# Equivalence of definitions of universal congruence condition

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

## Contents

## Statement

Suppose is a finite collection of finite -groups, groups of prime power order for the prime . The following conditions are equivalent for :

- For any finite -group that contains a subgroup isomorphic to an element of , the number of subgroups of isomorphic to elements of is congruent to modulo .
- For any finite -group that contains a subgroup isomorphic to an element of , the number of normal subgroups of isomorphic to elements of is congruent to modulo .
- For any finite -group and any normal subgroup of such that contains a subgroup isomorphic to an element of , the number of normal subgroups of isomorphic to elements of and contained in is congruent to modulo .
- For any finite -group that contains a subgroup isomorphic to an element of , the number of p-core-automorphism-invariant subgroups of isomorphic to elements of is congruent to modulo .
- For any finite group containing a subgroup isomorphic to an element of , the number of subgroups of isomorphic to an element of is congruent to modulo .

## Facts used

- Size of conjugacy class of subgroups equals index of normalizer
- Lagrange's theorem
- Sylow subgroups exist
- Sylow implies order-dominating
- Product formula
- 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 is divisible by .

**Given**: A finite -group , a collection of finite -groups.

**To prove**: The number of non-normal subgroups of isomorphic to an element of is divisible by .

**Proof**:

Step no. | Assertion/construction | Facts used | Given data used | Previous steps used | Explanation |
---|---|---|---|---|---|

1 | The set of non-normal subgroups of isomorphic to an element of is a disjoint union of conjugacy classes of subgroups in , each having size more than one. | [SHOW MORE] | |||

2 | The size of any conjugacy class of non-normal subgroups in is divisible by . | Facts (1), (2) | [SHOW MORE] | ||

3 | The number of non-normal subgroups of isomorphic to an element of is divisible by . | Steps (1), (2) | [SHOW MORE] |

### Equivalence of (1) and (3)

For this, it suffices to show that the number of subgroups of isomorphic to elements of and *not* normal inside is divisible by . This is because:

Total number of subgroups of isomorphic to elements of = (Number of subgroups of isomorphic to elements of that are *not normal* in ) + (Number of subgroups of isomorphic to elements of that are normal in )

**Given**: A finite -group that is a normal subgroup of a group . A collection of finite -groups.

**To prove**: The number of subgroups of that are not normal in and are isomorphic to elements of is divisible by .

**Proof**:

Step no. | Assertion/construction | Facts used | Given data used | Previous steps used | Explanation |
---|---|---|---|---|---|

1 | The set of subgroups of isomorphic to an element of and not normal in is a disjoint union of conjugacy classes of subgroups in (the subgroups are all in , but conjugacy is relative to ), each having size more than one. | is normal in . | [SHOW MORE] | ||

2 | The size of any conjugacy class of non-normal subgroups in is divisible by . | Facts (1), (2) | [SHOW MORE] | ||

3 | The number of non-normal subgroups of isomorphic to an element of is divisible by . | Steps (1), (2) | [SHOW MORE] |

### Equivalence of (3) and (4)

This follows by taking as the semidirect product of and the -core of .

### Equivalence of (2) and (5)

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

**To prove**: The number of subgroups of isomorphic to an element of is congruent to 1 mod .

**Proof**:

Step no. | Assertion/construction | Facts used | Given data used | Previous steps used | Explanation |
---|---|---|---|---|---|

1 | Let be a -Sylow subgroup of . | Fact (3) | |||

2 | contains a subgroup isomorphic to an element of . | Fact (4) | contains a subgroup isomorphic to an element of | Step (1) | [SHOW MORE] |

3 | The number of normal subgroups of isomorphic to an element of is congruent to 1 mod . | Assumption of congruence condition | Steps (1), (2) | Step-given combination direct. | |

4 | A subgroup of of prime power order is normalized by if and only if it is contained in and normal in . | Facts (2), (5) | Step (1) | [SHOW MORE] | |

5 | The number of subgroups of isomorphic to an element of and not normalized by is divisible by . |
Fact (6) | Step (1) | [SHOW MORE] | |

6 | The total number of subgroups of isomorphic to an element of is congruent to 1 mod . | Steps (3), (4), (5) | [SHOW MORE] |