Congruence condition relating number of normal subgroups containing minimal normal subgroups and number of normal subgroups in the whole group

Statement
Suppose $$p$$ is a prime number and $$G$$ is a group of prime power order with underlying prime $$p$$.

Suppose $$\mathcal{S}$$ is a collection of normal subgroups of $$G$$. Denote by $$\nu(G)$$ the total size of $$\mathcal{S}$$ and by $$\nu(G,H)$$ the number of elements of $$\mathcal{S}$$ that contain a given subgroup $$H$$ of $$G$$. Then, if $$\mathcal{S}$$ does not contain the trivial subgroup of $$G$$, we have:

$$\nu(G) \equiv \sum_N \nu(G,N) \pmod p$$

where the summation ranges over all the minimal normal subgroups of $$G$$.

If $$\mathcal{S}$$ contains the trivial subgroup, we have:

$$\nu(G) \equiv 1 + \sum_N \nu(G,N) \pmod p$$

Related facts

 * Congruence condition relating number of subgroups in maximal subgroups and number of subgroups in the whole group
 * Congruence condition on number of subgroups of given prime power order

Facts used

 * 1) uses::Minimal normal implies central in nilpotent, equivalently uses::minimal normal implies contained in Omega-1 of center for nilpotent p-group
 * 2) uses::Central implies normal
 * 3) uses::Equivalence of definitions of size of projective space

Proof
Note that the case for the trivial subgroup being in $$\mathcal{S}$$ follows from the case for the trivial subgroup not being in $$\mathcal{S}$$. We thus restrict ourselves to proving the case that the trivial subgroup is not in $$\mathcal{S}$$.

Given: $$G$$ is a group of prime power order with underlying prime $$p$$. $$\mathcal{S}$$ is a collection of normal subgroups of $$G$$. Denote by $$\nu(G)$$ the total size of $$\mathcal{S}$$ and by $$\nu(G,H)$$ the number of elements of $$\mathcal{S}$$ that contain a given subgroup $$H$$ of $$G$$. Suppose $$\mathcal{S}$$ does not contain the trivial subgroup of $$G$$.

To prove: $$\! \nu(G) \equiv \sum_N \nu(G,N) \pmod p$$

Proof: