I am trying to prove that the connecting homomorphism between $\frac{1}{n}\mathbb{Z}/\mathbb{Z}$ and $\mathbb{Z}/n\mathbb{Z}$ is the multiplication with $n$.
From the exact sequence $0\rightarrow\mathbb{Z}\stackrel{i}{\rightarrow}\mathbb{Q}\stackrel{j}{\rightarrow}\mathbb{Q}/\mathbb{Z}\rightarrow0$ one gets the inifinite exact cohomology sequence
$\dots \rightarrow H^{-1}(G,\mathbb{Z})\stackrel{\overline{i}}{\rightarrow}H^{-1}(G,\mathbb{Q})\stackrel{\overline{j}}{\rightarrow}H^{-1}(G,\mathbb{Q}/\mathbb{Z})\stackrel{\delta}{\rightarrow}H^0(G,\mathbb{Z})\rightarrow\dots$.
Here $G$ is the Galois group of a normal extension $L/K$ and $\delta$ is the connecting homomorphism. Since $\mathbb{Q}$ is cohomologically trivial, $H^0(G,\mathbb{Z})$ and $H^{-1}(G,\mathbb{Q}/\mathbb{Z})$ are isomorphic. One can compute that $H^0(G,\mathbb{Z})=\mathbb{Z}/n\mathbb{Z}$ and $H^{-1}(G,\mathbb{Q}/\mathbb{Z})=\frac{1}{n}\mathbb{Z}/\mathbb{Z}$.
How can I prove that $\delta$ is given by multiplication with $n$?