Thanks for any help or comments

How can I find the list of all non abelian simple groups (particularly simple lie type) such that all $p$-Sylow subgroups are non abelian for odd prime $p$?

  • 1
    $\begingroup$ Keep in mind that if a prime $p$ divides the group order just once, a Sylow $p$-subgroup is cyclic. $\endgroup$ Nov 11, 2016 at 16:35
  • 3
    $\begingroup$ Have a look at mathoverflow.net/questions/109611/… $\endgroup$ Nov 11, 2016 at 17:44
  • 3
    $\begingroup$ @YCor There should be more interesting things to do on a Friday evening than searching for simple groups with no Sylow subgroups of prime order, but you will hopefully be pleased to learn that the group ${\rm PSU}(3,19)$ has order $16938986400 = 2^53^25^27^319^3$ and ${\rm PSp}(4,7)$ has order $138297600=2^83^25^27^4$. I suppose you might ask whether there are only finitely many such examples. I guess yes. $\endgroup$
    – Derek Holt
    Nov 11, 2016 at 22:13
  • 1
    $\begingroup$ @DerekHolt thanks! at least I've seen less worthwhile queries on this site. At this very time, Sloane's EIS doesn't recognize the sequence 138297600,16938986400. $\endgroup$
    – YCor
    Nov 12, 2016 at 0:18
  • 2
    $\begingroup$ Another pair: $|{\rm PSp}(4,41)| = 6707334818822400 = 2^83^25^27^229^241^4$, and $|{\rm PSp}(4,239)| = 304047481612332847334400=2^{10}3^25^27^213^417^2239^4$. $\endgroup$
    – Derek Holt
    Nov 12, 2016 at 9:45

1 Answer 1


Perhaps I will write an answer ( which I will mark as Community Wiki) to highlight some of the issues. From the Classification of Finite Simple Groups it is known that every finite simple group $G$ has at least one cyclic Sylow $p$-subgroup ( for an odd prime $p$ which really does divide $|G|$), which is a stronger assertion than required to answer the question. The Alternating Groups are easy to check as noted in comments, and the Sporadic Groups can be checked by easy inspection ( in the Atlas, for example).

Others are better qualified than I am to give a case by case check of the groups of Lie Type, but an analysis of the groups ${\rm PSL}(n,q)$ illustrates both why a positive answer might be expected and why it is not clear that a cyclic Sylow $p$-subgroup would necessarily be expected to have prime order ( and Derek Holt's examples of various Symplectic Groups indeed show that this need not be the case).

If we consider $H = {\rm GL}(n,q)$, then $H$ contains a Singer cycle of order $q^{n}-1,$ which gives, in particular, a cyclic subgroup of that order. By Zsigmondy's Theorem, except when $n =2$ or $n =6$ and $q = 2$, there is a prime $r$ which divides $q^{n}-1$ but does not divide $q^{i}-1$ for $ 1 \leq r < n.$ From this it follows ( apart from the exceptional cases) that $H$ has a cyclic Sylow $r$-subgroup and that ${\rm PSL}(n,q)$ has a non-trivial cyclic Sylow $r$-subgroup.

If $n =2$, then any simple ${\rm PSL}(2,q)$ has cyclic Sylow $r$-subgroups for any odd prime $r$ which divides $q^{2}-1,$ and there is at least one such prime $r$ as $q >3$ when ${\rm PSL}(2,q)$ is simple.

If $n = 6$ and $q = 2,$ then ${\rm GL}(6,2)$ has a cyclic non-trivial Sylow $31$-subgroup ( this is not as ad hoc as it might appear- in general when $n >2$ it can be checked that ${\rm GL}(n,q)$ always has a cyclic Sylow $r$-subgroup for some "Zsigmondy prime divisor" $r$ of one of $q^{n}-1$ or $q^{n-1}-1).$

Notice however, that ( at least without some deeper number theoretic analysis) this "Zsigmondy method" of producing cyclic Sylow $r$-subgroups does not a priori guarantee that $r$ will divide the group order to the first power. For example, ${\rm GL}(5,3)$ has a cyclic Sylow $11$-subgroup of order $11^{2}.$ Looking at ${\rm GL}(4,3)$ shows that ${\rm GL}(5,3)$ does have a cyclic Sylow $5$-subgroup, but in general, I see no obvious reason why all Zsigmondy prime divisors of both $q^{n}-1$ and $q^{n-1}-1$ should not occur to at least the second power.

Later edit: continuing this methodology, or by a direct argument, it is easy to see that for each prime $r$ which divides $|{\rm GL}(n,q)|,$ but not $|{\rm GL}(\lfloor \frac{n}{2} \rfloor, q) |,$ the Sylow $r$-subgroup of ${\rm GL}(n,q)$ is cyclic, and for large $n$, it would seem very improbable that each such Sylow subgroup had order greater than $r$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.