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As a graded $\mathbb{Z}$-module, the structure of the group cohomology $H^{*}(\mathbb{Z}/n\mathbb{Z};\mathbb{Z})$ is extremely well-known. Yet, I am having difficulty finding a reference concerning its cup product structure. I assume this is also well-known, but I would appreciate a reference containing a precise statement of the $\mathbb{Z}$-algebra structure.

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3 Answers 3

up vote 7 down vote accepted

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 $H^*(\mathbb{Z}/n,\mathbb{Z}/n) \cong \mathbb{Z}/n [ \alpha, \beta]/(\alpha^2-k\beta)$ (where $k=0$ if $n$ is odd, and $k=n/2$ if $n$ is even, see Hatcher's comment below, and this is an graded-commutative 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 $n$).

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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 1-dimensional 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 $H^\ast({\mathbb Z}/n,{\mathbb Z}/n)={\mathbb Z}/n[\alpha,\beta]/(\alpha^2-k\beta)$. –  Allen Hatcher Jun 18 '13 at 14:07
Comment on my previous comment: It is an interesting little exercise to determine which powers of $\alpha$ are nonzero. –  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 odd-degree 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 Hilton-Stammbach, 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 $\zeta^p$.

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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 $\cdots\rightarrow\mathbb{Z}G\stackrel{t-1}{\rightarrow}\mathbb{Z}G\stackrel{N}{\rightarrow}\mathbb{Z}G\stackrel{t-1}{\rightarrow}\mathbb{Z}G\stackrel{\varepsilon}{\rightarrow}\mathbb{Z}\rightarrow 0$, where $N=\sum_{i=0}^{n-1} t^i$ 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
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\sum_{0\le i < j\le n-1}t^i\otimes t^j\;\;\;\;$ 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 $-\sum_{0\le p < q\le n-1} t^pm\otimes t^qm'$ when $i$ and $j$ are both odd.

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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 250-252) 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). –  Ricardo Andrade Jun 18 '13 at 23:56

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