I'm trying to understand the Bockstein morphism in cohomology, and one of the points is that $\delta : H^*(G,\mathbb{F}_p)\to H^*(G,\mathbb{F}_p)$ is a derivation that squares to $0$.
I could already prove that its square is $0$ by relating it to the other Bockstein morphism $\beta : H^*(G,\mathbb{F}_p)\to H^*(G,\mathbb{Z})$, but I could not prove that it is a derivation.
I did manage to prove that for $\alpha, \beta$, $\delta\alpha \cup \beta + (-1)^{|\alpha|}\alpha\cup\delta\beta$ is a $\delta$-cocycle, but of course that's far from enough.
I know that for $u$ coming from $\mathbb{Z/p^2Z}$, $\delta (\alpha \cup u) = \delta\alpha \cup u$ (and a symmetric formula with a sign for $u\cup\alpha$) by general knowledge on connecting morphisms; but that's not enough either obviously
I'd like to have an indication on how to prove it (if possible not a complete solution)
EDIT : I actually thought of something: the Künneth formula gives us a morphism $H^*(C)\otimes H^*(C')\to H^*(C\otimes C')$, so in particular here we have $\displaystyle\bigoplus_{p+q=n}H^p(G,\mathbb{F}_p)\otimes H^q(G,\mathbb{F}_p) \to H^n(G\times G, \mathbb{F}_p)$; and here we would have to get into the details of the construction of the connecting morphism, but it seems clear that it should "commute" with this Künneth morphism, which would imply the desired result. So is it true that the Künneth morphism commutes with the connecting morphism ?
EDIT 2 : Trying to detail what I wrote in the first edit, so expliciting the connecting morphism, I think I found what made it work : if $f$ denotes the inclusion $\mathbb{F}_p\to \mathbb{Z/p^2Z}$, and $g$ the reduction mod $p$, we have $f(a)b= f(ag(b))$. Then using this and actually computing the connecting morphism on the cross-product of cocycles, using the well-known results in constructing the connecting morphism that state that the result doesn't depend on the choices we make, we see that it actually commutes. Unless someone provides a better answer in a short time, I will at some point write this as an answer to remove this question from the unanswered queue (and provide a reference for future people who would want to know this)
