Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04
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If the field has characteristic p and p divides |G|, then Maschke's theorem still makes kG semisimple

Statement

If char⁡k=p and p∣∣G∣, then Maschke's theorem still implies that k[G] is semisimple.

Facts & Assumptions

Given: A finite group G and a field k with char⁡k=p∣∣G∣.

[L1]

In defining characteristic, Maschke can fail and k[G] need not be semisimple (When the characteristic divides the group order, Maschke can fail and kG need not be semisimple).

Refutation

technique · direct
1.1

The remark [L1] records precisely that the defining-characteristic situation is the failure regime for Maschke's theorem.

L1given
2.1

Therefore the claimed semisimplicity conclusion does not hold in general when p divides ∣G∣.

step 1.1
3.1

So the statement is false.

step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

2 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