# Generation in finite simple groups of Lie type

Let $S$ be a finite simple group of Lie type and $p$ be a prime such that $|S|_p=p$. I want to get some restrictions on $S$ with the conditions that $S$ is generated by elements of order $p$ and the number of Sylow p-subgroups of S divides $(p-2)!$. I expect that there is no such a group.

On one hand I have some ideas how to use the condition on the number of Sylow p-subgroups. For instance, unless $S=PSL(2,p)$ we know that $p$ is not the defining characteristic of $S$, so every elements of order $p$ in $S$ is semisimple. Then I can use the information on the centralizers of semisimple elements in groups of Lie type.

On the other hand, I am not familiar with generation in simple groups of Lie type. I am wondering if we can obtain anything on $S$ and $p$ such that $|S|_p=p$ and $S$ is generated by elements of order $p$. Thanks.

• What do you mean exactly with $|S|_p$? The $p$-valuation? I.e. $|S|_p=p$ means that $p$ divides the order of $S$, but $p^2$ doesn't? But then wouldn't $PSL_2(7)$ be an example? For $p=7$ contains 8 Sylow subgroups, and $(p-2)!=5!=120$ is divisible by 8. And of course it is generated by the elements of order 7. Jul 29, 2016 at 15:36
• To Max Horn: Yes, $|S|_p=p$ means that $p$ divides the order of $|S|$, but $p^2$ doesn't. Thank you for the example. So my question remains the same: can we obtain any information on $S$ and $p$ such that $|S|_p=p$ and $S$ is generated by elements of order $p$.
– Uep
Jul 29, 2016 at 16:47
• Indeed, I would expect most $PSL(2,p)$ to satisfy this. Indeed, a quick computation with GAP shows that your property fails for $PSL(2,8)$, but it holds for $p\in\{9,11,13,16,17,19,23,25,27\}$. It also holds for $PSL(3,3)$, but not $PSL(3,4)$. Jul 29, 2016 at 16:52
• If a non-Abelian finite simple group $G$ has order divisible by a prime $p$, then $G$ is generated by its elements of order $p$. The elements of order $p$ in $G$ form a union of conjugacy classes, so they generate a (non-trivial) normal subgroup which must be all of $G$ as $G$ is simple. Jul 29, 2016 at 19:00
• I think you should reword your question, as you seem to be aware that ${\rm PSL}(2,p)$ usually does satisfy your conditions ( for example, ${\rm PSL}(2,5)$ and ${\rm PSL}(2,7)$ already do, and it is rather rare that $p+1$ does not divide $(p-2)!$. Some prime power divisor $q^{r}$ of $p+1$ has to be greater than $(p-2)$ while we have $q^{r} \leq \frac{p+1}{2}$ unless $p$ is a Mersenne prime. This shows that the condition holds unless maybe $p$ is a Mersenne prime. But even when $p$ is Mersenne, the condition holds ( $p>3$ as $G$ is simple , so $(p-2)!$ has factors $2$ and $\frac{p+1}{2}$). Jul 29, 2016 at 19:20

You can easily find many examples of such groups using GAP. The following short program does it (it skips over the numerous examples of type $A_1$, and I did not bother to filter out alternating and sporadic examples)

it:=SimpleGroupsIterator();
for G in it do
if IsPSL(G) and ParametersOfGroupViewedAsPSL(G) = 2 then continue; fi;
n := Size(G);
primes := Set(Factors(n));
for p in primes do
if n mod p^2 = 0 then continue; fi;
N := Normalizer(G, SylowSubgroup(G, p));
numSyl := n / Size(N);
if Factorial(p-2) mod numSyl = 0 then
Print(G, "\n");
break;
fi;
od;
od;


Running this in GAP 4.8.4, I get this list (note that it lists $PSL(2,7)$, even though the code should skip $PSL(2,p)$ -- I think that's because the undocumented attribute ParametersOfGroupViewedAsPSL I used detects it as $PSL(3,2)$, which is of course isomorphic) :

PSL(2,7)
A7
PSL(3,3)
M11
Sz(8)
PSU(3,4)
M12
J_1
PSL(3,5)
M22
PSp(4,4)
PSU(3,8)
PSU(3,7)
PSL(5,2)
M23
PSL(3,8)
A11
Sz(32)
PSU(3,9)
J3
PSU(3,11)
O-(8,2)
M24
PSL(3,13)
PSU(3,13)
PSL(4,4)
PSU(4,4)
PSL(3,16)
PSp(4,9)
A13
PSU(3,16)
PSL(3,19)
G_2(5)
PSL(3,17)
PSL(4,5)
Ree(27)
PSp(4,11)
PSL(6,2)
Sz(128)
PSL(3,25)
PSL(3,23)
Ru
PSU(3,25)
PSU(3,29)
PSU(5,3)
PSL(3,27)
PSU(3,27)
PSL(3,31)
PSU(3,32)


The computation would go on, but I aborted it at this point.

• Hi Max. In your codes I don't see how $S$ is generated by elements of order $p$. Could you please double check it?
– Uep
Jul 29, 2016 at 17:19
• The code doesn't check for it, because this is automatic: Take a Sylow $p$-subgroup $P$, then the elements of order $p$ are contained in the conjugates of $P$. So they generate the normal closure of $P$ in $G$, which is $G$, since $G$ is simple. Jul 29, 2016 at 17:35
• @MaxHorn: this is a neat observation. +1 for this. Something I learnt today through this. Though the main question is not something I could get a hang of. Jul 30, 2016 at 12:26