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 $n_p(G)$, the number of Sylow $p$-subgroups.
By Sylow's theorem, we know that $n_p(G) = kp + 1$ for some integer $k \geq 0$. If $k = 0$, then it is clear that $f_p(G) = p^n$. If $k = 1$, we can show that $f_p(G) = p^{n+1}$. Furthermore, if $k > 1$, then by a theorem of Miller (see this question) we have $f_p(G) > p^{n+1}$, so by Frobenius theorem [*] $f_p(G) \geq p^{n+1} + p^n$. We can also improve this with Frobenius theorem to $f_p(G) \geq 2p^{n+1} – p^n$ by noticing that the number $f_p(G) – 1$ is divisible by $p-1$.
My question is this:
Can we find a better lower bound for $f_p(G)$ when $k > 1$?
I guess it would probably make sense that when $G$ has many Sylow subgroups, then there are many distinct elements among the subgroups. Thus I am also interested in the following question:
Can we show that $f_p(G) \rightarrow \infty$ as $k \rightarrow \infty$?
To make this precise, what I am asking here is for a function $g$ satisfying the following:
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For any group $G$ with Sylow $p$-subgroups of order $p^n$ and $n_p(G) = kp + 1$, we have $f_p(G) \geq g(k)$.
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$g(k) \rightarrow \infty$ as $k \rightarrow \infty$
Of course, for both questions the case $n = 1$ is easy, because then we know the value of $f_p(G)$ precisely. If $n = 1$, then $f_p(G) = (kp+1)(p-1)+1$ so for both questions we have a positive answer.
I think the following example shows that $f_p(G)$ gets arbitrarily large values. 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.
[*] Frobenius' Theorem says that when $G$ is a finite group with order divisible by $k$, the number of solutions to $x^k = 1$ in $G$ is a multiple of $k$. It is easy to see that $f_p(G)$ is the number of solutions to $x^{p^n} = 1$ in $G$.
LATER EDIT: Okay, the second question is not so difficult if I have this right. Any Sylow $p$-subgroup consists of $p^n$ elements from the $f_p(G)$ elements in the union, so we have the inequality $f_p(G)^{p^n} \geq n_p(G) = kp + 1$, and thus $$f_p(G) \geq (kp+1)^{p^{-n}}$$
which goes to infinity as $k \rightarrow \infty$. For huge $k$ this is better than the lower bound $f_p(G) \geq 2p^{n+1} – p^n$. However, this is an extremely weak bound and not so interesting, it seems to me useful only for showing that $f_p(G) \rightarrow \infty$ as $k \rightarrow \infty$.
Best Answer
For $p$ prime, $n \geq 1$ and $r \equiv 1 \mod{p}$, let $f(p,n,r)$ be the smallest possible value of $f_p(G)$ among groups $G$ where the largest power of $p$ dividing $|G|$ is $p^n$ and $n_p(G) = r$. This only makes sense if at least one such group exists, so when we talk about $f(p,n,r)$ we assume that this is the case.
Using this notation, the original question could be formulated as follows:
According to Miller's theorem, $f(p,n,1) = p^n$, $f(p,n,p+1) = p^{n+1}$ and $f(p,n,r) \geq (2p-1)p^n$ when $r \geq 2p+1$. It is also possible to show that $f(p,n,2p+1) = (2p-1)p^n$. See my answer to this question on MO. Also, it's easy to show that $f(p,1,r) = (p-1)r + 1$. Calculating the exact value (even finding a better lower bound) for $f(p,n,r)$ seems difficult in other cases.
Here is an example which suggests that the growth of $f(p,n,r)$ is slow in general. It implies that the growth is slower than linear when $p = 2$ and $n > 1$.
Let $q \equiv 3 \mod{8}$ be prime and let $G = \operatorname{PSL}(2,q)$. Then the Sylow $2$-subgroups of $G$ have order $4$. In this case $n_2(G) = \frac{q(q-1)(q+1)}{24}$ and
$$f_2(G) = \frac{q(q-1)}{2} + 1 = 4 \cdot (n_2(G)\frac{3}{q+1} + \frac{1}{4})$$
For $n \geq 2$, define $H = G \times C_{2^{n-2}}$. Now $n_2(H) = n_2(G)$ and $f_2(H) = 2^{n-2} f_2(G) = 2^n (n_2(G) \frac{3}{q+1} + \frac{1}{4})$.
This shows that $f(2,n,r) \leq 2^n(\frac{3r}{q+1} + \frac{1}{4})$ when $r = \frac{q(q-1)(q+1)}{24}$.
Hence there is no constant $C > 0$ such that $f(2,n,r) \geq Cr$ for all $r \equiv 1 \mod{p}$, since there are infinitely many primes $\equiv 3 \mod{8}$.
Questions: Is there a similar example for $p > 2$? For $n > 1$, is it possible to do any better than $$f(p,n,r) \geq r^{p^{-n}}$$
in general? That is, can we find a lower bound that grows faster than $r^{p^{-n}}$?
(The above inequality follows from the fact that every Sylow $p$-subgroup contains $p^n$ elements from the union of Sylow $p$-subgroups, so $f_p(G)^{p^n} \geq n_p(G)$.)