Corollary of centralizer product theorem for rank at least three

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Let q,r,s be primes such that q \ne r, q \ne s (we can have r = s).

Suppose Q is a finite abelian q-group of rank at least three, R is a finite r-group and S is a finite s-group. Suppose further that Q acts as automorphisms on both R and S. Then, there exists a non-identity element u \in Q such that C_R(u) and C_S(u) are both nontrivial.

Related facts

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Facts used

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 q, possibly equal primes r,s both distinct from q. A finite abelian q-group Q of rank at least three (in particular, Q is non-cyclic). A finite r-group R and a finite s-group S.

To prove: There exists a non-identity element u of Q such that both C_R(u) and C_S(u) are nontrivial.

Proof: Without loss of generality, we assume that Q is elementary abelian of order q^3. This is because we can always replace Q by an elementary abelian subgroup of order q^3 in Q, which exists because Q 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 x \in Q such that C_R(x) is nontrivial. Fact (1) Q is a non-cyclic abelian q-group , R is a r-group, q \ne r. Apply Fact (1) to the action of Q on R. We're starting with a rank three group Q, 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 R_0 = C_R(x). Then, R_0 is nontrivial and Q-invariant Q is abelian. Step (1) [SHOW MORE] Process of quotienting out begins
3 Consider Q_1 = Q/\langle x \rangle. Then, Q_1 has rank two (being elementary abelian of order q^2), and the action of Q on R_0 descends to an action of Q_1 on R_0. Q is abelian Step (2) [SHOW MORE] Process of quotienting out continues
4 There exists \overline{y} \in Q_1 such that C_{R_0}(\overline{y}) 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 R. We can still use Fact (1) on this quotient of Q, to get a nontrivial centralizer within this.
5 Let y be an element of Q that maps to \overline{y}. Then, C_{R_0}(y) = C_{R_0}(\overline{y}) is nontrivial. Let R_1 = C_{R_0}(y). Then, if A is the elementary abelian subgroup \langle x,y \rangle of Q, C_R(u) contains the nontrivial group R_1 for all non-identity elements u of A, and is hence nontrivial. Steps (2)-(4) [SHOW MORE] Now combining the previous steps to obtain a subgroup of Q of rank two in which every element fixes a nontrivial subgroup of R.
6 There exists a non-identity element u \in A such that C_S(u) is nontrivial. Fact (1) S is a s-group, q \ne s. Step (5) [SHOW MORE] We can use Fact (1) on this subgroup of Q because it has rank at least 2, to find an element with nontrivial centralizer.
7 The u \in A that works for Step (6) has the property that u \in Q and both C_R(u) and C_S(u) are nontrivial. Steps (5), (6) [SHOW MORE] Combine steps.


Textbook references

  • Finite Groups by Daniel Gorenstein, ISBN 0821843427, Page 292, Lemma 5.5, Chapter 8 (p-constrained and p-stable groups), Section 5 (The Thompson Transitivity Theorem), More info