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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:

  1. For any group $G$ with Sylow $p$-subgroups of order $p^n$ and $n_p(G) = kp + 1$, we have $f_p(G) \geq g(k)$.

  2. $g(k) \rightarrow \infty$ 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 $k \rightarrow \infty$.

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Nice question. I guess a start would be to try to identify properties of $G$ that make the Sylow subgroups intersect a lot (given the number of Sylow subgroups), since this will be where the number of smallest. –  Tobias Kildetoft Jan 14 '13 at 15:16
    
I asked this question on MO: link. –  Mikko Korhonen Jan 20 '13 at 20:11
    
My specific research was directed at giving sharp classifications of $\liminf_{G} \log(f_p(G)/|G|_p) / \log(n_p(G)) = 1- \tfrac1p$ where the Sylow $p$-subgroup of $G$ (and its fusion) is fixed. So basically, $f_p(G) \geq \sqrt{n_p(G)} |G|_p$ or so when rephrased. –  Jack Schmidt Aug 4 '13 at 21:48
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2 Answers

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:

For fixed $p$ and $n$, find a good lower bound for $f(p,n,r)$ in terms of $r$. How does $f(p,n,r)$ grow as $r \rightarrow \infty$?

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 $f_p(G)^{p^n} \geq n_p(G)$.)

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One gets the nice relation $$\frac{f_p(G)-1}{|G|_p-1} \leq n_p(G)$$ because the $f_p(G)-1$ non-identity $p$-elements each lie in a subset of size $|G|_p-1$ consisting of the non-identity elements of some Sylow $p$-subgroup. In case the Sylow $p$-subgroups intersect trivially, we get equality and the fraction exactly counts the Sylow $p$-subgroups.

In terms of the longer $f$, this is $f(p,n,r) \leq r(p^n-1)+1$. This is obviously an upper bound, but unless the Sylow $p$-subgroup is cyclic, this is quite rare (Suzuki).

In general the the Sylow $p$-subgroups can overlap, and I studied how much they can overlap. If the Sylow $p$-subgroup is restricted to be elementary abelian of order 4, then one can show that $$\frac{f_p(G)-1}{|G|_p-1} \geq n_p(G)^{2/3} \quad \text{if } p=2, |G|_p=4$$ with equality obtained for a group of order $4(2k+1)^3$ and $n_p(G)=(2k+1)^3$.

A specific group realizing the bound is given by $$\scriptsize G(2k+1) = \left\langle \newcommand{\m}[1]{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]} \m{ -1 & . & . & . \\ . & -1 & . & . \\ . & . & 1 & . \\ . & . & . & 1 }, \m{ 1 & . & . & . \\ . & -1 & . & . \\ . & . & -1 & . \\ . & . & . & 1 }, \m{ 1 & . & . & 1 \\ . & 1 & . & . \\ . & . & 1 & . \\ . & . & . & 1 }, \m{ 1 & . & . & . \\ . & 1 & . & 1 \\ . & . & 1 & . \\ . & . & . & 1 }, \m{ 1 & . & . & . \\ . & 1 & . & . \\ . & . & 1 & 1 \\ . & . & . & 1 } \right\rangle \leq \operatorname{GL}(4,\mathbb{Z}/(2k+1)\mathbb{Z}) $$ Here $f_p(G(2k+1))=3(2k+1)^2+1$ and $n_p(G)=(2k+1)^3$. In particular for $k\geq 2$ there are fewer $2$-elements than Sylow $2$-subgroups; this cannot happen with cyclic Sylow $p$-subgroups. The bound itself is demonstrated by considering centralizers of involutions as is done in Gorenstein–Walter.

In terms of the longer $f$ we get $$f(2,2,(2k+1)^3) =r^{2/3}(p^n-1)+1.$$ (Equality follows from a result of Wielandt, most easily understood in the Gorenstein-Walter papers; note that cyclic groups of order 4 don't enter into it.)

In general we get $$f\left(p,n,(pk+1)^{(p^n-1)/(p-1)}\right) \leq r^{ \left((p-1)p^{(n-1)}\right)/\left(p^n -1\right)}(p^n-1)+1$$ with equality conjectured (because surely the minimum is obtained for elementary abelian $p$-groups; this is the minimum for $P \cong(\mathbb{Z}/p\mathbb{Z})^n$ by an argument similar to Gorenstein–Walter's).

Rather than using “$n$” as the second parameter, my studies used the Sylow $p$-subgroup $P$, or more precisely, the fusion system induced on $P$ by $G$. For abelian $P$, this is just a $p'$-subgroup $N_G(P)/C_G(P)$ of $\newcommand{\Aut}{\operatorname{Aut}}\Aut(P)$, but in general it may include subgroups $N_G(Q) /C_G(Q)$ of $\Aut(Q)$ for $Q \leq P$ as well. At any rate, in these cases the ideas of Wielandt, some linear optimization, and some explicit examples provided tight bounds. None were better than elementary abelian with trivial fusion; this is expected since overlaps imply a large degree of uniformity in $P$, but non-trivial power structure, non-trivial commutator structure, or non-trivial fusion would all restrict how copies of $P$ could overlap.

The specific groups realizing the bound for general $P\cong \mathbb{Z}/p\mathbb{Z}^n$ are “obvious” generalizations of $G(2k+1)$. By obvious I mean it only took me a few months back in Spring 2011 to figure it out, but then it was obvious.

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If it isn't clear, the exponent on $r$ is basically $1-\tfrac1p$, so basically you get $f_p \approx n_p^{(1-1/p)} p^n$. –  Jack Schmidt Aug 5 '13 at 0:52
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