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The space $K(\mathbb{Z}/n,1)$ is a classifying space $B(\mathbb{Z}/n)$. The periodic resolution of a finite cyclic group givesThus positive odd cohomology vanishes and every positive even group is $\mathbb{Z}/n$.

Let $p$ be odd. If $p\nmid n$, transfer makes multiplication by $n$ both zero and invertible on positive-degree cohomology, soIf $p\mid n$, restriction to the cyclic Sylow $p$-subgroup and transfer giveFor $n=p$, one may take $v=\beta u$, where $\beta$ is the mod-$p$ Bockstein homomorphism.

For the second part put $A=\mathbb{Z}/p$ and $X=K(A,1)\times K(A,1)$. The long exact homotopy sequence of the homotopy fibre of $f:X\to K(A,2)$ givesand ${\pi }_k(F)=0$ for $k>1$. Therefore $F$ is a $K(G,1)$ and $|G|=p^3$.

Regard it as the fibrationWriteand write $w,t$ for the degree-one and degree-two generators of the fibre. The fibration is classified by $u_1{u}_2$, so in its cohomological Serre spectral sequenceThe Kudo transgression theorem and the mod-$p$ Bockstein giveConsequently $H^1(F;{\mathbb{F}}_p)$ has basis $u_1,u_2$, while $H^2(F;{\mathbb{F}}_p)$ has the four surviving classes represented byIf $G$ were abelian, an abelian group of order $p^3$ mapping onto $A^2$ would be either $A^3$ or $\mathbb{Z}/p^2\times A$. The first has three-dimensional $H^1(-;{\mathbb{F}}_p)$; the second has two-dimensional $H^1$ but three-dimensional $H^2$. Both contradict the dimensions just calculated. Hence $G$ is nonabelian.

Let $i:F\hookrightarrow E$. A pairis transgressive pair when, in the long exact sequence of the pair $(E,F)$,where $H^{r+1}(B,b_0)$ is identified with reduced cohomology. In the Serre spectral sequence, this says that $x$ survives to the transgression andunder the edge identifications, modulo the usual earlier-differential indeterminacy.

The Kudo transgression theorem says that if $(x,y)$ is transgressive and $0\le j\le r$, thenis transgressive. To prove it, use relative Steenrod squares. Naturality givesand stability, equivalently compatibility with the suspension isomorphism, makes squares commute with the connecting map:Applying ${\mathrm{Sq}}^j$ to $\delta x={\pi }^{\ast }y$ proves the theorem. The properties used are naturality, stability, additivity, and the instability conditions ${\mathrm{Sq}}^{j}z=0$ for $j>|z|$ and ${\mathrm{Sq}}^{|z|}z=z^2$.

Let $x\in H^1({\mathbb{RP}}^{\infty };{\mathbb{F}}_2)$ be the generator. Instability givesThe Cartan formula says the total square is multiplicative, soComparing components yields the complete formulawith the binomial coefficient reduced modulo two.

Finally, ${\mathrm{Sq}}^1$ is the Bockstein homomorphism associated withIf a mod-two cocycle representing $y$ is lifted to an integral cochain $A$, write $dA=2B$. Then $B$ modulo two represents ${\mathrm{Sq}}^{1}y$. But $dB=0$, because integral cochains are torsion-free and $2dB=d^{2}A=0$. Thus $B$ itself is a cocycle lift, so its Bockstein vanishes. Thereforefor every space $Y$ and every $y\in H^{\ast }(Y;{\mathbb{F}}_2)$.