I have a discrete group $G$ and classes $x,y\in H^1(G;\mathbb{Q})$ (group cohomology with coefficients in the rationals viewed as a trivial $G$-module) such that the Massey product $$\alpha:=\langle x, x, y\rangle\in H^2(G;\mathbb{Q})$$ is defined and has zero indeterminacy. 


I would like to find a short exact sequence of $G$-modules $0\to \mathbb{Q}\to M\to N\to 0$ and a class $a\in H^1(G;N)$ which maps to $\alpha$ under the associated Bockstein operator $\beta\colon\thinspace H^1(G;N)\to H^2(G;\mathbb{Q})$.



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> Is this always possible? If so, is there a procedure for doing this? If not, are there known necessary or sufficient conditions (on the group $G$, or the classes $x$ and $y$)?
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**What I know**


I'm attempting to use the interpretations of low-dimensional group cohomology described in Chapter IV of the book of [Brown](http://books.google.com/books?id=6x0mfOUFYgsC&printsec=frontcover&dq=brown+cohomology+of+groups&source=bl&ots=ux7WmwIZTv&sig=CD2wP9jU6fD4q_Ydqwh4mrvh7Hg&hl=en&ei=QdqdTeqRPKGX0QG4xOnWBA&sa=X&oi=book_result&ct=result&resnum=2&ved=0CBwQ6AEwAQ#v=onepage&q&f=false). Using the standard resolution, the classes $x$, $y$ are represented by derivations $d\_x,d\_y\colon\thinspace G\to \mathbb{Q}$, which I can write down explicitly. I am also able to write down cochains $\lambda,\mu\in C^1(G;\mathbb{Q})$ (essentially just functions $G\to \mathbb{Q}$) satisfying $\delta\lambda = d\_x\cup d\_x$ and $\delta\mu=d\_x\cup d\_y$. This enables me to write down explicitly the central extension $$0\to\mathbb{Q}\to E\to G\to 1$$ which represents the class $\alpha$. But I'm not sure where to go from here.


One approach would to look for a $G$-module $M$ which has $\mathbb{Q}\hookrightarrow M$ as a trivial submodule, and such that $\alpha$ is in the kernel of the induced map $H^2(G;\mathbb{Q})\to H^2(G; M)$. But again I'm not sure how to go about doing this.


Any general suggestions would be appreciated. 



When is a Massey product the image of a Bockstein operator?
If $\mathbb{Q}G$ is the group algebra, group cohomology is just $H^{\ast}(G,\mathbb{Q})=\operatorname{Ext}^{\ast}\_{\mathbb{Q}G}(\mathbb{Q},\mathbb{Q})$. Now take any ring $R$, any pair of $R$-modules $A$ and $B$, and any class $\alpha\in \operatorname{Ext}^{n}(A,B)$. Take a Yoneda $n$-extension representing $\alpha$:


$$B\hookrightarrow M\_n\rightarrow M\_{n-1}\rightarrow\cdots\rightarrow M\_2\rightarrow M\_1 \twoheadrightarrow A$$


Factor the homomorphism $M\_n\rightarrow M\_{n-1}$ as 


$$M\_n\twoheadrightarrow N \hookrightarrow M\_{n-1}$$


You can split the Yoneda $n$-extension representing $\alpha$ into a Yoneda $1$-extension and a Yoneda $(n-1)$-extension representing $\beta\in \operatorname{Ext}^{n-1}(A,N)$


$$B\hookrightarrow M\_n\twoheadrightarrow N$$


$$N\hookrightarrow M\_{n-1}\rightarrow\cdots\rightarrow M\_2\rightarrow M\_1 \twoheadrightarrow A$$


This $\beta\in \operatorname{Ext}^{n-1}(A,N)$ is kicked by $\alpha\in \operatorname{Ext}^{n}(A,B)$ in the long exact sequence associated to $B\hookrightarrow M\_n\twoheadrightarrow N$


$$\cdots\rightarrow \operatorname{Ext}^{n-1}(A,N) \rightarrow \operatorname{Ext}^{n}(A,B) \rightarrow\cdots$$


