= Solution
The space $K(\mathbb Z/n,1)$ is a classifying space $B(\mathbb Z/n)$. The periodic resolution of a finite cyclic group gives
$$
H^*(K(\mathbb Z/n,1);\mathbb Z)
\cong\mathbb Z[c]/(nc),
\qquad |c|=2.
$$
Thus 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, so
$$
H^*(K(\mathbb Z/n,1);\mathbb F_p)=\mathbb F_p.
$$
If $p\mid n$, restriction to the cyclic Sylow $p$-subgroup and transfer give
$$
H^*(K(\mathbb Z/n,1);\mathbb F_p)
\cong\Lambda(u)\otimes\mathbb F_p[v],
\qquad |u|=1,\quad |v|=2.
$$
For $n=p$, 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 $f:X\to K(A,2)$ gives
$$
1\longrightarrow A\longrightarrow G:=\pi_1(F)
\longrightarrow A^2\longrightarrow1
$$
and $\pi_k(F)=0$ for $k>1$. Therefore $F$ is a $K(G,1)$ and $|G|=p^3$.
Regard it as the fibration
$$
K(A,1)\longrightarrow F\longrightarrow K(A,1)^2.
$$
Write
$$
H^*(K(A,1)^2;\mathbb F_p)
=\Lambda(u_1,u_2)\otimes\mathbb F_p[v_1,v_2]
$$
and write $w,t$ for the degree-one and degree-two generators of the fibre. The fibration is classified by $u_1u_2$, so in its cohomological <Serre spectral sequence>
$$
d_2(w)=u_1u_2.
$$
The <Kudo transgression theorem> and the mod-$p$ Bockstein give
$$
d_3(t)=\beta(u_1u_2)=v_1u_2-u_1v_2\ne0.
$$
Consequently $H^1(F;\mathbb F_p)$ has basis $u_1,u_2$, while $H^2(F;\mathbb F_p)$ has the four surviving classes represented by
$$
v_1,\quad v_2,\quad u_1w,\quad u_2w.
$$
If $G$ 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.
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