Convergence of the Lyndon-Hochschild-Serre spectral sequence as an algebra Consider a short (not necessarily split) exact sequence of groups
$1 \rightarrow N \rightarrow G \rightarrow Q \rightarrow 1$
and suppose we wish to find the cohomology of $G$ with coefficients in a ring $R$. Then, it is known that there is a first quadrant cohomological spectral sequence of algebras converging as an algebra:
$E_2^{p,q} = H^p(Q;H^q(N;R)) \implies H^{p+q}(G;R)$. 
Suppose $R$ is replaced by a $Q$-module $M$. (A specific example of interest to me is when $Q = \mathbb Z_2$ and $M = \mathbb Z$ with the inversion action of $Q$.) Then, can we still define a spectral sequence of algebras with an $E_2$-page as given above, so that it converges to $H^{p+q}(G,M)$ as an algebra?
(Knowing the answer for the specific example I mentioned is sufficient.)  
 A: As Joshua Grochow mentioned in a comment, there is not necessarily an algebra structure on this spectral sequence. (In particular, $H^0(G;M) = M^G$ does not necessarily have a ring structure.) Generally, an equivariant pairing $M \otimes N \to P$ gives rise to a multiplication map on spectral sequences.
In your case of $\Bbb Z$ with the sign action, here is a handy trick that does elaborate on the structure. Let $R$ be the ring $\Bbb Z[x]/(x^2 -1)$, where $G$ acts on $x$ through the quotient $Q$ by sending it to $-x$. Then $R$ is a $G$-equivariant ring, but as a module it decomposes as $\Bbb Z \cdot 1 \oplus \Bbb Z^{sgn} \cdot x$. As a result, the spectral sequence naturally additively decomposes as a direct sum:
$$
H^p(Q; H^q(N;R)) \cong H^p(Q;H^q(N;\Bbb Z)) \oplus H^p(Q;H^q(N;\Bbb Z^{sgn}))
$$
The multiplication on $R$, however, gives this spectral sequence a multiplication. This both makes the sign-spectral sequence into a module over the trivial-action spectral sequence and gives you a bilinear pairing
$$
H^p(Q;H^q(N;\Bbb Z^{sgn})) \otimes H^{p'}(Q;H^{q'}(N;\Bbb Z^{sgn})) \to 
H^{p+p'}(Q;H^{q+q'}(N;\Bbb Z))
$$
that is compatible with the differentials in the spectral sequence. These converge to a natural module structure
$$
H^n(G;\Bbb Z) \otimes H^{n'}(G;\Bbb Z^{sgn}) \to 
H^{n+n'}(G;\Bbb Z^{sgn})
$$
and a pairing
$$
H^n(G;\Bbb Z^{sgn}) \otimes H^{n'}(G;\Bbb Z^{sgn}) \to 
H^{n+n'}(G;\Bbb Z).
$$
(This trick of making a ring out of modules can be used to determine extra structure in a number of other cases.)
