When the number of $p$-Sylows in a Simple group is odd and $p$ is odd

Question

From a paper by Hall, if $n$ is the number of $p$-Sylow subgroups in some finite group $G$ for some prime $p$ dividing $|G|$ then $n$ is a product of numbers of two kinds

• Those numbers which are the number of $p$-Sylows in a simple group
• The set of primes or prime powers congruent to $1 \pmod p$

The latter of these cases is not as particularly interesting as the former. For example, we could always construct a semidirect product $(\mathbb Z_q)^k \rtimes \mathbb Z_p$ Where $q^k = 1 \pmod p$ is some prime power.

The paper also says:

Any odd number may be the number of Sylow 2-subgroups in a finite group. But for every prime p > 2, not every integer n = 1 (mod p) is the n_p of a finite group. In particular Theorem 3.2 shows that there is no finite group with n_3 = 22, n_5 = 21, or n_p = 1 + 3p for p > 7. As with so many questions, the mysterious part as to the possible values for n_p lies in the study of the finite simple groups.

However, assuming no flaws in the classification of finite simple groups, we should be able to derive a full classification of the possible $n_p$'s (the possible values of the number of $p$-Sylows for some finite group). In particular, it appears in most cases, $n_p$ is even, and also highly composite. For example, $A_9$ has $4320=2^5*3^3*5$, 7-Sylow subgroups. The Mathieu group $M_{22}$ has $22176=2^5*3^2*7*11$, 5-Sylow subgroups. Upon further inspetion, it appears the following is true when $n_p$ is odd:

Let $G$ be a simple group.

My Claim: If $p$ is an odd prime dividing $|G|$, and $n_p$ is the number of $p-$Sylows in $G$, then $n_p \text{ is odd} \implies \exists k \text{ } n_p = k(k+1)/2$. (Main conjecture)

Additionally:

• One of $k$ or $k+1$ is a prime power congruent to $\pm1 \pmod p$.
• G = PSL$(2,q)$ for some prime power $q=\pm 1 \pmod p$ ($q=k$ or $k+1$).

For example, it appears that:

PSL$(2,q)$ has $q(q+1)/2$ $\text{ }$ $p-$Sylows if $q=1 \pmod p$

PSL$(2,q)$ has $q(q-1)/2$ $\text{ }$ $p-$Sylows if $q=-1 \pmod p$

(I suspect this is not hard to prove, especially when $p=3$)

---

I ran a GAP script to iterate over Simple groups and determine the number of $p$-Sylows for each group, (e.g. $p = 3,5,7,11,13, ...$) and didn't find a single counter example.

For example, for simple groups up to order $PSL(2,521) = 70710120$, the $n_p$'s (sorted in order) are $\{10, 28, 52, 55, 70, 82, 91, 136, 160, 190, 244, 253, 280, 325, 406, 496, 703, 730, 820, 880, 910, 946, ...\}$

In order of the sizes of the simple groups they are:

Group: A5
Order: 60
Number of 3-Sylow subgroups: 10
Group: PSL(2,7)
Order: 168
Number of 3-Sylow subgroups: 28
Group: A6
Order: 360
Number of 3-Sylow subgroups: 10
Group: PSL(2,8)
Order: 504
Number of 3-Sylow subgroups: 28
Group: PSL(2,11)
Order: 660
Number of 3-Sylow subgroups: 55
Group: PSL(2,13)
Order: 1092
Number of 3-Sylow subgroups: 91
Group: PSL(2,17)
Order: 2448
Number of 3-Sylow subgroups: 136
Group: A7
Order: 2520
Number of 3-Sylow subgroups: 70
Group: PSL(2,19)
Order: 3420
Number of 3-Sylow subgroups: 190
Group: PSL(2,16)
Order: 4080
Number of 3-Sylow subgroups: 136
Group: PSL(3,3)
Order: 5616
Number of 3-Sylow subgroups: 52
Group: PSU(3,3)
Order: 6048
Number of 3-Sylow subgroups: 28
Group: PSL(2,23)
Order: 6072
Number of 3-Sylow subgroups: 253
Group: PSL(2,25)
Order: 7800
Number of 3-Sylow subgroups: 325
Group: M11
Order: 7920
Number of 3-Sylow subgroups: 55

The first few $n_{p=3}$'s that are odd appear to be:

$\{55,91,253,325,703,1081,1225,1711,1891,2485,2701,3403,4753,5671,5995,7381,8515,...\}$

but these are all triangular numbers.

Similarly, the ones for $p=5$ are
Group: A5
Order: 60
Number of 5-Sylow subgroups: 6
Group: A6
Order: 360
Number of 5-Sylow subgroups: 36
Group: PSL(2,11)
Order: 660
Number of 5-Sylow subgroups: 66
Group: A7
Order: 2520
Number of 5-Sylow subgroups: 126
Group: PSL(2,19)
Order: 3420
Number of 5-Sylow subgroups: 171
Group: PSL(2,16)
Order: 4080
Number of 5-Sylow subgroups: 136
Group: PSL(2,25)
Order: 7800
Number of 5-Sylow subgroups: 26
Group: M11
Order: 7920
Number of 5-Sylow subgroups: 396

We can see the smallest example of an odd $n_{p=5}$ being $171 = (19*18)/2$ where the corresponding subgroup is PSL(2,19).

The next smallest example appears to be PSL(2,41) with $n_{p=5} = 861 = (41*42)/2$ 5-Sylows.

I'm not quite expecting a full answer, but I am wondering if all known Simple groups (particularly the groups of Lie Type) admit a nice classification to the number of corresponding $p$-Sylows. I expect the $n_p$'s for sporadic groups to be "highly composite" (generally). Thanks for any feedback, even if it's not a direct answer!

Answer 1

Consider $PSL(3,29)$. It has order $499631102880=2^5*3*5*7^2*13*29^3*67$, so a Sylow $7$-subgroup has order $7^2$. If I didn't make a mistake, $n_7=106214095$, which is not a triangular number.

$n_p$ is equal to the index of the normaliser of a Sylow $p$-subgroup. These are quite well understood for simple groups. In particular, if this is to be odd, the simple group will have a subgroup of odd index, and thus a maximal subgroup that happens to have odd index. This is somewhat rare already and this situation has more or less been classified. Of course having a maximal subgroup of odd index is not sufficient, but this could serve as a starting point if you want to understand this situation better.

Answer 2

Although as @verret showed, the conjecture is not correct in general, in the simplest case $G=PSL(2,q),$ your values for $n_p(G)$ are correct and follow from well-known properties of $G.$ We have $|G|=q(q^2-1)/d$ where $d=\gcd(2,q+1),$ and $G$ contains cyclic subgroups $N,$ $D$ and $C$ of orders $|N|=q,$ $|D|=(q-1)/d$ and $|C|=(q+1)/d.$ (The subgroups are the images in $G$ of respectively the group of upper unitriangular matrices, the group of determinant one diagonal matrices, and the group generated by the matrices, with respect to a chosen vector space basis of $F_{q^2}$ over $F_q,$ of the subgroup of $F_{q^2}^\times$ of order $q+1$ acting by multiplication.)

These subgroups have respectively $n(N)=q,n(D)=q(q+1)/2$ and $n(C)=q(q-1)/2$ conjugates in $G$ and all have the property that $H\cap H^g=H$ or $\{1\}$ of $g\in G$ (``trivial intersection'') where $H=N,D$ or $C.$

No odd prime divides $|G|/(|N||D||C|)$ or two of $|N|,|D|$ or $|C|,$ so if $P\subseteq G$ is a Sylow $p$-subgroup for $p$ odd, then up to conjugacy $P$ is contained in one of $N,D$ or $C.$ Because the subgroups are cyclic and have the trivial intersection property mentioned above, then $n_p=n(N),n(D)$ or $n(C)$ accordingly. Hence one of the following occurs, as expected in the question:

• $p$ divides $q$ and $n_p=n(N)=q+1$ (a triangle number only if $q=9,$ but $pn_p$ is even in this case)

• $p$ divides $q-1$ and $n_p=n(D)=q(q+1)/2,$ a triangle number

• $p$ divides $q+1$ and $n_p=n(C)=q(q-1)/2,$ a triangle number

The case of the alternating groups $G=A_n$ is also straightforward for the disappointing reason that for odd $p,$ $n_p(G)=n_p(S_n)$ is even except for $n<p$ or $n=p=3,$ when $n_p=1.$ If $n=n_0+\cdots n_kp^k$ with $0\le n_i<p$ is the base $p$ expansion of $n,$ then $$n_p(S_n)=\frac{(n!)_{p^\prime}}{n_0!\cdots n_k!\cdot (p-1)^{n_1+2n_2+\cdots+kn_k}}.$$ Since $p$ is odd it's enough to show evenness of $$(n!)_pn_p(S_n)=\binom{n}{n_0,\dots,p^kn_k}\prod_{i=1}^k\binom{p^in_i}{n_i}\prod_{i=1}^k\frac{((p^i-1)n_i)!}{(p-1)^{in_i}}.$$ We claim that the final product on the right is an even integer unless either $k=0$ or $k=n_1=1.$ For any $i$ with $n_i>0$ the numerator of the $i^{th}$ factor in this product, contains factors $(p^j-p^k)l$ for $1\le l\le n_i,$ $1\le j\le i$ and $0\le k<j.$ These are distinct since $p^jl+p^{k^\prime}l^\prime=p^{j^\prime}l^\prime+p^kl$ are two base $p$ expansions, so if they are equal with $k^\prime\neq j^\prime,$ we must have $k^\prime=k$ and $l^\prime =l.$ Each is divisible by $p-1$ and there are $n_ii(i+1)/2$ of them. Hence the second product on the right is an integer, and is even (as $p-1$ is even) unless we always have $i(i+1)/2n_i=in_i$ and no even $l's$ appear, i.e. either $k=0,$ i.e. $n<p,$ or $k=n_1=1.$ In that last case $n=n_0+p\le 2p-1$ and $n_p(S_n)=\binom{n}{n_0}(p-2)!$ is even unless $p=3,$ which requires $n=3,4$ or $5$ and we check $n_3(S_n)=1,4$ or $10.$