What's the ring structure of $H^*(C_2,\mathbb{Z})$ with a nontrivial group action? Right now, I am trying to understand better the cup product structure. I am interested in deriving the ring structure of the group cohomology $H^*(C_2,\mathbb{Z})$ with a nontrivial group action. The group action of a nontrivial element of $C_2$ sends an element $e\in \mathbb{Z}$ to $-e$. The cohomology groups are easily to obtain. The result is that $H^n(C_2,\mathbb{Z})=\mathbb{Z}/2\mathbb{Z}$ for $n$ an odd number and $H^n(C_2,\mathbb{Z})=0$, otherwise.
I think, basically, my question can be boiled down to some smaller question.


*

*Is there a cup product structure for nontrivial group actions? Since people normally talk about group cohomology with trivial group actions, I am wondering whether it works when a group action is nontrivial. Do you need to modify the cup product formula?

*Is there any good procedure to follow to obtain a cohomology ring from cohomology groups?

*It seems that in this case it could be a ring with infinite number of generators with odd degrees and the cup product of any two generators vanishes. Am I correct?
 A: Let me start with a brief correction. For the graded group $H^*(G; M)$ to carry a ring structure, you should demand that $M$ is a ring and, most frequently, that the action of $G$ on $M$ is trivial. However, you do have the following structure more generally. Given $G$-modules $M$ and $N$, there is always a map $$H^*(G;M) \otimes H^*(G; N) \to H^*(G; M \otimes N).$$ So you can ask about what this map is for various values of $M$ and $N$. (If the action of $G$ on $M$ is non-trivial, but there is still a $G$-map $M \otimes M \to M$ satisfying the usual associativity rule, then $H^*(G;M)$ is still a ring. 
Of interest to you are $C_2$ and what I will write as $\Bbb Z_+$ and $\Bbb Z_-$, where the nontrivial element acts either as $\times (+1)$ or $\times (-1)$; that is, the action on $\Bbb Z_+$ is trivial and the action on $\Bbb Z_-$ is negation. For computational reasons, we will also be interested in the case of $M = \Bbb Z/2$, the field with two elements and (necessarily) trivial action. Note that the three things we get are (ignoring ordering, which is essentially inconsequential): 
\begin{align*}H^*(C_2; \Bbb Z_+) \otimes H^*(C_2; \Bbb Z_+) &\to H^*(C_2; \Bbb Z_+)\\
H^*(C_2; \Bbb Z_+) \otimes H^*(C_2; \Bbb Z_-) &\to H^*(C_2; \Bbb Z_-)\\
H^*(C_2; \Bbb Z_-) \otimes H^*(C_2; \Bbb Z_-) &\to H^*(C_2; \Bbb Z_+)\\
\end{align*}
So $H^*(C_2; \Bbb Z_+)$ is a ring, $H^*(C_2; \Bbb Z_-)$ is a module over that ring, and there is a pairing $$H^*(C_2; \Bbb Z_-) \otimes H^*(C_2; \Bbb Z_-) \to H^*(C_2; \Bbb Z_+),$$ which is compatible with all of the module structures in an appropriate sense, but don't worry about that.

I take as a given the following fact: as a ring, we have $H^*(C_2; \Bbb Z/2) = \Bbb Z/2[x]$, where $x$ lives in degree $1$. In particular, $H^n(C_2; \Bbb Z/2) = \langle x^n\rangle$.
Next, allow me to make a point. The tensor product up above is natural in the modules $M$ and $N$, in the sense that if you have a $G$-module homomorphism $M \to M'$ and $N \to N'$, it induces maps $H^*(G; M) \to H^*(G; M')$ and similarly with $N'$, and the diagram you get comparing the two products commutes. 
OK! Now let's do some computation. Using the resolution given in this post, you can compute the chain complexes computing $H^*(C_2; \Bbb Z_+)$ and $H^*(C_2; \Bbb Z_-)$ are the following, in that order: 
$$\cdots \to \Bbb Z \xrightarrow{0} \Bbb Z \xrightarrow{2}\Bbb Z \xrightarrow{0}\Bbb Z \xrightarrow{2}\Bbb Z \xrightarrow{0}\Bbb Z \to 0 $$
$$\cdots \to \Bbb Z \xrightarrow{2}\Bbb Z \xrightarrow{0}\Bbb Z \xrightarrow{2}\Bbb Z \xrightarrow{0}\Bbb Z \xrightarrow{2} \Bbb Z \to 0 $$
Therefore, at the level of cohomology we have, for each $n \geq 0$,
\begin{align*}H^{2n+1}(C_2; \Bbb Z_-) &= \Bbb Z/2\\
H^{2n+2}(C_2; \Bbb Z_+) &= \Bbb Z/2\\
H^0(C_2; \Bbb Z_+) &= \Bbb Z
\end{align*}
and otherwise zero. 
Now, we also need to know the map $H^*(C_2; \Bbb Z_\pm) \to H^*(C_2; \Bbb Z/2)$ induced by $\Bbb Z_\pm \to \Bbb Z/2$ given by reduction by $2$. Using the same (given) resolution, we find that the induced map between chain complexes defining these is precisely given by reducing each $\Bbb Z$ to $\Bbb Z/2$ (which has the result of giving us a complex with no differentials). In particular, the maps $H^*(C_2; \Bbb Z_\pm) \to H^*(C_2, \Bbb Z/2)$ has precisely the result of reduction mod 2. Therefore, the map $H^{2n+1}(C_2; \Bbb Z_-) \to H^{2n+1}(C_2; \Bbb Z/2)$ is an isomorphism for all $n \geq 0$, and similarly $H^{2n+2}(C_2; \Bbb Z_+) \to H^{2n+2}(C_2; \Bbb Z/2)$.

What is this good for? As a warm up, observe that it implies that if we write $y_k \in H^{2k}(C_2; \Bbb Z_+)$ for the nonzero element, we see that $y_k$ reduces to $x^{2k}$; therefore, because $y^k$ reduces to $(x^2)^k$ mod $2$ (because the product structure defined at the top of this post is natural), we see that $y_k = y^k$, and therefore $H^*(C_2; \Bbb Z_+) = \Bbb Z[y]/(2y)$, where $|y| = 2$. 
Our next goal is to understand the action of $y$ on $H^*(C_2; \Bbb Z_-)$. Write $z_k \in H^{2k+1}(C_2; \Bbb Z_-)$ for the nontrivial element, as before. Then our computation from the previous section showed that $z_k$ reduces to $x^{2k+1}$, and compatibility of the product structure means that we know $(y^\ell) \cdot (z_k)$ must reduce to $(x^{2\ell}) \cdot (x^{2k+1}) = x^{2\ell + 2k + 1}$; the only thing that reduces to this is $z_{k+\ell}$, so we see that $y^\ell \cdot z_k = z_{k+\ell}$. 
Inspired by this, we rewrite $z := z_0$, the element generating $H^1(C_2; \Bbb Z_-)$, and we see that as an $H^*(C_2; \Bbb Z_+)$-module, we may write $H^*(C_2; \Bbb Z_-) = \Bbb Z/2[y] \cdot z;$ that is, every element is of the form $y^k z$ for some $k$, and the action of $y^\ell$ is simply to multiply that power.
Lastly, we want to compute the pairing $$H^*(C_2; \Bbb Z_-) \otimes H^*(C_2; \Bbb Z_-) \to H^*(C_2; \Bbb Z_+).$$
Now we already know that $y^k z$ reduces to $x^{2k+1}$ and $y^\ell z$ reduces to $x^{2\ell+1}$. So by compatibility of product structure, we see that $(y^k z) \cdot (y^\ell z)$ should reduce to $x^{2k+2\ell + 2}$, and so $$(y^k z) \cdot (y^\ell z) \mapsto y^{k+\ell+1}.$$ If you like, you could write this as $z^2 = y$, but this is a little sketchy, since $z$ and $y$ live in different rings, so you're using the pairing above while not making it explicit in notation.
