Let $G$ be a finite group and $H$ a subgroup of $G$. We say that $H$ is a trivial intersection (for short T.I.) subgroup of $G$ if $H\cap H^x=1$ for each $x\in G-N_G(H)$.

I read the next result in the paper The p-local ranks of finite simple groups with abelian Sylow p-subgroups.

Theorem Let $G$ be a non-abelian simple group with a non-cyclic T.I. Sylow $p$-subgroup $P$. Then $G$ is isomorphic to one of the following groups:

(a) $\mathrm{PSL}_2(q)$, where $q=p^n$ and $n \geq 2$;

(b) $\operatorname{PSU}_3(q^2)$, where $q^2=p^n$;

(c) $p=2$ and $G \cong{ }^2 B_2(2^{2 m+1})$;

(d) $p=3$ and $G \cong{ }^2 G_2(3^{2 m+1})$ and $m \geq 1$;

(e) $p=3$ and $G \cong \mathrm{PSL}_3(4)$ or $M_{11}$;

(f) $p=5$ and $G \cong { }^2 F_4(2)^{\prime}$ or $McL$;

(g) $p=11$ and $G \cong J_4$.

I am wondering whether there is a similar version for non-abelian simple groups with a cyclic T.I. Sylow $p$-subgroup?

Note that the class of non-abelian simple groups with $p$-part $p$ is one of the classes. But, this is not the only one. For instance, $\mathrm{PSL}_2(17)$ and $\mathrm{PSL}_3(8)$ both have a cyclic T.I. Sylow $3$-subgroup of order $9$.


1 Answer 1


This is answered in the paper of Harvey Blau, "On Trivial Intersection of Cyclic Sylow Subgroups" Proc AMS 1985. Whenever a Sylow $p$-subgroup of a finite simple group is cyclic, it is T.I. The proof uses the classification, of course. The proof shows that if a finite simple group has a cyclic Sylow $p$-subgroup of order greater than $p$ then the group has Lie type with characteristic not equal to $p$. There are very many like this.

  • $\begingroup$ Thank you very much! That is very helpful. By the way, Is there a classification of finite simple groups with cyclic Sylow $p$-subgroup (of order larger than $p$)? $\endgroup$
    – user44312
    Commented Feb 1 at 8:56
  • $\begingroup$ For every Lie type there are many. Random example: ${^3}\!D_4(19)$ at the prime $13$. $\endgroup$ Commented Feb 1 at 9:42
  • $\begingroup$ In this example, $19^4-19^2+1$ is divisible by $13^2$ but no other irreducible factor in the group order formula is divisible by $13$. This gives you a clue as to how to find many such examples. The same happens for $89^4-89^2+1$, as a random example. $\endgroup$ Commented Feb 1 at 10:02
  • 1
    $\begingroup$ I see. Thanks again for your detailed explanation! $\endgroup$
    – user44312
    Commented Feb 1 at 10:28
  • $\begingroup$ If you want a more extreme example, $\operatorname{\rm PSL}(3,2819)$ has a cyclic T.I. Sylow $19$-subgroup of order $19^4$. $\endgroup$ Commented Feb 1 at 11:50

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