As a graded $\mathbb{Z}$module, the structure of the group cohomology $H^{*}(\mathbb{Z}/n\mathbb{Z};\mathbb{Z})$ is extremely wellknown. Yet, I am having difficulty finding a reference concerning its cup product structure. I assume this is also wellknown, but I would appreciate a reference containing a precise statement of the $\mathbb{Z}$algebra structure.
See Example 3.41, p. 251 of Hatcher's Algebraic Topology. This example computes the cohomology ring of finite Lens spaces. Since the infinite Lens space is a $K(\mathbb{Z}/n,1)$, and is a increasing union of finite Lens spaces, the same computation holds for the infinite Lens spaces.
Edit: My original answer was incomplete, since Example 3.41 computes the ring $H^*(\mathbb{Z}/n,\mathbb{Z}/n) \cong \mathbb{Z}/n [ \alpha, \beta]/(\alpha^2k\beta)$ (where $k=0$ if $n$ is odd, and $k=n/2$ if $n$ is even, see Hatcher's comment below, and this is an gradedcommutative ring with $\alpha$ of degree $1$, and $\beta$ of degree $2$), whereas you want $H^*(\mathbb{Z}/n,\mathbb{Z})$.
We have $H^{2i}(\mathbb{Z}/n,\mathbb{Z}) \cong \mathbb{Z}/n$ for $i>0$, $H^0(\mathbb{Z}/n,\mathbb{Z})\cong \mathbb{Z}$, and $H^{2i+1}(\mathbb{Z}/n,\mathbb{Z}) \cong 0$ by universal coefficients (or Poincare duality). Let $\beta\in H^2(\mathbb{Z}/n,\mathbb{Z})$ be a generator, then the map from $H^*(\mathbb{Z}/n,\mathbb{Z})\to H^*(\mathbb{Z}/n,\mathbb{Z}/n)$ is a ring isomorphism for even degrees above zero, so we see that $H^*(\mathbb{Z}/n,\mathbb{Z})\cong \mathbb{Z}[\beta]/(n\beta)$, where $\beta$ has degree $2$ (and is torsion of order $n$).

3$\begingroup$ Just for the record, here's a minor correction which doesn't affect the overall argument in this answer. In $H^\ast({\mathbb Z}/n,{\mathbb Z}/n)$ the 1dimensional generater $\alpha$ satisfies $2\alpha^2=0$ by commutativity of cup product, which forces $\alpha^2=0$ when $n$ is odd but not when $n$ is even. In fact $\alpha^2\neq 0$ when $n$ is even. This is well known for $n=2$, and for larger even $n$ this is shown in Example 3.9 in my book. Thus for $n=2k$ we have $H^\ast({\mathbb Z}/n,{\mathbb Z}/n)={\mathbb Z}/n[\alpha,\beta]/(\alpha^2k\beta)$. $\endgroup$ – Allen Hatcher Jun 18 '13 at 14:07

$\begingroup$ Comment on my previous comment: It is an interesting little exercise to determine which powers of $\alpha$ are nonzero. $\endgroup$ – Allen Hatcher Jun 18 '13 at 14:09
I don t recall seeing the computation, but you can almost immediately obtain a description as follows. Let me write $C$ for the infinite cyclic group, so as not to have too many $\mathbb Z$s.
Consider the Lyndon–Hochschild–Serre spectral sequence corresponding to the extension of groups $$0\to C\to C\to\mathbb Z/n\mathbb Z\to0$$ whose $E_2$ page looks like $H^p(\mathbb Z/n\mathbb Z,H^q(C,\mathbb Z))\Rightarrow H^\bullet(C,\mathbb Z)$. It is very easy to see that $H^q(C,\mathbb Z)$ is $\mathbb Z$ for $q\in\lbrace 0,1\rbrace$ and zero otherwise.
It follows now from convergence of the spectral sequence and the distribution of the zeroes in the $E_2$ page (we only have two rows) and in the limit, that the differential $d_2^{p,1}:H^p(\mathbb Z/n\mathbb Z,\mathbb Z)\to H^{p+2}(\mathbb Z/n\mathbb Z,\mathbb Z)$ is surjetive for $p=0$ and an isomorphism for $p>0$. Now $d_2$ is given by the cup product with the class $\zeta=d_2^{0,1}(1)\in H^2(\mathbb Z/n\mathbb Z,\mathbb Z)$ (Here the $1$ is that of $H^0(\mathbb Z/n\mathbb Z,\mathbb Z)$)
This is enough to get the whole ring structure. Indeed, the spectral sequence degenerates at $E_2$, so gives us an exact sequence (values in $\mathbb Z$ everywhere) $$0\to H^1(\mathbb Z/n\mathbb Z)\to H^1(C)\to H^0(\mathbb Z/n\mathbb Z)\xrightarrow{\zeta\cup(\mathord)}H^2(\mathbb Z/n\mathbb Z)\to0$$ and isomorphisms $$\zeta\cup(\mathord):H^p(\mathbb Z/n\mathbb Z))\to H^{p+2}(\mathbb Z/n\mathbb Z)$$ for all $p\geq1$. We know that $H^1(\mathbb Z/n\mathbb Z)$ is torsion, and according to the exact sequence it is also a subgroup of $H^1(C)=\mathbb Z$, so $H^1(\mathbb Z/n\mathbb Z)$, and along with it all the odddegree groups, is zero.
The map $H^1(C)\to H^0(\mathbb Z/n\mathbb Z)$ is easily computed (it is one of the maps appearing in the usual «$5$term sequence», as in HiltonStammbach, Th. VI.8.1) to be multiplication by $n$, so $H^2(\mathbb Z/n\mathbb Z)\cong \mathbb Z/n\mathbb Z$ generated by the class $\zeta$. The isomorphisms above then imply that $H^{2p}(\mathbb Z/n\mathbb Z)$ is cyclic of order $n$, generated by $\zeta^p$.
Well, it's kind of a homework exercise. In particular, in Ken Brown's "Cohomology of Groups" it's Exercise V.3.2 which recommends using diagonal approximation. Here is my solution for completeness, using only the basic machinery:
Let $G=\langle t\rangle$ be a finite cyclic group of order $n$ and let $F$ be the periodic resolution $\cdots\rightarrow\mathbb{Z}G\stackrel{t1}{\rightarrow}\mathbb{Z}G\stackrel{N}{\rightarrow}\mathbb{Z}G\stackrel{t1}{\rightarrow}\mathbb{Z}G\stackrel{\varepsilon}{\rightarrow}\mathbb{Z}\rightarrow 0$, where $N=\sum_{i=0}^{n1} t^i$ is the norm element. Let $\Delta:F\rightarrow F\otimes F$ be the diagonal approximation map whose $(p,q)$component $\Delta_{pq}:F_{p+q}\rightarrow F_p\otimes F_q$ is given by $\Delta_{pq}(1) =$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;1\otimes 1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ for $p$ even
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;1\otimes t\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ for $p$ odd and $q$ even
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\sum_{0\le i < j\le n1}t^i\otimes t^j\;\;\;\;$ for $p$ and $q$ odd
(Check that this is a diagonal approximation!)
Consider the cohomology groups $H^{2r}(G,M)\cong M^G/NM$ and $H^{2r+1}(G,M')\cong Ker(N:M'\rightarrow M')/IM'$ where $I=\langle t  1\rangle$ is the augmentation ideal of $G$.
The cup product in $H^*(G,M\otimes M')$ is given by $u\smallsmile v=(u\times v)\circ\Delta$ with $\langle u\times v,x\otimes x'\rangle=(1)^{deg(v)\cdot deg(x)}\langle u,x\rangle\otimes\langle v,x'\rangle$.
Choose representatives $\langle u,x\rangle=m\in M$ of $H^i(G,M)$ with $m\in M^G$ for $i$ even and $m\in Ker(N:M\rightarrow M)$ for $i$ odd, and choose representatives $\langle v,x'\rangle=m'\in M'$ of $H^j(G,M')$ with $m'\in M'^G$ for $j$ even and $m'\in Ker(N:M'\rightarrow M')$ for $j$ odd.
If $i$ is even then $(1)^{deg(v)\cdot deg(x)}=(1)^{deg(v)\cdot i}=1$, and if $j$ is even then $(1)^{deg(v)\cdot deg(x)}=(1)^{j\cdot deg(x)}=1$, and if both $i$ and $j$ are odd then $(1)^{deg(v)\cdot deg(x)}=(1)^{j\cdot i}=1$.
Thus the cup product element of $H^{i+j}(G,M\otimes M')$ is represented by $m\otimes m'$ for $i$ or $j$ even and is represented by $\sum_{0\le p < q\le n1} t^pm\otimes t^qm'$ when $i$ and $j$ are both odd.

4$\begingroup$ Since the question asked for a reference, I will say that Chris Gerig's calculation esssentially appears with some more elaborations in section 7 of chapter XII (pages 250252) of the book "Homological algebra" by Cartan and Eilenberg. There the calculation is phrased in terms of Tate cohomology, which agrees with group cohomology in positive degrees (including the ring structure). $\endgroup$ – Ricardo Andrade Jun 18 '13 at 23:56

$\begingroup$ I get stuck with this computation, especially about the sign convention. Given this diagonal approximation, do we succeed to find a commutative differential graded algebra representing $\mathbb Z^{hG}\simeq\operatorname{RHom}_{\mathbb ZG}(\mathbb Z,\mathbb Z)$? $\endgroup$ – Yai0Phah May 14 '19 at 18:30