The following sentence is quoted from the paper *ON THE COHOMOLOGY OF SPLIT EXTENSIONS* by D. J. BENSON AND M. FESHBACH:


**In general, the differentials in the Lyndon-Hochschild-Serre spectral sequence
for a group extension are difficult to understand. In the case of a central extension,
the $E\_2$ page is easy to calculate and the transgressions may be computed
using the fact that they commute with the Steenrod operations on the base and
fiber.**


May I know what is the complete statement of the result "the transgressions commute with the Steenrod operations on the base and fiber" and which reference can I refer to for it? And why this makes calculation easy? Can anyone give some examples illustrating that?



Transgressions commute with the Steenrod operations on the base and fiber in a central group extension?
The statement in question refers to the Kudo-Serre transgression theorem: If $E\_r$ is, for example, the LHS spectral sequence of a group extension and $x \in E\_{2k+1}^{0,2k}$ is transgressive with $d\_{2k+1}(x)=y$, then $$d\_i(x^p)=0\;\;(i=2k+1,...,2kp)\quad \text{and }\;\; d\_{2kp+1}(x^p)=P^ky$$ where $P^k$ denotes the Steenrod power operation (a textbook reference is McCleary: A User's guide to spectral sequences, Theorem 6.14). 


I'll give an example that is related to the Benson-Feshbach paper: Let $p$ be a prime and let $$1 \to C \to G \to Q \to 1$$ be a central extension with $C \cong (\mathbb{Z}/p)^n$ elementray abelian. If $k$ is a field of char. $p$ (with trivial $G$-action), than $Q$ acts trivially on $H^\ast(C;k)$ and by the universal coefficient theorem, the LHS spectral sequence satisfies 
$$E\_2^{ij} \cong H^i(Q;k) \otimes\_k H^j(C;k).$$
Moreover, $E\_2 \cong H^\ast(Q;k) \otimes\_k H^\ast(C;k)$ is an isomorphism of $k$-algebras (up to a sign). This is what is ment by B-F when they say the $E\_2$-page is easy to compute. 


The cohomology of $C$ is given by $H^\ast(C;k) = k[x\_1,...,x\_n,y\_1,...,y\_n]$ with $\deg(x\_i)=1, \deg(y\_j)=2$ and the relations $x\_i^2=y\_i$ if $p=2$ and $x\_i^2=0$ for $p$ odd. Note that $y\_i = \beta(x\_i)$ where $\beta$ is the Bockstein homomorphism. 


In terms of the spectral sequence, we have $x\_i \in E\_2^{0,1}$. Hence $d\_2(x\_i) =: z\_i \in E\_2^{2,0}=H^2(Q;k)$, i.e. $x\_i$ is transgressive. Now Kudo's theorem says that $d\_2, \beta$ commute, i.e.  

$$d\_2(y\_i) = d\_2(\beta(x\_i))=\beta(d\_2(x\_i))=\beta(z\_i) \in H^3(Q;k).$$
Hence $y\_i$ is again transgressive and by Kudo's theorem $d\_5(y\_i^2)=P^1\beta(z\_i).$ Continuing this way, one obtains 
$$d\_{2p^k+1}(y\_i^{p^k})=P^{p^{k-1}}P^{p^{k-2}}\cdots P^1\beta(z\_i).$$
As a result, the differentials on the fiber $E\_\ast^{0,\ast}$ are completely determined by the values of $z\_i \in H^2(Q;k)$ (the base) under the power operations. 


