Let $k$ be an imaginary quadratic field and $p$ a prime such that $p = \mathfrak{p}\overline{\mathfrak{p}}$ in $k$. Let $k_n$ be an extension of $k$ such that $Gal(k_n/k) \cong \mathbb{Z}/p^n\mathbb{Z}$ and suppose that $\mathfrak{p}$ and $\overline{\mathfrak{p}}$ are totally ramified. Let $L_n$ be the genus field for $k_n/k$ (i.e. the maximal unramified extension of $k_n$ such that $L_n$ is Abelian over $k$). Let $A_n$ be the $p$-primary part of the class group of $k_n$ and $Gal(k_n/k) = \langle \sigma \rangle$. Also, assume that the $p$-primary class group for $k$ is trivial.
Here is my question: Why is it that $[L_n:k_n] = [A_n:A_n^{\sigma -1}]$? The paper I am reading says that this is well known, but doesn't give any references.
I know that $L_n$ is contained in the class field $\bar{L}_n$ of $k_n$ since $L_n$ is unramified Abelian over $k_n$, but I'm having trouble seeing how $L_n$ being Abeian over $k$ implies $[L_n:k_n] = [A_n:A_n^{\sigma -1}]$. The total ramification of $\mathfrak{p}, \overline{\mathfrak{p}}$ in $k_n/k$ must also play a role. Any hints or references are appreciated!