pi}_L$, we see immediately that the valuation map $w:L^* \to \mathbf Z$ induces a surjection from the set of subextensions of $L/K$ to the set of subgroups of $w(L^*)/v(K^*)\cong \mathbf Z /e\mathbf Z$. The difficulty is to show that this map is injective @. To this end, let us go to Galois to catch a clear picture.
Introduce the normal closure $N$ of $L$, which is the compositum of all the conjugates of $L$. Recall that the compositum of two tamely ramified extensions is again tame, but the compositum of two totally ramified extensions may contain an unramified one (see e.g. Abhyankar's lemma). Denote by $f$ the inertia index of $N/K$, and $e'$ its ramification index (so $e'$is a multiple of $e$, not divisible by $p$). The structure of $N/K$ is classically known (Cassels-Fröhlich, chap. 1, prop. 8-1) : let $\zeta$ be a prmitive $e'$-th root of 1; the maximal unramified subextension $M$ of $N/K$ is cyclic of degree $f$, contains $K(\zeta)$, and the totally tamely ramified extension $N/M$ is cyclic of degree $e'$, a Kummer extension of the form $N=M(\sqrt [e'] {\pi_K})$, where $\pi_K$ is an uniformizer of $K$ (recall that $M/K$ is unramified). Let us show further that $e=e'$, i.e. $LM=N$. Since $M/K$ and $L/K$ are linearly disjoint (because of ramification), $LM/M$ is the unique subextension of $N/M$ having degree $e$ (because of cyclicity). This holds also for any conjugate $L'$ of $L$ over $K$, so $L'M=LM$, which means that $LM$ is stabilized by $Gal(N/K)$, hence $LM/K$ is normal, and $LM=$the normal closure $N$. Note then that the exact sequence of groups $1\to Gal(N/M) \to Gal(N/K) \to Gal(M/K) \to 1$ is split, i.e. $Gal(N/K)$ is a semi-direct product.
The cyclicity of $N/M$ clearly implies that the injectivity property @ holds when replacing $K$ by $M$ (see also @Matmo123). In the general case, any subextension $L'$ of $L/K$ is the subfield of $N$ fixed by $Gal(N/L'M)$ and $Gal(L'N/L')\cong Gal(M/K)$. The uniqueness of $L'M$ implies that of $L'$, and the injectivity property @ is proved.