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 $H^*(\mathbb{Z}/n,\mathbb{Z}/n) \cong \mathbb{Z}/n [ \alpha, \beta]/(\alpha^2)$, 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 $n$).

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 $n$).

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.

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 $H^*(\mathbb{Z}/n,\mathbb{Z}/n) \cong \mathbb{Z}/n [ \alpha, \beta]/(\alpha^2)$, 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 $n$).