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Write $u=u_n\in H^n(K(\mathbb{Z}/2,n);{\mathbb{F}}_2)$ for the fundamental class. We use the path-loop Serre spectral sequence, its multiplicative structure, and the Kudo transgression theorem: transgression of the fundamental classes and their compatible Steenrod squares is natural, equivalently cohomology suspension commutes with these stable operations. The initial ring is $H^{\ast }(K(\mathbb{Z}/2,1);{\mathbb{F}}_2)={\mathbb{F}}_2[t_1]$.

Here is the low-degree calculation. In the first path fibration, $t$, $t^2$ and $t^4$ transgress respectively to $u_2$, ${\mathrm{Sq}}^1{u}_2$ and ${\mathrm{Sq}}^2{\mathrm{Sq}}^1{u}_2$, in degrees $2,3,5$. These supply all indecomposable base classes through degree five; the other classes there are products. Repeating the path-loop calculation raises the degree of the transgressive classes by one. For $n=3$, the class ${\mathrm{Sq}}^2{u}_3$ transgresses from $u_2^2$, and ${\mathrm{Sq}}^2{\mathrm{Sq}}^1{u}_3$ from the degree-five generator. For $n=4$, ${\mathrm{Sq}}^3{u}_4$ transgresses from $u_3^2$. For larger $n$ all four displayed low-degree operations are transgressive indecomposables. Acyclicity of the path-space total cohomology forces these transgressions and excludes additional classes in this range.

More systematically, this is the range up to $n+3$ of the Serre polynomial generators for mod-two Eilenberg–MacLane cohomology:For an admissible sequence of Steenrod squares, $I=(i_1,\dots ,i_r)$ has $i_j\ge 2i_{j+1}$; its degree increment is $\sum i_j$, and its Steenrod excess is $e(I)=i_1-i_2-\cdots -i_r$. Include the empty sequence. Up to increment three the nonempty possibilities are $(1),(2),(3),(2,1)$; the strict excess bound and possible products explain precisely the small-$n$ exceptions.

The low-degree mod-two cohomology of an Eilenberg–MacLane space isNegative-degree cohomology is zero. For $n=2$, ${\mathrm{Sq}}^{2}u=u^2$ and ${\mathrm{Sq}}^{3}u=0$ by instability, so the product in degree five must not be omitted. For $n=3$, ${\mathrm{Sq}}^{3}u=u^2$ is a product, still independent from ${\mathrm{Sq}}^2{\mathrm{Sq}}^{1}u$. For $n\ge 4$ the listed classes are indecomposable. The Adem relations include ${\mathrm{Sq}}^1{\mathrm{Sq}}^1=0$ and ${\mathrm{Sq}}^1{\mathrm{Sq}}^2={\mathrm{Sq}}^3$, so no further increment-three class comes from reversing the two squares.

Now use the space actually printed in the PDF,the converted TeX dropped the projective-space $P$'s. This stunted real projective space has one cell in dimensions $7,8,9,10$, besides its basepoint, and is $6$-connected. The integral cellular boundary is $2$ in even dimensions and zero in odd dimensions. HenceThe Hurewicz theorem gives ${\pi }_7(Y)=\mathbb{Z}/2$.

For the next two groups, let $K=K(\mathbb{Z}/2,7)$ and choose $f:Y\to K$ representing the generator of $H^7(Y;{\mathbb{F}}_2)$. It induces an isomorphism on ${\pi }_7$, and the target's higher homotopy groups vanish.

We need the low-degree integral homology of a mod-two Eilenberg–MacLane space. Here it isTo justify the torsion orders, the Serre class theorem first makes every positive-degree integral homology group of $K$ a finite 2-group. The universal coefficient theorem for cohomology with ${\mathbb{F}}_2$ and the low-degree dimensions give one cyclic summand in degrees seven, nine and ten and none in degree eight. The mod-two Bockstein homomorphism is ${\mathrm{Sq}}^1$. Its relevant nonzero actions areThe last target is nonzero: $(3,1)$ is admissible with excess two, below seven. It also follows by transgressing the nonzero square $({\mathrm{Sq}}^1{u}_2{)}^2$ through successive path fibrations. Each nonzero Bockstein pairs the mod-two classes associated with a cyclic integral summand of order exactly two; for a summand of order $2^r$ with $r>1$, this first Bockstein would be zero. In degree ten, ${\mathrm{Sq}}^{3}u$ is already the Ext class from $H_9$, so the independent class ${\mathrm{Sq}}^2{\mathrm{Sq}}^{1}u$ detects $H_{10}$. This proves all four integral groups without confusing them with mod-two Betti numbers.

Let $x\in H^1({\mathbb{RP}}^{10};{\mathbb{F}}_2)$ be its usual generator. The quotient classes in degrees $7$ through $10$ identify with $x^7,\dots ,x^{10}$ via the pair's cohomology. The Steenrod squares on real projective space satisfyThussince $\left(\genfrac{}{}{0ex}{}{7}{2}\right)$ and $\left(\genfrac{}{}{0ex}{}{7}{3}\right)$ are odd, while $\left(\genfrac{}{}{0ex}{}{8}{2}\right)$ is even. In degree nine, the nonzero cohomology map detects the map on $H_9$, as $H_8=0$ on both sides. Consequently $f_{\ast }:H_9(Y;\mathbb{Z})\to H_9(K;\mathbb{Z})$ is an isomorphism.

Treat $f$ as a mapping-cylinder pair $(K,Y)$. It is $8$-connected: both spaces are $6$-connected, ${\pi }_7$ is an isomorphism, and ${\pi }_8(K)=0$. The relative homology sequence gives $H_9(K,Y)=0$, since $H_9(Y)\to H_9(K)$ is an isomorphism and $H_8(Y)=0$. The Relative Hurewicz theorem then gives ${\pi }_9(K,Y)=0$, and the relative homotopy sequence identifies this group with ${\pi }_8(Y)$. Thus ${\pi }_8(Y)=0$ and the pair is now $9$-connected.

Next,because $H_{10}(Y)=0$ and the degree-nine map is an isomorphism. Apply the Relative Hurewicz theorem again and use ${\pi }_{10}(K)={\pi }_9(K)=0$:Therefore the homotopy groups of the stunted projective space through degree nine are

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 $y\in H^{\ast }(Y;{\mathbb{F}}_2)$.