Hall subgroups of $\mathrm{PSL}$

The following is an exercise in Peter Cameron's notes on classical groups.

Exercise 2.10 (a) Show that $$\mathrm{PSL}(2,5)$$ fails to have a Hall subgroup of some admissible order.

(b) Show that $$\mathrm{PSL}(2,7)$$ has non-conjugate Hall subgroups of the same order.

(c) Show that $$\mathrm{PSL}(2,11)$$ has non-isomorphic Hall subgroups of the same order.

(d) Show that each of these groups is the smallest with the stated property.

I know how to solve these problems.

My questions:

(1) First consider $$\mathrm{PSL}(2,p)$$ where $$p>3$$ is a prime number. If $$\mathrm{PSL}(2,p)$$ satisfies each of the above conditions, what can I say about $$p$$?

(2) What can I say about $$p$$ if $$\mathrm{PSL}(2,p)$$ meets ALL the conditions (a)-(c)? Does such $$p$$ exist? (This is just (1), my bad. But how can I find $$p$$ such that (b) holds but (c) does not?)

(3) For a more general case, what if $$p$$ my first question is replaced by $$q$$, where $$q>3$$ is a prime power?

(4) How can we generalize (d) to $$\mathrm{PSL}(n,p)$$ or $$\mathrm{PSL}(n,q)$$ for some fixed $$n$$?

Any idea is a help. Thank you in advance!

• Well condition c) implies b), and for $p\ge 5$ we have $\operatorname{PSL}(2,p)$ simple (hence not solvable) so by the converse to Hall's theorem there are missing Hall subgroups so a) is always true for $p\ge 5$ (and if $n \ge 3$ there is no condition on $p$ and $\operatorname{PSL}(n,p)$ is always non-solvable). So the answer to question 2) is 11 still: see groupprops.subwiki.org/wiki/… which states in particular that there is no Hall subgroup of order 44 inside $\operatorname{PSL}(2,11)$. – Alex J Best May 12 at 5:08
• @AlexJBest Yes $p$ exists, but how can I find all? Or some conditions? – Hongyi Huang May 12 at 5:13
• A group with hall subgroups of all admissible orders is solvable, so (a) holds for all prime powers greater than $3$. – Derek Holt May 12 at 21:53
• Yes I understand it. Thanks for pointing out. Actually my main question is: what prime or prime power satisfies (b) but does not satisfy (c)? Are there some results?@DerekHolt – Hongyi Huang May 13 at 1:39

There are infinitely many prime powers $$q$$ for which (b) but not (c) can happen for $$G := {\rm PSL}(2,q)$$. This just depends on conditions of congruences of $$|G|$$ modulo powers of $$2$$, $$3$$, and $$5$$.

There are two mutually exclusive ways in which this can happen.

$${\mathbf 1}$$. If $$8$$ divides $$|G|$$, then $$G$$ has two conjugacy classes of subgroups isomorphic to $$S_4$$, but for some $$q$$ there are also subgroups isomorphic to the dihedral group $$D_{24}$$ of order $$24$$.

To get (b) but not (c), we need $$S_4$$ to be Hall subgroup of $$G$$. For this, we need $$3$$ but not $$9$$ to divide either $$q-1$$ or $$q+1$$, and similarly we need $$8$$ but not $$16$$ to divide $$q-1$$ or $$q+1$$. To avoid the $$D_{24}$$, we need either (a) $$3|q-1$$ and $$8|q+1$$; or (b) $$3|q+1$$ and $$8|q-1$$.

This happens, for example, when $$q = 7,23,41,103,137,151,281,\ldots$$.

$${\mathbf 2}$$. If $$60$$ divides $$|G|$$ then $$G$$ has two conjugacy classes of subgroups isomorphic to $$A_5$$. To get (b) but not (c), we need these to be Hall subgroups of $$G$$ and also for there to be no subgroup $$D_{60}$$.

I haven't written down the precise conditions for this in terms of congruences, bu it happens for example when $$q = 11,29,131,139,211,229,331,\ldots$$.

• Can this depend on congruences of $|G|$ modulo powers of $7$? How do you know that all the Hall subgroups of same order are isomorphic in your $q$'s? – Hongyi Huang May 13 at 11:42
• Unless I have made a mistake, the anwer to your first question is no. Why do you think it should depend on the congruence mod 7? As I said, in the two cases I mentioned, there are two Hall subgroups of that order, and they are isomorphic to $S_4$ in the first case, and to $A_5$ in the second case. The only other possible Hall subgroups of the same order would be dihedral groups and, as I said, you have to choose $q$ such that there are none of those. – Derek Holt May 13 at 11:57
• I think, for you first case, the condition $8$ divides $|G|$ but not $16$ implies $q\equiv 7$ or $9\pmod{16}$, while $3$ divides $|G|$ but not $9$ implies $q\equiv 2,4,5$ or $7\pmod{16}$. Only these $q$ are such that $S_4$ is Hall. But I still don't know how can avoid $D_{24}$ by your (a) and (b). For the second case I think it should be similar. But until now we can only figure out these two cases that satisfy (b) but not (c), what about other $\pi$'s? – Hongyi Huang May 13 at 12:57
• Oh $q\equiv 2,4,5$ or $7\pmod{9}$. – Hongyi Huang May 13 at 13:06
• We get $D_{24}$ or $D_{60}$ if and only if either $q-1$ or $q+1$ is divisible by $24$ or $60$, respectively. $S_4$ and $A_5$ are the only isomorphism types of subgroups of ${\rm PSL}(2,q)$ which could be Hall subgroups, and for which there is more than one conjugacy class of subgroups having that isomorphism type. To prove that you need to consider the completely list of subgroups of ${\rm PSL}(2,q)$. – Derek Holt May 13 at 14:29