Equivalence of definitions of universal congruence condition

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

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:

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:

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: