If a finite group has $\geq p+1$ Sylow $p$-subgroups (of order $p^n$), then there are $\geq p^{n+1}$ elements in the Sylow $p$-subgroups Suppose $G$ is a finite group with Sylow $p$-subgroups of order $p^n$. I am trying to show that if there are $\geq p+1$ Sylow $p$-subgroups, then the union of all these subgroups has order $\geq p^{n+1}$.
In the case where there are exactly $p+1$ Sylow $p$-subgroups, we can find a homomorphism $G \rightarrow S_{p+1}$, which has kernel $K$ of size $p^{n-1}$ since $(p+1)!$ is divisible by $p$ but not by $p^2$. Then if $P$ is a Sylow $p$-subgroup, the subgroup $PK$ has order power of $p$, hence $|PK| \leq p^n$ and by the order formula for $PK$ we find that $|P \cap K| \geq p^{n-1}$. Therefore $K \leq P$. Since $K$ is contained in every Sylow $p$-subgroup, the intersection of all Sylow $p$-subgroups has order $p^{n-1}$. Thus there are $(p+1)(p^n - p^{n-1}) + p^{n-1} = p^{n+1}$ elements in the union of all Sylow $p$-subgroups.
Now I'm stuck. How can I handle the case where there are $> p+1$ Sylow $p$-subgroups?
 A: There is a proof of this fact in Theory and Applications of Finite Groups by G. A. Miller, H .F. Blichfeldt and L. E. Dickson, in the chapter about Frobenius' theorem (pages 79-80). They actually prove a stronger result as well: when there are $> p+1$ Sylow $p$-subgroups, then the number of elements in the union of Sylow $p$-subgroups is $> p^{n+1}$. But here's a proof of just the fact asked about, using the ideas in Miller/Blichfeldt/Dickson (See here and here for the relevant pages from the book).
Let $P_1$ and $P_2$ be distinct Sylow $p$-subgroups such that their intersection $D = P_1 \cap P_2$ has largest possible order, say $|D| = p^k$. Let $g \in N_{P_1}(D)$ such that $g \not\in D$. The existence of such $g$ is guaranteed since $p$-groups satisfy the normalizer condition.
Then $g$ has order $\geq p$ and $g \not\in P_2$, so there are at least $p$ distinct conjugates of the form $g^i P_2 g^{-i}$. Each of these conjugates contains $D$ since $g^i D g^{-i}  = D$, thus by the maximality of $D$ the intersection of these conjugates is equal to $D$.
When we include $P_1$ among these conjugates, we get a family of at least $p+1$ distinct Sylow subgroups which have the common intersection $D$. Thus there are $$(p+1)(p^n - p^k) + p^k = p^{n+1} + p^n - p^{k+1} \geq p^{n+1}$$ elements among this family.
