Difference between revisions of "Classification of cyclicity-forcing numbers"

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(The definitions that we have to prove as equivalent)
(The definitions that we have to prove as equivalent)
 
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# Any group of that order is a direct product of cyclic Sylow subgroups.
 
# Any group of that order is a direct product of cyclic Sylow subgroups.
 
# It is a product of distinct primes <math>p_i</math>, such that <math>p_i</math> does not divide <math>p_j - 1</math> for any prime divisors <math>p_i, p_j</math> of the order.
 
# It is a product of distinct primes <math>p_i</math>, such that <math>p_i</math> does not divide <math>p_j - 1</math> for any prime divisors <math>p_i, p_j</math> of the order.
 +
# It is relatively prime to the value of its [[Euler totient function]].
  
 
== Related facts ==
 
== Related facts ==

Latest revision as of 16:45, 24 July 2017

This article gives a proof/explanation of the equivalence of multiple definitions for the term cyclicity-forcing number
View a complete list of pages giving proofs of equivalence of definitions

The definitions that we have to prove as equivalent

The following are equivalent for a natural number:

  1. There exists exactly one isomorphism class of groups of that order.
  2. Any group of that order is a cyclic group.
  3. Any group of that order is a direct product of cyclic Sylow subgroups.
  4. It is a product of distinct primes p_i, such that p_i does not divide p_j - 1 for any prime divisors p_i, p_j of the order.
  5. It is relatively prime to the value of its Euler totient function.

Related facts

Facts used

  1. Finite abelian implies direct product of Sylow subgroups
  2. Finite non-abelian and every proper subgroup is abelian implies not simple: This is the meat of the proof.
  3. Description of automorphism group of cyclic group
  4. Homomorphism between groups of coprime order is trivial
  5. Cyclic over central implies abelian
  6. Lagrange's theorem
  7. Order of quotient group divides order of group

Proof

Equivalence of definitions (1) and (2)

This follows from two basic observations:

  • For any natural number n, there exists a cyclic group of order n.
  • Two cyclic groups of the same order are isomorphic.

Thus, the existence of only one isomorphism class of groups of a given order is equivalent to asserting that every group of that order is cyclic.

Equivalence of definitions (2) and (3)

This follows from the Chinese remainder theorem.

(3) implies (4)

For this, we first prove that the number must be square-free, i.e., it is a product of distinct primes.

Suppose we have a prime factorization as follows:

n = p_1^{k_1}p_2^{k_2} \dots p_r^{k_r}.

Consider the group G that is a direct product of elementary abelian groups of order p_i^{k_i}. Then, G is an abelian group of order n. Further, if any of the k_i is greater than one, then the p_i-Sylow subgroup is not cyclic, so G is not cyclic. Thus, if n has a square factor, there is a non-cyclic group of order n. Thus, any cyclicity-forcing number must be square-free.

We thus have:

n = p_1p_2 \dots p_r

where all the p_i are distinct primes.

Now, suppose there exist primes p_i and p_j such that p_i | p_j - 1. Then, there exists a non-abelian group H of order p_ip_j, given as the semidirect product of a cyclic group of order p_j, and a cyclic subgroup of order p_i in its automorphism group. Let G be the direct product of H with a cyclic group of order n/(p_ip_j). Then, G is a group of order n. However, G is not cyclic since it has a subgroup isomorphic to H, a non-cyclic group.

Thus, we're forced to have:

n = p_1p_2 \dots p_r

with p_i not dividing p_j - 1 for any two prime divisors p_i, p_j of n.

(4) implies (3)

Given: A group G, whose order is n = p_1p_2 \dots p_r, where the p_i are distinct primes and p_i does not divide p_j - 1 for i \ne j.

To prove: G is cyclic.

Proof: We prove this claim by induction on n. First, note that any divisor of n also satisfies the condition of being square-free as well as the condition that no prime divisor of it divides any other prime divisor minus one.

Base case of induction: The base case of induction, n = 1, is trivial.

Proof of inductive step:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Every proper subgroup of G is cyclic Fact (6) inductive hypothesis, arithmetic condition on n [SHOW MORE]
2 If G is non-abelian, G is not simple Fact (2) Step (1) [SHOW MORE]
3 Any proper normal subgroup of G is central Fact (4) For any two prime divisors p_i, p_j of n, p_i does not divide p_j - 1 Step (1) [SHOW MORE]
4 If G is non-abelian, the center Z(G) of G is nontrivial Steps (2), (3) [SHOW MORE]
5 If G is non-abelian, the quotient G/Z(G) is cyclic Fact (7) inductive hypothesis Step (4) [SHOW MORE]
6 G is abelian Fact (5) Step (5) [SHOW MORE]
7 G is cyclic n is square-free Step (6) [SHOW MORE]

References

Textbook references

  • Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, Page 149, Exercises 54-55, Section 4.5 (Sylow's theorem), (hints given in exercise)More info