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If $char k ∣ ∣ G ∣$ , then $k [G]$ is not semisimple

Statement

Let $G$ be a finite group and let $k$ be a field with $char k ∣ ∣ G ∣$ . Then the group algebra $k [G]$ is not a semisimple ring.

Facts & Assumptions

Given: A finite group $G$ and a field $k$ with $char k ∣ ∣ G ∣$ .

[L1]

Under the same characteristic hypothesis, the augmentation ideal of $k [G]$ has no complement as a left $k [G]$ -submodule of the regular module (If $char k ∣ ∣ G ∣$ , the augmentation ideal of $k [G]$ has no $k [G]$ -module complement in the regular representation).

[L2]

A unital ring is semisimple exactly when its left regular module is semisimple (A semisimple ring as a ring whose left regular module is semisimple).

[L3]

Every finitely generated semisimple module is a finite direct sum of simple modules (A finitely generated semisimple module is a finite direct sum of simple modules).

[L4]

For a finite-length module, the direct-sum, sum-of-simples, and complement characterizations of semisimplicity are equivalent without Choice (Choice-free semisimple characterizations for finite-length modules).

Proof

technique · contradiction

1.1

Assume, for contradiction, that $k [G]$ is semisimple. Then [L2] makes the left regular module $_{k [G]} k [G]$ semisimple. It is generated by $[e]$ , so [L3] makes it a finite direct sum of simple submodules. In particular it is a finite-length module.

L2L3givenassume-contra

2.1

Applying [L4] to that finite-length semisimple module shows that every submodule of $_{k [G]} k [G]$ has a complementary submodule. In particular the augmentation ideal $I_{G}$ has a complement.

step 1.1L4given

3.1

Step 2.1 contradicts [L1]. Therefore $k [G]$ is not semisimple.

step 2.1L1discharge-contradiction∎

Depends on

Used by

False statement: every finite-dimensional algebra over a field is semisimple False statement

Dependency tree · two levels

12 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

Pavel Etingof et al., Introduction to Representation Theory, Proposition 3.2 (standard reference, not scraped)
Peter Webb, A Course in Finite Group Representation Theory, Example 1.1.7 (standard reference, not scraped)

If char⁡k∣∣G∣chark∣∣G∣, then k[G]k[G] is not semisimple