Choose a generator \(g\) of \(G\text{.}\) We start with the four-term exact sequence of \(G\)-modules
\begin{equation*}
0 \to \ZZ \to \ZZ[G] \to \ZZ[G] \to \ZZ \to 0
\end{equation*}
in which the first map is \(1 \mapsto \sum_{g \in G} [g]\text{,}\) the second map is \([h] \mapsto [hg] - [h]\text{,}\) and the third map is \([h] \mapsto 1\text{.}\) Since everything in sight is a free abelian group, we can tensor over \(\ZZ\) with \(M\) and get another exact sequence:
\begin{equation*}
0 \to M \to M \otimes_\ZZ \ZZ[G] \to M \otimes_\ZZ \ZZ[G] \to M \to 0\text{.}
\end{equation*}
The terms in the middle are just \(\Ind^G_{1} M\text{,}\) where we first restrict \(M\) to a module for the trivial group and then induce back up. Thus their Tate groups are all zero. The desired result now follows from the following general fact: if
\begin{equation*}
0 \to A \stackrel{f}{\to} B \stackrel{g}{\to} C \stackrel{h}{\to} D \to 0
\end{equation*}
is exact and \(B\) and \(C\) have all Tate groups zero, then there is a canonical isomorphism \(H^{i+2}_T(G, A) \to H^i_T(G, D)\text{.}\) To see this, apply the long exact sequence to the short exact sequences
\begin{gather*}
0 \to A \to B \to B/\im(f) \to 0\\
0 \to B/\ker(g) \to C \to D \to 0
\end{gather*}
to get
\begin{equation*}
H^{i+2}(G, A) \cong H^{i+1}(G, B/\im(f)) = H^{i+1}(G, B/\ker(g)) \cong H^i(G,D).
\end{equation*}
