ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passverified 2026-09-26 (gpt-6-sol)

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The C_p^2 extension as a nonzero two-cocycle

Example

For every integer $p \geq 2$ , the nonsplit extension

$0 \to C_{p} \to C_{p^{2}} \to C_{p} \to 0$

with trivial action on the kernel, a normalized section yields the cocycle

$f (i, j) = {\begin{cases} 0, & i + j < p, \\ 1, & i + j \geq p, \end{cases}$

for $0 \leq i, j < p$ . This cocycle is nonzero in $H^{2} (C_{p}, C_{p})$ .

Facts & Assumptions

Given: An integer $p \geq 2$ , the quotient map $Z / p^{2} Z \to Z / p Z$ , and the normalized section $s (i) = i$ for $0 \leq i < p$ .

[L1]

An extension determines a well-defined class in $H^{2}$ (An extension determines a well-defined H^2 class).

[L2]

The zero class in $H^{2} (C_{p}, C_{p})$ consists of two-coboundaries, with trivial-action coboundary $δ b (i, j) = b (i) + b (j) - b (i + j)$ (Second cohomology by factor sets).

Verification

technique · direct

1.1givenalgebra

In $Z / p^{2} Z$ , adding two chosen lifts $i$ and $j$ either stays below $p$ or crosses the first multiple of $p$ . Therefore $s (i) + s (j) - s (i + j m o d p) = p f (i, j),$ with $f (i, j)$ given by the carry function above.

2.1L1L2step 1.1algebra

The kernel is central, so [L1] identifies the carry function with the extension class. If $[f] = 0$ , [L2] gives a normalized cochain $b : C_{p} \to C_{p}$ with $f = δ b$ . Subtract the kernel element $p b (i)$ from each chosen lift: $s^{'} (i) = s (i) - p b (i) \in C_{p^{2}}$ . Its factor set is $f - δ b = 0$ , so $s^{'}$ is a group-homomorphic section of the quotient. But $s^{'} (1)$ would have order dividing $p$ and map to $1 \in C_{p}$ ; every element of $C_{p^{2}}$ killed by $p$ lies in the kernel $p C_{p^{2}}$ and maps to $0$ . This contradiction shows $[f] \neq 0$ without any global Choice premise.

3.1step 2.1∎

Hence the displayed cocycle represents a nonzero class in $H^{2} (C_{p}, C_{p})$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Clara Loh, Group Cohomology, SS 2019 (standard reference, not scraped)
Caroline Lassueur, Cohomology of Groups, SS 2021 (standard reference, not scraped)