The local invariant map is a group homomorphism

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Let $ (K, \nu) $ be a nonarchimedian local field. I have read that the Brauer group, $ \text{Br}(K) $ (which for me, is defined by the similarity classes of CSAs with group operation as tensor product) is canonically isomorphic to $ \mathbb{Q}/\mathbb{Z} $ via the local invariant map defined as follows:

Let $ c $ be a class in $ \text{Br}(K) $. Then $ c $ is represented by a central division algebra, $ D $. It is possible to prove that the valuation $ \nu $ extends to a (nonarchimedian) valuation on $ D $ (also called $ \nu $) and that there is an unramified subfield $ L $ of $ D/K $ with $ [D:K]=[L:K]^2 $. Let $ \sigma_L $ be the Frobenius automorphism of $ L/K $. By Skolem-Noether, there is an invertible $ a \in D $ such that $ \sigma_L (x)= axa^{-1} $ for all $ x \in L $. Finally define, $$ \text{inv}_K(c) = \nu(a) \pmod {\mathbb{Z}} $$ and it is easy to show that this is well defined.

My question is: Why does the invariant map defined in this way become a group homomorphism?

I have no idea how to show this. The definition above is from Milne's notes on CFT, Chapter 4 and he doesn't really prove it there. On a side note, one has $$ \text{Br}(K) \cong H^2(\text{Gal}(K^{un}/K), (K^{un})^{\times})$$ and there is an isomorphism $ H^2(\text{Gal}(K^{un}/K), (K^{un})^{\times}) \cong \mathbb{Q}/\mathbb{Z} $ using pure group cohomology.