I posted this question on Math.SE (link), but it didn't get any answers so I'm going to ask here. This is an edited version of the question.

Let $p$ be a prime and $n \geq 1$ some integer. Furthermore, let $G$ be a finite group where $p$-Sylow subgroups have order $p^n$. Denote by $n_p(G)$ the number of Sylow $p$-subgroups of $G$. Denote the number of elements in the union of all Sylow $p$-subgroups of $G$ by $f_p(G)$. I am interested in finding lower bounds for $f_p(G)$ that do not depend on the group $G$, but only on $p$, $n$ and $n_p(G)$.

By Sylow's theorem, we know that $n_p(G) = kp + 1$ for some integer $k \geq 0$. What I know so far:

- If $k = 0$, then $f_p(G) = p^n$.
- If $k = 1$, then $f_p(G) = p^{n+1}$.
- If $k \geq 2$, then $f_p(G) \geq 2p^{n+1} - p^n$.

This is a theorem due to G. A. Miller, see also this question from Math.SE. To prove the inequality in the case $k \geq 2$, you first prove that then $f_p(G) > p^{n+1}$. Then observe that $f_p(G) - 1$ is divisible by $p-1$, then the inequality follows from Frobenius theorem (*). Details are in a book of Miller, Blichfeldt and Dickson (Theory and Applications of Finite Groups) and a paper of Miller ("Some deductions from Frobenius Theorem").

My main question is the following:

What is a better lower bound for the case $k > 2$?

The case $n = 1$ is easy, because then we know the value of $f_p(G)$ precisely. If $n = 1$, then $f_p(G) = n_p(G)(p-1)+1$. What about when $n > 1$? Answers regarding particular $n$ or particular $k$ are also welcome.

If the Sylow $p$-subgroups are cyclic, then we have $f_p(G) \geq n_p(G)\varphi(p^n) + p^{n-1}$ and this bound is okay. But most $p$-groups are not cyclic..

I think the following example shows that $f_p(G)$ gets arbitrarily large values for fixed $p$ and $n$ (not surprising). By Dirichlet's theorem, there exist arbitrarily large primes $q$ such that $q \equiv 1 \mod{p}$. Then in a direct product $G = C_{p^{n-1}} \times H$, where $H$ is a non-abelian group of order $pq$, the Sylow subgroups of $G$ have $C_{p^{n-1}}$ as their common intersection. There are exactly $q$ Sylow $p$-subgroups, because otherwise $G$ would be nilpotent but its subgroup $H$ is not. Therefore the number of elements in the $p$-Sylow subgroups is $f_p(G) = q(p^{n} - p^{n-1}) + p^{n-1}$, and this goes to infinity as $q$ goes to infinity. Thus there exist groups $G$ with Sylow $p$-subgroups of order $p^n$ such that $f_p(G)$ is arbitrarily large.

Also, $f_p(G) \rightarrow \infty$ as $k \rightarrow \infty$. This is seen by noticing that $f_p(G)^{p^n} \geq n_p(G)$, so

$$f_p(G) \geq (kp + 1)^{p^{-n}}$$

which goes to infinity as $k \rightarrow \infty$.

One more observation: not all integers $\equiv 1 \mod{p}$ are possible amounts of Sylow $p$-subgroups. For example, there does not exist a group with exactly $22$ Sylow $3$-subgroups, although $22 \equiv 1 \mod{3}$. I don't know if this complicates things.

(*) Frobenius Theorem says that when $G$ is a finite group with order divisible by $s$, the number of solutions to $x^s = 1$ in $G$ is a multiple of $s$. We know that $f_p(G)$ is the number of solutions to $x^{p^n} = 1$ in $G$.