Focal subgroup theorem
This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., Sylow subgroup) must also satisfy the second subgroup property (i.e., subgroup whose focal subgroup equals its intersection with the commutator subgroup)
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Get more facts about Sylow subgroup|Get more facts about subgroup whose focal subgroup equals its intersection with the commutator subgroup
In other words, is the focal subgroup of in .
In other words, is a subgroup whose focal subgroup equals its intersection with the commutator subgroup.
For the proof using linear representation theory
For the proof using the transfer homomorphism
Given: A finite group , a -Sylow subgroup . is the focal subgroup of , defined by:
Further, we have:
To prove: .
|Step no.||Assertion/construction||Facts used||Given data used||Previous steps used||Explanation|
|2||is normal in and is abelian.||Step (1)||This follows from the part of Step (1) that asserts that .|
|3||Let be the distinct prime divisors of with . For each , pick a -Sylow subgroup such that .|
|4||Any can be expressed as a product where each is conjugate to some and the commute pairwise.||Fact (2)||Step (3)||[SHOW MORE]|
|5||If (continuing notation from Step (4), though it does not really matter) is conjugate to and also conjugate to then .||[SHOW MORE]|
|6||Let be a linear character (i.e., the character of a one-dimensional representation) of whose kernel contains . The function , where is as described in Step (4), is well-defined||Steps (4), (5)||[SHOW MORE]|
|7||Continuing with the notation of Step (6), is a class function on .||[SHOW MORE]|
|8||Continuing with the notation of Step (6), is a linear character of . In other words, for all .||This critical step is unclear|
|9||For any linear character of whose kernel contains , there is a linear character of that extends , i.e., the restriction of to is .||Steps (6)-(8)||Step-combination direct, plus: [SHOW MORE]|
|11||.||Steps (1), (10)||Step-combination direct.|
Proof using the transfer homomorphism
We have , with normal in and abelian. In particular, we can construct the transfer homomorphism . Let be the quotient map.
Claim: The restriction of to is surjective to .
Proof: For any , we have, by fact (1), a collection of elements and nonnegative integers such that , and we have:
Since we chose , and is abelian, we can rearrange terms to obtain:
Every term in the second product is of the form , which is of the form with conjugate in (here ). In particular, every term in the second product is in , and we obtain:
which, since , reduces to:
Note now that since is -Sylow, is relatively prime to , and the map is a bijection from to itself. Thus, , restricted to , surjects to .
Proof using the claim: Let be the kernel of .
Since is a kernel of a map to an abelian group, . Thus, . The restriction of to is a bijective map from to (surjective by the claim, and injective because we've quotiented by the kernel). Thus, the size of equals the size of forcing . In particular, .
- Focal series in finite groups by Donald Gordon Higman, , Volume 5, Page 477 - 497(Year 1953): More info