Question is as stated in the title.
What is cup product in group cohomology is all about?
I am looking at the following books
- Class field theory - Bonn lectures Neukirch
- Cohomology of Number fields - Neukirch
- Cohomological methods in Group theory - Ararat Babakhanian
Let $G$ be a finite group and $A$ be a $G$ module.
I was told by some one that cup products gives a graded ring structure on Homology groups $$\{H^q(G,A)\}_{q\geq 0}$$
As we have more structre, we can understand the group $G$ and module $A$ better.
Given $G$ modules $A$ and $B$ we have $G$ module $A\otimes B$ and for each $p,q\geq 0$ we have a bilinear map (a cup product) as follows
$$H^p(G,A)\times H^q(G,B)\xrightarrow{\cup} H^{p+q}(G,A\otimes B)$$
Though the definitions are not same in these three books, let us assume we do have a product with some properties.
So, how does this give a graded ring structure on cohomology groups?
As $p^{\rm{th}}$ cohomology (of some module) and $q^{\rm{th}}$ cohomology (of another module) is mapping to $(p+q)^{\rm{th}}$ cohomology (of some another module) there is some hope to get some graded structure.
As we are looking for ring structure on $\{H^q(G,A)\} $ we expect to consider special case when $A=B$. So, we have
$$H^p(G,A)\times H^q(G,A)\xrightarrow{\cup} H^{p+q}(G,A\otimes A)$$
So, we have a product into module $A\otimes A$ but we want a map into module $A$.
I do not see any obvious map from $H^{p+q}(G,A\otimes A)$ to $H^{p+q}(G,A)$
We do have some obvious map $A\rightarrow A\otimes A$ but this gives a map from $H^{p+q}(G,A)$ to $H^{p+q}(G,A\otimes A)$ but not the other way.
So, how do we make this into a ring? And, What is this cup product in group cohomology all about?
I have seen some questions related to cup products in stack exchange but could not understand it better.