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

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] Conjugate subgroups are isomorphic, so if any element of a conjugacy class is isomorphic to an element of ${\mathcal {S}}$, all elements of the conjugacy class are. Further, any conjugate to a non-normal subgroup is non-normal. Thus, we get a union of conjugacy classes. Further, since the subgroups are consideration are non-normal, all conjugacy classes have size greater than 1.
2	The size of any conjugacy class of non-normal subgroups in $P$ is divisible by $p$.	Facts (1), (2)			[SHOW MORE] By Fact (1), the size equals the index of the normalizer of the subgroup, and by Fact (2), this divides the order of the group and is hence a power of $p$. Because of non-normality, this number is not 1, hence is a power of $p$, so it is divisible by $p$.
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] By Step (1), the total number of non-normal subgroups of $P$ isomorphic to an element of ${\mathcal {S}}$ is a sum of sizes of conjugacy classes of subgroups, each of which is bigger than 1. By Step (2), each summand is a multiple of $p$, so the total sum is divisible by $p$.

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] Conjugate subgroups in $Q$ are isomorphic, so if any element of a conjugacy class is isomorphic to an element of ${\mathcal {S}}$, all elements of the conjugacy class are. Further, any $Q$-conjugate of a subgroup contained inside $P$ is also contained in $P$, because $P$ is normal. Further, any conjugate to a non-normal subgroup is non-normal. Thus, we get a union of $Q$-conjugacy classes. Further, since the subgroups are consideration are non-normal, all conjugacy classes have size greater than 1.
2	The size of any conjugacy class of non-normal subgroups in $Q$ is divisible by $p$.	Facts (1), (2)			[SHOW MORE] By Fact (1), the size equals the index of the normalizer of the subgroup, and by Fact (2), this divides the order of the group and is hence a power of $p$. Because of non-normality, this number is not 1, hence is a power of $p$, so it is divisible by $p$.
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] By Step (1), the total number of subgroups of $P$ not normal in $Q$ and isomorphic to an element of ${\mathcal {S}}$ is a sum of sizes of conjugacy classes of subgroups, each of which is bigger than 1. By Step (2), each summand is a multiple of $p$, so the total sum is divisible by $p$.

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] By Fact (4), every $p$-subgroup of $G$ has a conjugate in $P$. Since conjugation preserves the isomorphism class, we can pass from a subgroup isomorphic to an element of ${\mathcal {S}}$ to a new subgroup contained in $P$.
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] Suppose $H$ is a subgroup of prime power order in $G$. Then, if $P$ normalizes $H$, we get that $PH$ is a group. By Fact (5), it has order $|P||H|/|P\cap H|$. This is a power of $p$ dividing the order of $G$, and using that $p$ is $p$-Sylow, we obtain it must be equal to $|P|$, so $H=P\cap H</mah>and<math>H\leq P$.
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] Consider the action of ${\mathcal {S}}$ on the collection of all subgroups of $G$ isomorphic to an element of ${\mathcal {S}}$ and not normalized by $P$. The action is well defined because conjugation by elements of $P$ preserves the isomorphism class and also the property of not being normalized by $P$. By Fact (6), each orbit has size dividing the order of $P$. Also, the orbits are nontrivial, since $P$ does not normalize the subgroups. Thus, the size of each orbit is a power of $p$ greater than 1, so is divisible by $p$. The total size is thus also divisible by $p$.
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] The subgroups normalized by $P$ have a count congruent to 1 mod $p$ by combining Steps (3) and (4). The subgroups not normalized by $P$ have a count divisible by $p$ by Step (5). Adding up, we get a total congruent to 1 mod $p$.