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For a $p$-group $P$, the number of maximal subgroups is $\sum_{k=0}^r p^k$ where $r$ is the minimum size of a generating set for $P$. You can see this from looking at the maximal subgroups of $P/\Phi(P)$, which is elementary abelian of order $p^r$.

What I can tell you is that there is at least one normal subgroup for every power of $p$ up to the order of the group. Sylow theory style orbit counting gives us that the number of normal subgroups of each order $p^k$ is going to be congruent to $1 \mod{p}$, so the total number of normal subgroups in a $p$-group of order $p^n$ will then be congruent to $n+1 \mod{p}$.

EDIT: I thought of a bound.

$n+1$ is the lower bound, attained by the cyclic group of order $p^n$. There must be at least one normal subgroup for every prime power divisor, so this is the lowest it can go.

On the other hand, I claim that elementary abelian groups $E_{p^n}$ contain the largest number of normal subgroups. This is because it has the maximum rank of all groups of order $p^n$. Thinking of $E_{p^n}$ as an $\mathbb{F_p}$-vector space, we count obtain the number of subspaces by $$\mathcal{N}(E_{p^n})=\sum_{m=0}^{n}\prod_{k=0}^{m-1}\frac{p^n-p^k}{p^m-p^k}.$$ Here we count the number of ordered combinations of $m$ linearly independent vectors in $\mathbb{F_p}^n$, then divide by the number of possible bases of an $m$-dimensional subspace. Summing over $m$ we have the total number of normal subgroups in $E_{p^n}$.

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For a $p$-group $P$, the number of maximal subgroups is $\sum_{k=0}^r p^k$ where $r$ is the minimum size of a generating set for $P$. You can see this from looking at the maximal subgroups of $P/\Phi(P)$, which is elementary abelian of order $p^r$.

What I can tell you is that there is at least one normal subgroup for every power of $p$ up to the order of the group. Sylow theory style orbit counting gives us that the number of normal subgroups of each order $p^k$ is going to be congruent to $1 \mod{p}$, so the total number of normal subgroups in a $p$-group of order $p^n$ will then be congruent to $n+1 \mod{p}$.Beyond that,

EDIT: I can't think thought of a good way to count normal subgroups bound.

$n+1$ is the lower bound, attained by the cyclic group of order $p^n$. There must be at least one normal subgroup for every prime power divisor, so this timeis the lowest it can go.

On the other hand, I claim that elementary abelian groups $E_{p^n}$ contain the largest number of normal subgroups. This is because it has the maximum rank of all groups of order $p^n$. Thinking of $E_{p^n}$ as an $\mathbb{F_p}$-vector space, we count the subspaces by $$\mathcal{N}(E_{p^n})=\sum_{m=0}^{n}\prod_{k=0}^{m-1}\frac{p^n-p^k}{p^m-p^k}.$$ Here we count the number of ordered combinations of $m$ linearly independent vectors in $\mathbb{F_p}^n$, then divide by the number of possible bases of an $m$-dimensional subspace. Summing over $m$ we have the total number of normal subgroups in $E_{p^n}$.

6 Corrected the wording to make this more accurate and reduce confusion about where the sylow theorems come into this.

For a $p$-group $P$, the number of maximal subgroups is $\sum_{k=0}^r p^k$ where $r$ is the minimum size of a generating set for $P$. You can see this from looking at the maximal subgroups of $P/\Phi(P)$, which is elementary abelian of order $p^r$.

What I can tell you is that there is at least one normal subgroup for every power of $p$ up to the order of the group. The Sylow theory style orbit counting gives us that the number of normal subgroups of each order $p^k$ is going to be congruent to $1 \mod{p}$, from Sylow theory, so the total number of normal subgroups in a $p$-group of order $p^n$ will then be congruent to $k+1 n+1 \mod{p}$. Beyond that, I can't think of a good way to count normal subgroups at this time.

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