Corollary of centralizer product theorem for rank at least three
Let be primes such that (we can have ).
Suppose is a finite abelian -group of rank at least three, is a finite -group and is a finite -group. Suppose further that acts as automorphisms on both and . Then, there exists a non-identity element such that and are both nontrivial.
The table below lists key facts used directly and explicitly in the proof. Fact numbers as used in the table may be referenced in the proof. This table need not list facts used indirectly, i.e., facts that are used to prove these facts, and it need not list facts used implicitly through assumptions embedded in the choice of terminology and language.
|Fact no.||Statement||Steps in the proof where it is used||Qualitative description of how it is used||What does it rely on?||Difficulty level||Other applications|
|1||Centralizer product theorem||(1), (4), (6)||At each step, we have at hand a non-cyclic abelian group of prime power order acting on a group of a different prime power order, and we use the fact to show that the centralizer of at least one of the acting elements in the acted upon group is nontrivial.||centralizer product theorem for elementary abelian group||click here|
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Given: A prime , possibly equal primes both distinct from . A finite abelian -group of rank at least three (in particular, is non-cyclic). A finite -group and a finite -group .
To prove: There exists a non-identity element of such that both and are nontrivial.
Proof: Without loss of generality, we assume that is elementary abelian of order . This is because we can always replace by an elementary abelian subgroup of order in , which exists because has rank at least three.
|Step no.||Assertion/construction||Facts used||Given data used||Previous steps used||Explanation||Commentary|
|1||There exists a non-identity element such that is nontrivial.||Fact (1)||is a non-cyclic abelian -group , is a -group, .||Apply Fact (1) to the action of on .||We're starting with a rank three group , which is more than what we need at this stage, since we only need rank at least two to guarantee an element with nontrivial centralizer. We have used up one piece of slack.|
|2||Let . Then, is nontrivial and -invariant||is abelian.||Step (1)||[SHOW MORE]||Process of quotienting out begins|
|3||Consider . Then, has rank two (being elementary abelian of order ), and the action of on descends to an action of on .||is abelian||Step (2)||[SHOW MORE]||Process of quotienting out continues|
|4||There exists such that is nontrivial.||Fact (1)||Step (3)||[SHOW MORE]||Having used up 1 of the rank 3 in Step (1), we are left with a rank 2 quotient acting on a subgroup of . We can still use Fact (1) on this quotient of , to get a nontrivial centralizer within this.|
|5||Let be an element of that maps to . Then, is nontrivial. Let . Then, if is the elementary abelian subgroup of , contains the nontrivial group for all non-identity elements of , and is hence nontrivial.||Steps (2)-(4)||[SHOW MORE]||Now combining the previous steps to obtain a subgroup of of rank two in which every element fixes a nontrivial subgroup of .|
|6||There exists a non-identity element such that is nontrivial.||Fact (1)||is a -group, .||Step (5)||[SHOW MORE]||We can use Fact (1) on this subgroup of because it has rank at least 2, to find an element with nontrivial centralizer.|
|7||The that works for Step (6) has the property that and both and are nontrivial.||Steps (5), (6)||[SHOW MORE]||Combine steps.|