# Higher Bockstein maps in group cohomology

Let $$p$$ be an odd prime and $$n>1$$. I am trying to understand why the cohomology ring $$H^{\ast}(\mathbb{Z}/p^n;\mathbb{F_p})$$ is given by $$\mathbb{F}_p[y]\otimes\Lambda(x),$$ with $$|x|=1,|y|=2$$ and $$\beta_n(x)=y$$, where $$\beta_n:H^{1}(\mathbb{Z}/p^n;\mathbb{F_p})\to H^{2}(\mathbb{Z}/p^n;\mathbb{F_p})$$ is the $$n$$th Bockstein map. I have seen a few abstract definitions of $$\beta_n$$ as a higher cohomology operation or as a higher differential in the Bockstein spectral sequence. The first questions that come to my mind are:

-Can we describe explicitly $$\beta_n(x)=y$$ as a $$2$$-cocycle?

-Is the group $$E=\mathbb{Z}/p\times_{y}\mathbb{Z}/p^n$$ determined by $$y$$ abelian?

Of course, the latter question would follow from the former.

• I think you have $x$ and $y$ reversed. The polynomial generator should have degree two, while the exterior generator should have degree one. Commented Aug 8 at 10:27
• @DaveBenson you are right, sorry for the typo. I edited it. Commented Aug 8 at 10:32
• So now yes, you can write $y$ as a cocycle using the usual carry digit for addition, reduced mod $p$. I'm not sure it's very helpful in understanding the Bockstein, but the corresponding extension is $\mathbb{Z}/p^{n+1}$. Commented Aug 8 at 10:36
• @DaveBenson so it works exactly the same like the case $n=1$, thanks! Commented Aug 8 at 11:19
• @DaveBenson In other words, $B\mathbb{Z}/p^n$ and $B\mathbb{Z}/p$ are indistinguishable through the eyes of mod-$p$ cohomology, unless we consider higher structure. Commented Aug 8 at 11:27

The homology of $$\mathbb{Z}/p^n$$ with integral coefficients is computed by the chain complex $$\dots 0 \to \mathbb{Z}\stackrel{0}{\to} \mathbb{Z}\stackrel{p^n}{\to} \mathbb{Z} \stackrel{0}{\to} \dots.$$ Let $$e_i$$ be the generator of the $$i$$-th term of this complex, and let $$\overline{e_i}$$ be its image in the reduction modulo $$p$$. Then $$\delta(e_1) = p^n e_2$$. The way $$\beta_n$$ is computed on the homology of $$C/p$$ for $$C$$ a chain complex with $$p$$-torsion free terms is by applying the differential $$\delta$$ and dividing by $$p^n$$. This shows that $$\beta_n(\overline{e_1}) = \overline{\frac{p^ne_2}{p^n}} = \overline{e_2}$$, which is the generator of $$H^2(\mathbb{Z}/p^n;\mathbb{F}_p)$$.
The central extension corresponding to this cocycle is $$\mathbb{Z}/p^{n+1}$$, and it is indeed abelian. This can be seen for example from the elementary fact that if the quotient of a group by a central subgroup is cyclic then the group is abelian; choose a lift of the generator of the quotient, it commutes with the kernel of the quotient and hence everything's commutes.