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Frobenius kernel theorem
Statement
Let be a finite Frobenius group with complement and let be the associated kernel set. Then is a normal subgroup of .
Facts & Assumptions
Given: A finite group with Frobenius complement and the kernel set with its associated family of nontrivial irreducible characters of .
for a nonempty family of irreducible complex characters of (Frobenius kernel is an intersection of character kernels).
If is a finite-dimensional complex representation of with character , then (The kernel of a complex character agrees with the kernel of any representation affording it).
The intersection of a nonempty family of normal subgroups of is again a normal subgroup of (The intersection of a nonempty family of normal subgroups is normal).
Proof
By [F1], is the intersection over the nonempty family of kernels of irreducible characters of .
Each occurring in [F1] is a normal subgroup of : it is the kernel of the representation affording the character , hence equals , which is normal by [F2].
Therefore is an intersection of a nonempty family of normal subgroups of , so by [F3] it is a subgroup of and is normal in . In particular the kernel set is closed under products and inverses, a fact that the counting argument of Frobenius kernel cardinality could not supply. ∎
Depends on
Used by
Dependency tree · two levels
14 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
- Alex Bartel, Introduction to Representation Theory of Finite Groups, §6.1 (standard reference, not scraped)