$\newcommand{\Tors}{{\rm Tors}}
\newcommand{\tf}{{\rm\, t.f.}}
\newcommand{\Gt}{{\Gamma\!,\,\Tors}}
\newcommand{\Gtf}{{\Gamma\!,\tf}}
\newcommand{\Q}{{\mathbb Q}}
\newcommand{\Z}{{\mathbb Z}}
\newcommand{\HH}{{\mathbb H}}$Let $A$ be an abelian group. We denote by $A\_\Tors$ the torsion subgroup of $A$.
We set $A\_\tf=A/A\_\Tors\,$, which is a torsion free group.


Let $\Gamma$ be a finite group, and let $M$ be a $\Gamma$-module, that is, an abelian group on which $\Gamma$ acts.
We denote by $M\_\Gamma$ the group of coinvariants of $\Gamma$ in $M$, that is,
$$ M\_\Gamma=M\, \big/\bigg\{\sum\_{\gamma\in \Gamma}(\,{}^\gamma y\_\gamma-y\_\gamma\,)
\ \big|\ y\_\gamma\in M\bigg\}.$$
We write $M\_\Gt:= (M\_\Gamma)\_\Tors$ (which is the torsion subgroup of $M\_\Gamma$),
$\ M\_\Gtf=M\_\Gamma/ M\_\Gt\ $ (which is a torsion free group).
We consider the functors:
\begin{align\*}
&(\Gamma,M)\,\rightsquigarrow\, M\_\Gt\,,\\
&(\Gamma,M)\,\rightsquigarrow\, M\_\Gtf\otimes(\Q/\Z)=M\_\Gamma\otimes (\Q/\Z).
\end{align\*}


**Theorem.**
Let $\Gamma$ be a finite group, and let
$$0\to M\_1\xrightarrow{i} M\_2\xrightarrow{j} M\_3\to 0$$
be a short exact sequence of $\Gamma$-modules.
Then there exists a natural homomorphism
$$\delta\colon (M\_3)\_\Gt\to (M\_1)\_\Gamma\otimes (\Q/\Z)$$
such that the following sequence is exact:
\begin{multline\*}
(M\_1)\_\Gt \xrightarrow{i\_\*} (M\_2)\_\Gt \xrightarrow{j\_\*} (M\_3)\_\Gt\xrightarrow{\delta}\\
(M\_1)\_\Gamma\otimes(\Q/\Z)\xrightarrow{i\_\*} (M\_2)\_\Gamma\otimes(\Q/\Z)\xrightarrow{j\_\*} (M\_3)\_\Gamma\otimes(\Q/\Z)\to 0.
\end{multline\*}



> 
> **Question.** Is this exact sequence known?
> If not, does this follow easily from some more general known exact sequence?
> 
> 
> 


I do have a proof, I am asking for a reference!


**Special cases.**
$\DeclareMathOperator\Gal{Gal}$Let $\Gamma=\Gal(E/F)$ be the Galois group
of a finite Galois extension $E/F$ of nonarchimedean local fields.
Assume that our $\Gamma$-modules $M\_i$ for $i=1,2,3$ are finitely generated and torsion-free, and for each $i=1,2,3$, let $T\_i$ be the corresponding algebraic $F$-torus, which splits over $E$, with cocharacter group $M\_i$.
We have a short exact sequence of $F$-tori
$$ 1\to T\_1\to T\_2\to T\_3\to 1$$
and the corresponding Galois cohomology exact sequence
$$
H^1(F,T\_1)\to H^1(F,T\_2)\to H^1(F,T\_3)\xrightarrow{\delta}
H^2(F,T\_1)\to H^2(F,T\_2)\to H^2(F,T\_3)\to 1,
$$
where $H^n(F,T\_i):=H^n(\Gal(F\_s/F), M\_i\otimes F\_s^\times)$
is the cohomology of the *profinite* group $\Gal(F\_s/F)$.
This is our sequence in this special case.


More generally, let again $\Gamma=\Gal(E/F)$ be the Galois group
of a finite Galois extension $E/F$ of nonarchimedean local fields.
Assume that our $\Gamma$-modules $M\_i$ for $i=1,2,3$ are finitely generated, but now we do not assume that they are torsion-free.
For each $i=1,2,3$, we choose a torsion-free resolution
$$ 0\to M\_i^{-1}\to M\_i^0\to M\_i\to 0.$$
We can choose these resolutions compatibly, so that we obtain a short exact sequence of complexes of torsion-free $\Gamma$-modules
$$0\to (M\_1^{-1}\to M\_1^0)\to (M\_2^{-1}\to M\_2^0) \to (M\_3^{-1}\to M\_3^0)\to 0$$
and a short exact sequence of complexes of $F$-tori
$$0\to (T\_1^{-1}\to T\_1^0)\to (T\_2^{-1}\to T\_2^0) \to (T\_3^{-1}\to T\_3^0)\to 0.$$
Write $T\_i^\bullet$ for the complex of tori $(T\_i^{-1}\to T\_i^0)$.
Then we know the hypercohomology groups $\HH^n(F,T\_i^\bullet)$:
$$ \HH^1(F,T\_i^\bullet)\cong(M\_i)\_\Gt,\quad\ \HH^2(F,T\_i^\bullet)\cong(M\_i)\_\Gamma\otimes(\Q/\Z);$$
see M. Borovoi, [Abelian Galois cohomology of reductive groups](https://doi.org/10.1090/memo/0626), Memoirs AMS 132(626), 1998, Proposition 4.1.
We have a hypercohomology exact sequence
\begin{multline\*}
\HH^1(F,T\_1^\bullet)\to \HH^1(F,T\_2^\bullet)\to \HH^1(F,T\_3^\bullet)\xrightarrow{\delta}
\HH^2(F,T\_1^\bullet)\to \HH^2(F,T\_2^\bullet)\to \HH^2(F,T\_3^\bullet)\to 1.
\end{multline\*}
This is our sequence in this special case.



Is this exact sequence known?
$\newcommand{\bQ}{\mathbb{Q}}\newcommand{\bZ}{\mathbb{Z}}\DeclareMathOperator{\Tor}{Tor}\newcommand{\Tors}{\mathrm{Tors}}$I tried to write up the computation with some level of details, please let me know if anything looks dubious.


Consider the long exact sequence $$\ldots\to H\_1(\Gamma, M\_2)\to H\_1(\Gamma,M\_3)\to M\_{1,\Gamma}\to M\_{2,\Gamma}\to M\_{3,\Gamma}\to 0$$ which we will view as an (acyclic) complex and denote its terms by $C^n, n\leq 0$, for brevity, so that $C^0=M\_{3,\Gamma},C^{-1}=M\_{2,\Gamma}$ etc. Consider the spectral sequence associated with the derived functor of $-\otimes\_{\bZ}\bQ/\bZ$ and the 'bête' filtration on the complex $\ldots C^1\to C^0$. I'll use the cohomological grading conventions so that the first page of the spectral sequence looks like $E\_{1}^{i,j}=\Tor\_{-j}^{\bZ}(C^{-i},\bQ/\bZ)$. Note that for an abelian group $M$ we have $\Tor^{\bZ}\_0(M,\bQ/\bZ)=M\otimes\bQ/\bZ,\Tor^{\bZ}\_1(M,\bQ/\bZ)=M\_{\mathrm{Tors}}$, and $\Tor\_{>1}(M,\bQ/\bZ)=0$, hence the $1$st page looks like this, having only $2$ potentially non-zero rows:


$$\begin{matrix}\dots & \color{red}{H\_1(\Gamma,M\_2)\otimes\bQ/\bZ} & \color{red}{H\_1(\Gamma,M\_3)\otimes\bQ/\bZ} & M\_{1,\Gamma}\otimes\bQ/\bZ & M\_{2,\Gamma}\otimes\bQ/\bZ & M\_{3,\Gamma}\otimes\bQ/\bZ & \\ \dots & H\_1(\Gamma,M\_2)\_{\mathrm{Tors}} & H\_1(\Gamma,M\_3)\_{\Tors} & M\_{1,\Gamma,\Tors} & M\_{2,\Gamma,\Tors} & M\_{3,\Gamma,\Tors}\end{matrix}$$


The differentials $d\_{i,j}:E\_1^{i,j}\to E^{i+1,j}\_1$ are simply the maps induced by the maps in the above exact sequence. Since we started with an acyclic complex, the spectral sequence must converge to zero.


Now, since $H\_i(\Gamma, M)$ is annihilated by $|\Gamma|$ for all $\Gamma$-modules $M$, the two entries that are highlighted in red are in fact zero (and all the hidden entries in the top row are zero likewise). This implies that the only possibly non-trivial differential on the second page is $$E\_{2}^{-2,0}=\ker (E\_1^{-2,0}\to E\_1^{-1,0})=\ker (M\_{1,\Gamma}\otimes\bQ/\bZ \xrightarrow{i\_\*} M\_{2,\Gamma}\otimes\bQ/\bZ )\to E\_2^{0,-1}=\mathrm{coker} (M\_{2,\Gamma,\Tors} \xrightarrow{j\_\*} M\_{3,\Gamma,\Tors})$$
This differential must be an isomorphism, while all other $E\_2^{i,j}$ must already be zero as otherwise something would survive to the abutment $E\_3^{i,j}=E\_{\infty}^{i,j}$ of the spectral sequence. But this is exactly saying that the $6$-term sequence in question is exact, with $\delta$ being defined as the inverse to the differential $E\_2^{-2,0}\to E\_2^{0,-1}$.


