Crossed Homomorphisms Complements and First Cohomology — Examples

By [L1], it is enough to compute homomorphisms $C_2\to C_3$.

The generator of $C_2$ must map to an element whose order divides $2$, but [L2] says the only such element of $C_3$ is the identity. So every homomorphism is trivial, and therefore $H^1(C_2,C_3)=0$.

Repeatedly applying [L1] gives $z(t^n)=z(t)(t\cdot z(t))\cdots (t^{n-1}\cdot z(t))$ for every integer $n\ge 1$. So all positive powers of $t$ are determined by $z(t)$.

Since $1=z(1)=z(t^n{t}^{-n})=z(t^n)(t^n\cdot z(t^{-n}))$, the value on $t^{-n}$ is determined by the already known value on $t^n$. Therefore every value of $z$ is determined by $z(t)$.

If $p=2$, then ${\mathbb{F}}_2^{\times }=\{1\}$ is trivial, so the only crossed homomorphism ${\mathbb{F}}_2^{\times }\to {\mathbb{F}}_2$ is the zero map. Hence $H^1({\mathbb{F}}_2^{\times },{\mathbb{F}}_2)=0$ by [L4].

Suppose $p>2$. By [L2] and [L3], choose a generator $u$ of the cyclic group ${\mathbb{F}}_p^{\times }$. Any crossed homomorphism $z:{\mathbb{F}}_p^{\times }\to {\mathbb{F}}_p$ is determined by $m=z(u)$, so it suffices to show that $m$ is always principal.

Because $p>2$, the generator $u$ is not $1$, so $u-1\mathrm{\ne }0$ in the field ${\mathbb{F}}_p$. By [L2] it is therefore invertible. Choose $a\in {\mathbb{F}}_p$ with $(u-1)a=m$. Then the principal cocycle $g\mapsto g\cdot a-a$ agrees with $z$ on the generator $u$, hence everywhere by [L3]. Thus every crossed homomorphism is principal, and [L4] gives $H^1({\mathbb{F}}_p^{\times },{\mathbb{F}}_p)=0$.

Steps 1.1 and 2.1 show that $H^1({\mathbb{F}}_p^{\times },{\mathbb{F}}_p)=0$ for every prime $p$. Now [L1] shows that there is exactly one ${\mathbb{F}}_p$-conjugacy class of complements to the translation subgroup in $\mathrm{AGL}(1,p)$.

Define $z_a$ on every $g\in C_2$ by the displayed formula. Fact [L1] gives $(-a,1){\mathrm{\Gamma }}_z(a,1)={\mathrm{\Gamma }}_{z_a}$. So the two complements are conjugate by the kernel element $(-a,1)$.

Therefore they determine the same class under the bijection of [L2], exactly because their defining cocycles differ by the principal crossed homomorphism attached to $a$.

Under the trivial action, a cocycle is just a homomorphism $C_2\to S_3$. The image of the generator is therefore either the identity or an element of order $2$, that is, a transposition.

All transpositions are conjugate in $S_3$, so [L1] leaves exactly two orbits of cocycles: the trivial one and the orbit of any transposition-valued cocycle. Thus $H_{\mathrm{n}\mathrm{a}\mathrm{b}}^1(C_2,S_3)$ has two points, with the trivial one as basepoint. This matches the complement picture from [L2].

With the negation action, $z(t^2)=0=1+(-1)=z(t)+t\cdot z(t)$, so $z$ is a crossed homomorphism.

But $z(t)=1$ has infinite order in $\mathbb{Z}$, so $z$ cannot be an ordinary homomorphism from the order-two group $C_2$. Therefore the displayed map is a counterexample to the claim.