Commuting of non-identity elements defines an equivalence relation between prime divisors of the order of a finite CN-group

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Suppose G is a finite CN-group (i.e., a finite group that is also a CN-group). Let \pi be the set of prime divisors of the order of G. Define a relation \sim on \pi as follows: for p,q \in \pi (possibly equal, possibly distinct), p \sim q if and only if either p = q or there exists a p-Sylow subgroup P of G and a q-Sylow subgroup Q of G such that the following equivalent conditions hold:

  1. There exists a non-identity element x of P and a non-identity element y of Q such that x and y commute.
  2. Every element of P commutes with every element of Q.

Note that the conditions are equivalent because Sylow subgroups for distinct primes in CN-group centralize each other iff they have non-identity elements that centralize each other. Note also that these conditions are not as strong as the statement that every element of p-power order in G commutes with every element of q-power order. It is simply a statement about being able to find two specific Sylow subgroups that centralize each other.

The claim is that \sim is an equivalence relation on \pi.

Related facts

Facts used

  1. Sylow implies order-conjugate
  2. Sylow satisfies intermediate subgroup condition


To prove that a relation is an equivalence relation, we need to prove that it is reflexive, symmetric, and transitive.


This follows by definition.


This follows by definition.


This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

We want to show that, for p,q,r (possibly equal, possibly distinct) primes, p \sim q, q \sim r implies p \sim r. Note that if p = r, the conclusion follows. If p = q or q = r, it again follows. Thus, we can assume that p,q,r are all distinct.

Given: A finite CN-group G, three distinct primes p,q,r such that p \sim q, q \sim r.

To prove: p \sim r


Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 There is a p-Sylow subgroup P and a q-Sylow subgroup Q of G such that every element in P centralizes every element in Q. p \sim q plus the interpretation of \sim from given
2 There is a q-Sylow subgroup Q_1 and a r-Sylow subgroup R_1 of G such that every element in Q_1 centralizes every element in R_1. q \sim r plus the interpretation of \sim from given
3 There exists a r-Sylow subgroup R of G such that every element of Q commutes with every element of R. Fact (1) Steps (1), (2) The q-Sylow subgroups Q,Q_1 of G are conjugate in G by Fact (1). Let \varphi be a conjugation operation in G such that \varphi(Q_1) = Q. Let R = \varphi(R_1). Since \varphi is an automorphism, it preserves centralizing, so the fact that Q_1 centralizes R_1 implies that Q = \varphi(Q_1) centralizes R = \varphi(R_1). Finally, by order considerations, R is also a r-Sylow subgroup of G.
4 Let x be a non-identity element of Q. q divides the order of G, Q is q-Sylow From the given, Q is nontrivial.
5 C_G(x) is a nilpotent subgroup of G. G is a CN-group Step (4) Direct from the given and x being non-identity by Step (4).
6 P and R are respectively the p-Sylow and r-Sylow subgroups of C_G(x). Fact (2) Steps (1), (3), (4) By Steps (1) and (3), P and R centralize Q, hence are both in C_G(x). Note that since they are already Sylow in G, Fact (2) gives that they are Sylow in C_G(x).
7 P and R are respectively p-Sylow and r-Sylow subgroups of G that centralize each other. Fact (3) Steps (1), (3), (5), (6) By Fact (3) and Step (5), the Sylow subgroups in C_G(x) form an internal direct product, hence centralize each other. Apply to Step (6) to get that P and R centralize each other. Note that they are Sylow by Steps (1) and (3).
8 p \sim r Definition of \sim Step (7) Step-direct


Textbook references

  • Finite Groups by Daniel Gorenstein, ISBN 0821843427, Page 408, Theorem 14.2.5(i), (The theorem is stated for minimal simple CN-groups of odd order, but this part of the theorem does not require those assumptions.)More info