For a short exact sequence of groups $1\rightarrow A\rightarrow B\rightarrow C\rightarrow 1$ there is an associated fibration $K(A,1)\rightarrow K(B,1)\rightarrow K(C,1)$, which can be constructed by realizing the homomorphism $B\rightarrow C$ by a map $K(B,1)\rightarrow K(C,1)$ and the convert it into a fibration. The fiber is $K(A,1)$ (from the associated long exact sequence of homotopy groups). 


For a fibration $F\rightarrow X\rightarrow B$, the differential $d\_n\colon E\_{n,0}^n\to E\_{0,n-1}^n$ in the Serre spectral sequence was shown to be equal to the transgression in Hatcher's book on Spectral Sequences (Proposition 1.13). The transgression was defined using (relative) homology groups.


My questions is: From the short exact sequence of groups $1\rightarrow A\rightarrow B\rightarrow C\rightarrow 1$, is there any method to directly compute the transgression of the associated fibration $K(A,1)\rightarrow K(B,1)\rightarrow K(C,1)$, at least for the case $n=2$, without constructing $K(G,1)$'s and considering their homologies?



How to Compute Transgressions in a Serre Spectral Sequence?
I can give a description in case of cohomology: Let $$1 \to H \to G \to G/H \to 1$$ be an extension of groups. Then we obtain an extension with abelian kernel 
$$1 \to H\_{ab} \to G/H' \to G/H \to 1$$ Let $\varepsilon \in H^2(G/H;H\_{ab})$ be its extension class. If $M$ is a trivial $G$-module, then the differential (which equals the transgression) 
$$d\_2^{0,1}: E\_2^{0,1}=H^1(H;M)^{G/H} \to H^2(G/H;M) = E\_2^{2,0}$$ is given as follows: Let $f \in H^1(H;M)^{G/H} \le Hom(H,M)=Hom(H\_{ab},M)$. Since $f$ is $G/H$-invariant, we have a hom. of $G/H$-modules $f:H\_{ab}\to M$ and an induced hom. $f\_\ast: H^2(G/H;H\_{ab}) \to H^2(G/H;M)$. Then: 



> 
> $\hspace{120pt}d\_2^{0,1}(f) = f\_\ast(\varepsilon)$ 
> 
> 
> 


A good reference for this is Theorem 2.1.8 in Neukirch et. al.: Cohomology of Number Fields. 


In case of $M=\mathbb{F}\_p$, Kudo's transgression theorem may also be of relevance. 


In case of homology you can try to dualize the result above. But I personally would always prefer to use cohomology, since here the cup product is available that is very helpful in computing spectral sequences. 


