Lower bounds on the number of elements in Sylow subgroups

Solution 1:

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

Solution 2:

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.