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If , then is a semisimple ring
Statement
Let be a finite group and let be a field with . Then the group algebra is a semisimple ring.
Facts & Assumptions
Given: A finite group and a field with .
If is finite, then (If is finite then ).
The regular representation of over is the action on by left multiplication, namely (The trivial representation, the regular representation, and permutation representations from finite -sets).
Under the characteristic hypothesis, every finite-dimensional representation of over is completely reducible (If , every finite-dimensional representation of is completely reducible).
A representation is completely reducible exactly when its underlying space is an internal direct sum of irreducible subrepresentations (A completely reducible representation as a finite direct sum of irreducible subrepresentations).
Under the dictionary, irreducible representations are exactly simple left -modules (Under the dictionary, subrepresentations are exactly submodules and irreducible representations are exactly simple modules).
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).
Proof
By [L1] and [L2], the regular representation of on is finite-dimensional. Therefore [L3] makes it completely reducible.
Expanding that term with [L4], the left regular representation is an internal direct sum of irreducible subrepresentations. By [L5], those are exactly simple left -submodules. So the left regular module is an internal direct sum of simple submodules.
By [L6], that is exactly the definition that is a semisimple ring.
Depends on
- If $\operatorname{char} k \nmid |G|$, every finite-dimensional representation of $G$ is completely reducible
- A completely reducible representation as a finite direct sum of irreducible subrepresentations
- The trivial representation, the regular representation, and permutation representations from finite $G$-sets
- Under the dictionary, subrepresentations are exactly submodules and irreducible representations are exactly simple modules
- A semisimple ring as a ring whose left regular module is semisimple
- If $G$ is finite then $\dim_k k[G]=|G|$
Used by
Dependency tree · two levels
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Sources
- Pavel Etingof et al., Introduction to Representation Theory, Theorem 3.1(i) (standard reference, not scraped)
- Peter Webb, A Course in Finite Group Representation Theory, Corollary 1.2.5 (standard reference, not scraped)