Equivalence of definitions of universal congruence condition

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This article gives a proof/explanation of the equivalence of multiple definitions for the term collection of groups satisfying a universal congruence condition
View a complete list of pages giving proofs of equivalence of definitions

Statement

Suppose \mathcal{S} is a finite collection of finite p-groups, groups of prime power order for the prime p. The following conditions are equivalent for \mathcal{S}:

  1. For any finite p-group P that contains a subgroup isomorphic to an element of \mathcal{S}, the number of subgroups of P isomorphic to elements of \mathcal{S} is congruent to 1 modulo p.
  2. For any finite p-group P that contains a subgroup isomorphic to an element of \mathcal{S}, the number of normal subgroups of P isomorphic to elements of \mathcal{S} is congruent to 1 modulo p.
  3. For any finite p-group Q and any normal subgroup P of Q such that P contains a subgroup isomorphic to an element of \mathcal{S}, the number of normal subgroups of Q isomorphic to elements of \mathcal{S} and contained in P is congruent to 1 modulo p.
  4. For any finite p-group P that contains a subgroup isomorphic to an element of \mathcal{S}, the number of p-core-automorphism-invariant subgroups of P isomorphic to elements of \mathcal{S} is congruent to 1 modulo P.
  5. For any finite group G containing a subgroup isomorphic to an element of \mathcal{S}, the number of subgroups of G isomorphic to an element of \mathcal{S} is congruent to 1 modulo p.

Facts used

  1. Size of conjugacy class of subgroups equals index of normalizer
  2. Lagrange's theorem
  3. Sylow subgroups exist
  4. Sylow implies order-dominating
  5. Product formula
  6. Fundamental theorem of group actions

Proof

Equivalence of (1) and (2)

It suffices to prove that the number of non-normal subgroups isomorphic to elements of \mathcal{S} is divisible by p.

Given: A finite p-group P, a collection \mathcal{S} of finite p-groups.

To prove: The number of non-normal subgroups of P isomorphic to an element of \mathcal{S} is divisible by p.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 The set of non-normal subgroups of P isomorphic to an element of \mathcal{S} is a disjoint union of conjugacy classes of subgroups in P, each having size more than one. [SHOW MORE]
2 The size of any conjugacy class of non-normal subgroups in P is divisible by p. Facts (1), (2) [SHOW MORE]
3 The number of non-normal subgroups of P isomorphic to an element of \mathcal{S} is divisible by p. Steps (1), (2) [SHOW MORE]

Equivalence of (1) and (3)

For this, it suffices to show that the number of subgroups of P isomorphic to elements of \mathcal{S} and not normal inside Q is divisible by p. This is because:

Total number of subgroups of P isomorphic to elements of \mathcal{S} = (Number of subgroups of P isomorphic to elements of \mathcal{S} that are not normal in Q) + (Number of subgroups of P isomorphic to elements of \mathcal{S} that are normal in Q)

Given: A finite p-group P that is a normal subgroup of a group Q. A collection \mathcal{S} of finite p-groups.

To prove: The number of subgroups of P that are not normal in Q and are isomorphic to elements of \mathcal{S} is divisible by p.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 The set of subgroups of P isomorphic to an element of \mathcal{S} and not normal in Q is a disjoint union of conjugacy classes of subgroups in Q (the subgroups are all in P, but conjugacy is relative to Q), each having size more than one. P is normal in Q. [SHOW MORE]
2 The size of any conjugacy class of non-normal subgroups in Q is divisible by p. Facts (1), (2) [SHOW MORE]
3 The number of non-normal subgroups of P isomorphic to an element of \mathcal{S} is divisible by p. Steps (1), (2) [SHOW MORE]

Equivalence of (3) and (4)

This follows by taking Q as the semidirect product of P and the p-core of \operatorname{Aut}(P).

Equivalence of (2) and (5)

Given: A finite group G. p is a prime number. \mathcal{S} is a collection of finite p-groups such that, in any finite p-group containing a subgroup isomorphic to an element of \mathcal{S}, the number of normal subgroups isomorphic to an element of \mathcal{S} is congruent to 1 mod p. G contains a subgroup isomorphic to an element of \mathcal{S}.

To prove: The number of subgroups of G isomorphic to an element of \mathcal{S} is congruent to 1 mod p.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Let P be a p-Sylow subgroup of G. Fact (3)
2 P contains a subgroup isomorphic to an element of \mathcal{S}. Fact (4) G contains a subgroup isomorphic to an element of \mathcal{S} Step (1) [SHOW MORE]
3 The number of normal subgroups of P isomorphic to an element of \mathcal{S} is congruent to 1 mod p. Assumption of congruence condition Steps (1), (2) Step-given combination direct.
4 A subgroup of G of prime power order is normalized by P if and only if it is contained in P and normal in P. Facts (2), (5) Step (1) [SHOW MORE]
5 The number of subgroups of G isomorphic to an element of \mathcal{S} and not normalized by P is divisible by p. Fact (6) Step (1) [SHOW MORE]
6 The total number of subgroups of G isomorphic to an element of \mathcal{S} is congruent to 1 mod p. Steps (3), (4), (5) [SHOW MORE]