Let $K$ be a $p$-adic field with Galois group $G$ and inertia subgroup $I\subset G$. Denote $(-)^\ast=\mathrm{Hom}_{cont}(-,\mathbb{Q}/\mathbb{Z})$. Using Tate local duality, we can compute $$H^2(G,\mathbb{Q}/\mathbb{Z})=\varinjlim H^2(G,\mathbb{Z}/n\mathbb{Z})=\varinjlim \mathrm{Hom}_G(\mathbb{Z}/n\mathbb{Z},\mu)^\ast=\mathrm{Hom}_G(\mathbb{Q}/\mathbb{Z},\mu)^\ast=\mathrm{Hom}(\mathbb{Q}/\mathbb{Z},\mu(K))^\ast = 0$$since $\mathbb{Q}/\mathbb{Z}$ has no finite quotient. On the other hand, the Hochschild-Serre spectral sequence for $H^2(G,\mathbb{Q}/\mathbb{Z})$ has $E_2$ page \begin{array}{ll} (I^\ast)^{\phi=1} & I^\ast/1-\phi \\ \mathbb{Q}/\mathbb{Z} & \mathbb{Q}/\mathbb{Z} \end{array} where $\phi$ is the Frobenius in $\widehat{\mathbb{Z}}$. I was under the impression that the action of the Frobenius on the Pontryagin dual of the inertia is trivial by cocycle computations of mine, but this contradicts the first computation as then $I^\ast/1-\phi=I^\ast$ is non-zero. Moreover $H^1(G,\mathbb{Q}/\mathbb{Z})=G^\ast$ is an extension of $\mathbb{Q}/\mathbb{Z}$ by $I^\ast$, which naively suggests that $(I^\ast)^{\phi=1}=I^\ast$ under the canonical inclusion; but this implies that $\phi$ acts trivially.

What is happening here ?