Only a solvable group can have 3 Sylow p-subgroups

Is there an easy proof that amongst finite groups, only a solvable group can have exactly 3 Sylow p-subgroups?

I have a proof, but it is a bit complex, and I'm looking to use this as an easy first example of how counting can influence global structure.

It also appears to be somewhat uniquely dramatic. If the first question is too easy, then:

Is n = 3 the only positive integer such that every finite group with exactly n Sylow p-subgroups is solvable? (where p depends on the group, and where there is at least one finite group with n Sylow p-subgroups)

I haven't found another n, even fixing p = 2. Of course, there are n such that no group has n Sylow p-subgroups, but I am not asking about such vacuous n.

• For the first question: We obviously need $p=2$ for it to be possible that there are exactly 3 Sylow $p$-subgroups. If $H$ is any group of odd order, then $G=H\times S_3$ has exactly three Sylow $2$-subgroups. Therefore any argument proving that $G$ is solvable will have Feit-Thompson as a Corollary, so is surely beyond me. – Jyrki Lahtonen Nov 7 '11 at 22:36
• Let $G$ be a group of odd order. Then $S_3 \times G$ has exactly three 2-sylows. Therefore, by your theorem $G$ is solveable! Thus, proving Feit–Thompson. – jspecter Nov 7 '11 at 22:38
• I'm not sure if I understand the question. I can think of two interpretations. 1. What are the pairs $(n,p)$ such that there exist groups with exactly $n$ Sylow $p$-subgroups and all such groups are solvable. 2. For which $n$ is it true that for all primes $p$ for which there exist groups with exactly $n$ Sylow $p$-subgroups, all such groups are solvable? – Derek Holt Nov 7 '11 at 22:49
• Perhaps I am reading this wrong, but if a group has 3 Sylow p-subgroups, one must have p=2, since $n_p(G)\equiv 1\pmod{p}$. The action on these Sylow 2-groups gives a map of $G$ into $S_3$, so it's enough to show $N_G(P)$ is solvable, for a Sylow 2-group $P$. But $N_G(P)$ has a normal Sylow 2-group, whose quotient is odd order, so by F-T - and the fact solvability is closed under extension - $N_G(P)$ (and hence $G$) is solvable. – user641 Nov 8 '11 at 4:46
• The Suzuki groups are nonabelian simple groups (in fact the only ones) with order not divisible by 3. So the direct product of $A_4$ with a Suzuki group is a nonsolvable group with $n=4$.In fact for any odd prime $p$ there are simple groups with order not divisible by $p$, so for $n$ to work there must be no groups at all with $n$ Sylow $p$-subgroups for any odd prime $p$ dividing $n-1$. – Derek Holt Nov 8 '11 at 15:37

How about $n=77$? The possible $p$ are $2$ and $19$. A group with 77 Sylow 19-subgroups would have to act primitively on them by conjugation. But using the library of primitive permutation groups, we find that there are just 4 of them, and only $A_{77}$ and $S_{77}$ are divisible by 19, and they have far too many Sylow subgroups. So there are no groups with exactly 77 Sylow 19-subgroups.

There do exist groups with 77 Sylow 2-subgroups, such as the dihedral group. But all of these appear to be solvable. Again consider the conjugation action on the set of Sylow 2-subgroups. This time the action could be imprimitive, but the blocksize could only be 7 or 11. So if there were an unsolvable group with 77 Sylow 2-subgroups, then there would have to be an unsolvable primitive group of degree 7, 11 or 77 for which the number of Sylow 2-subgroups was a divisor of 77. But checking the library of primitive groups shows that there is no such group.

$n=143$ might be another example. Perhaps there are infinitely many, but that might be hard to prove.

(This CW takes the ideas presented by Jyrki Lahtonen, JSpecter, Derek Holt, and Steve Dalton, and adjusts the question to something more reasonable.)

Original question 2:

No, n = 77 is another such positive integer.

Part of the solution involved proving that no group had n Sylow p-subgroups for p odd. This is necessary because:

Proposition: If a finite group has n Sylow p-subgroups for odd p, then a finite insoluble group has n Sylow p-subgroups for odd p.

The proof is silly: one just realizes that for every odd prime p there is a finite simple group whose order is coprime to p. For p = 3, one takes a Suzuki group, and for p ≥ 5, one takes an appropriate PSL(2,q).

Original question 1 and a better question 2:

Suppose G has n Sylow p-subgroups, one of them P. Let N = NG(P) be the normalizer of P in G, and let K be the normal core of N in G. Since P is normal in N, N is clearly p-solvable. One has that G/K is p-solvable if and only if G is p-solvable, hence we can assume the action of G on the Sylow p-subgroups is faithful as far as p-solvability is concerned.

This suggests we change "solvable" to "p-solvable" for several reasons:

• The solvable case has a weird "no odd p" requirement, and p = 3 can never be interesting.
• The solvable case for any particular n that works has the Feit–Thompson theorem as a corollary, so the proof must include, at least by reference, fairly difficult ideas.
• The p-solvable case in several instances involved no heavy machinery.
• 2-solvable = solvable by Feit–Thompson, so for p = 2 not much has changed

One thus has the more interesting, but probably much harder question:

What are the pairs (n, p) such that every finite group with n Sylow p-subgroups is p-solvable? (again where we ignore (n, p) such that there are no finite groups with n Sylow p-subgroups)

When n is small (n ≤ 32 for me, n ≤ 63 for some) one has a complete list of possible Sylow actions, and one can just check the list. The result is a much larger list now: (n, p) in

{ (3,2), (4,3), (7,2), (7,3), (11,2), (11,5), (13,2), (13,3), (16,3), (16,5), (19,2), (19,3), (23,2), (23,11), (25,3), (27,13), (29,2), (29,7), (31,2), (31,3), (31,5) }

or n ≥ 33. The original question only had n in { 3 } or n ≥ 33.

When n is medium (n ≤ 2499 for me) one has a complete list of primitive groups, and one can often leverage this to understand Sylow actions. Derek used this to find a larger example, n = 77 that works with the stronger requirement of solvability for all p.

When n = 3 and p = 2, then G/K ≅ S3 is p-solvable, and since KN is p-solvable, G itself must be p-solvable, using no difficult results. Of course, if n = 3, then p divides n − 1 = 2 by Sylow's theorem, but this seems like an unnecessary complication now that Derek has presented his answer.

When n = 77 and p = 2, then we consult the four primitive groups of degree 77 (all of which have at least 3465 Sylow 2-subgroups) to see that N is contained in a maximal subgroup M such that the indices $[G:M],[M:N]$ are either 7,11 or 11,7. Since N is also a Sylow p-normalizer in M, we get that M/K is either dihedral or affine of degree 7 or 11 (four possibilities, D7, D11, AGL(1,7), AGL(1,11)). In all cases M is p-solvable, so the G-core L of M is p-solvable. The possibilities for G/L are similarly restricted: a subgroup of an affine group in degree 7 or 11. In particular, both G/L and L are p-solvable, so G is p-solvable, and n = 77 is another example.

Of course, p must divide n−1, and the list of primitive groups of degree 77 rules out p = 19 (no intermediate M can exist by Sylow's theorem since neither 7 nor 11 is equivalent to 1 mod 19, but none of the primitive groups actually has 77 Sylow p-subgroups). Thus even when we had not specified p = 2, this still was an example.

When n = 143 and p = 71, one gets that N cannot be contained in M, so the action is primitive, but no primitive group of degree n has n Sylow p-subgroups, so this case does not occur.

When n = 143 and p = 2, the argument is more complicated (the action must be imprimitive, the only insoluble action on 11 or 13 points is by L2(11), but the number of Sylow p-subgroups of a normal section of a group divides the number of Sylow p-subgroups of the whole group, and n2(L2(11)) = 55 does not divide 143), but again is an example.

• Concerning $n=143$, the only non-solvable primitive group of degree 11, 13 or 143 with at most 143 Sylow 2-subgroups is $L_2(11)$ with 55 Sylow 2-subgroups. But since 55 does not divide 143, I don't believe that any group having $L_2(11)$ as a composition factor can have 143 Sylow 2-subgroups. – Derek Holt Nov 9 '11 at 13:29
• @Derek: Thanks, that is very clear. $n_p(G) = n_p(G/K) \cdot n_p(K) \cdot [ N_K(P\cap K):N_G(P) \cap K ]$. – Jack Schmidt Nov 12 '11 at 20:47