Let $K$ be a finite extension of $\mathbb{Q}_p$, and $L,M$ be finite Galois extensions of $K$ such that $L\subset M$. Let $\phi_L\in Z^2(\text{Gal}(L/K),L^\times)$ such that $[\phi_L]$ is a generator of $H^2(\text{Gal}(L/K),L^\times)$, and $\phi_M\in Z^2(\text{Gal}(M/K),M^\times)$ such that $[\phi_M]^{[M:L]}=[\phi_L]$.
Then by local class field theory there exists an isomorphism: $\text{Gal}(L/K)^{\text{ab}}\cong K^\times/N_{L/K}(L^\times)$ induced by $[\phi_L]$ (I think the concrete map is $g\mapsto \prod_{h\in \text{Gal}(L/K)}\phi_L(h,g)$) and an isomorphism $\text{Gal}(M/K)^{\text{ab}}\cong K^\times/N_{M/K}(M^\times)$ induced by $[\phi_K]$. The question is how to prove the following diagram commutes: \begin{equation*} \begin{array}[c]{ccc} \text{Gal}(M/K)^{\text{ab}} & \rightarrow K^\times/N_{M/K}(M^\times)\\ \downarrow & \downarrow \\ \text{Gal}(L/K)^{\text{ab}} & \rightarrow K^\times/N_{L/K}(L^\times) \end{array} \end{equation*}
I think it makes sense because the diagram is independent of the choice of $\phi_M$ such that $[\phi_M]^{[M:L]}=[\phi_L]$. But I don't know how to prove it.
Thank you!