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For a topological group $G$, a universal principal bundle $EG\to BG$ has contractible total space and classifies numerable principal $G$-bundles by pullback. Over a paracompact base, isomorphism classes correspond to homotopy classes of maps to $BG$. For a discrete group, its classifying space of a discrete group is a $K(G,1)$; for the circle, $B{S}^1\simeq {\mathbb{CP}}^{\infty }$.

A connected Eilenberg–MacLane space $K(G,n)$, $n\ge 1$, has ${\pi }_n\cong G$ and all other positive-degree homotopy groups trivial. The group $G$ must be abelian when $n\ge 2$, but may be nonabelian when $n=1$. For an abelian group, the Dold–Kan correspondence constructs a simplicial model from the chain complex concentrated in degree $n$. A classifying space of a discrete group gives the general $K(G,1)$ case.

Choose a contractible free $G$-space $EG$, with $BG=EG/G=K(G,1)$ the classifying space of a discrete group. The Borel construction gives the fibrationBecause the action on the sphere is free, the map $E\to S^{n-1}/G$ has contractible fibre $EG$ and is a homotopy equivalence. The quotient is a closed $(n-1)$-dimensional manifold, so $H^k(E;\mathbb{F})=0$ for $k>n-1$.

The hypothesis on integral sphere cohomology makes the $G$-action on $H^{n-1}(S^{n-1};\mathbb{F})$ trivial as well. Thus the Serre spectral sequence has just two rows, both copies of $H^{\ast }(BG;\mathbb{F})$, in fibre degrees zero and $n-1$. If $u$ is the fibre generator, the only possible differential is $d_n$, and its transgression definesThe multiplicative differential rule gives $d_n(au)=(-1{)}^{|a|}a⌣\Delta$. Up to the graded sign, this is multiplication by $\Delta$. For $i>0$, the terms $E_{\infty }^{i,n-1}$ and $E_{\infty }^{i+n,0}$ must both vanish because their total degrees exceed $n-1$. The first vanishing says the relevant multiplication map has zero kernel; the second says it has zero cokernel. ThereforeThis is periodic group cohomology from a free sphere action. Degree zero was excluded for a reason: $E_{\infty }^{0,n-1}$ can contribute to the quotient's top cohomology.

For $C_p=\mathbb{Z}/p$, use its periodic free resolution with alternating maps $g-1$ and $N=1+g+\cdots +g^{p-1}$. Applying ${\mathrm{Hom}}_{\mathbb{Z}[C_p]}(-,{\mathbb{F}}_p)$ makes every differential zero, so $H^i(B{C}_p;{\mathbb{F}}_p)$ is one-dimensional in every nonnegative degree. The two-step shift of the resolution supplies a nonzero degree-two class $u$ whose cup product gives the periodicity isomorphisms. For odd $p$, let $v$ be a nonzero degree-one class. Graded commutativity gives $v^2=0$, and the shift gives nonzero $v{u}^k$ and $u^k$ in every degree. For $p=2$, the two resolution maps agree over ${\mathbb{F}}_2[C_2]$, giving a one-step shift whose degree-one class has nonzero powers. Hence the cohomology ring of a finite cyclic group over its prime field isHere the exterior factor is an exterior algebra and the polynomial factor is a polynomial ring. For odd $p$, $u$ may be chosen as the Bockstein homomorphism of $v$.

If $G$ contained $C_p\times C_p$, restriction would give that subgroup the same free, cohomologically trivial sphere action. It would therefore also have period $n$ in positive-degree cohomology. But $B(C_p\times C_p)\simeq B{C}_p\times B{C}_p$, and the Künneth theorem givesThese dimensions strictly increase, so the groups in degrees $i$ and $i+n$ cannot be isomorphic. Thus $G$ contains no subgroup isomorphic to $\mathbb{Z}/p\times \mathbb{Z}/p$ for any prime $p$.