1998 / 2007

No group of order 720 is simple:

By Sylow's theorem, np, the number of Sylow p-groups in a finite group G, is 1 mod p. Also np divides the order of the group.

So for order 720, n3 can be 1, 4, 10, 16, 40. Looking at the different cases -

1) If n3 = 1 we are done.

2) If n3 = 4 we're done since there's a homomorphism mapping G into S4, and its kernel is normal.

3) If n3 = 10, G can be mapped into S10 by multiplication on the cosets of the normalizer of a Sylow-3 subgroup.

If G isn't contained in A10 it has a normal subgroup of index 2, so one can assume it's contained in A10.

Suppose each Sylow-3 subgroup of G has an intersection of order 3 with another. Let P1 and P2 be Sylow-3 subgroups with an intersection of order 3.

From Sylow's theorem, N(P1 ∩ P2), the normalizer of P1 ∩ P2 in G, can contain 1,4 or 10 Sylow-3 subgroups.

It obviously doesn't have only one.

Suppose N(P1 ∩ P2) has 4 Sylow-3 subgroups. Two of them are P1 and P2. The order of N(P1) ∩ N(P2) is divisible by 3, but not by 9. So [N(P1) : N(P1) ∩ N(P2)] is divisible by 3, but not by 9.  It's not more than 9, so it could be 3 or 6.

N(P1) ∩ N(P2) < N(P1 ∩ P2) ∩ N(P1), the normalizer of P1 in N(P1 ∩ P2).  So [N(P1 ∩ P2) : N(P1) ∩ N(P2)] is divisible by 4 and 3. So the order of N(P1) ∩ N(P2) is a factor of 60. Since [G : N(P1)] = 10, N(P1) ∩ N(P2) actually has order a factor of 12. Since [N(P1) : N(P1) ∩ N(P2)] = 3 or 6, N(P1) ∩ N(P2) has order 12.  But then [G : N(P1 ∩ P2)] is a factor of 5, which leads to a nontrivial normal subgroup of G. So N(P1 ∩ P2) can't have 4 Sylow-3 subgroups.

Suppose N(P1 ∩ P2) has 10 Sylow-3 subgroups. Then N(P1 ∩ P2) is at least of order 90. If it's of order 90, it's generated by its Sylow-3 subgroups. Since it also contains all the Sylow-3 subgroups in G, it's normal in G.   And if the order of N(P1 ∩ P2) is a multiple of 90, G has a non-trivial normal subgroup.

Suppose that no Sylow-3 subgroup intersects with another. Then the orbit of a Sylow-3 subgroup acting by conjugation on the other Sylow-3 subgroups is of length 9, so it's transitive on the other Sylow-3 subgroups.   So if P1 and P2 are two Sylow-3 subgroups, N(P1) ∩ N(P2) is of order 8. A Sylow-3 subgroup can't be a 9-cycle, because the normalizer of a 9-cycle in A10 is of order 54, and in G a normalizer of a Sylow-3 subgroup is of order 72.

So, a P1 is of the form < (a1a2a3)(b1b2b3)(c1c2c3), (a1b1c1)(a2b2c2)(a3b3c3) >, since this is the only kind of transitive group on 9 letters with elements of order 3. This subgroup of A9 is centralized only by itself and is isomorphic to C3 x C3.   N(P1) ∩ N(P2) is a Sylow-2 subgroup of N(P1), and is isomorphic to a subgroup of Aut(C3 x C3).

What does N(P1) ∩ N(P2) look like? Aut(C3 x C3) has a Sylow-2 subgroup of order 16, which contains an element of order 8 which can't be realized as a permutation in A9. So N(P1) ∩ N(P2) is uniquely determined.

Let x = (a1a2a3)(b1b2b3)(c1c2c3), y = (a1b1c1)(a2b2c2)(a3b3c3). Then N(P1) ∩ N(P2) contains

α: α(x) = y, α(y) = x2, and powers of α;
β: β(x) = xy2, β(y) = x2y2, and powers of β;
γ: γ(x) = xy, γ(y) = xy2, and powers of γ.

By finding square roots I divined that

α = (a1c3b1c2)(a2a3b3b2)
β = (a1b3b1a2)(b2c3a3c2)
γ = (a1b2b1a3)(a2c3b3c2).

If another permutation in A9 had the same action as α, say, on P1, it would be equal to α multiplied by an element of C(P1) = P1.   So it wouldn't fix P2.   So, α, β and γ are uniquely determined.

No non-identity element of N(P1) ∩ N(P2) fixes any Sylow-3 subgroup other than P1 or P2.  So G has 45 conjugates of N(P1) ∩ N(P2), which don't intersect each other.   So each conjugate is in its own Sylow-2 subgroup of G, and there are 45 Sylow-2 subgroups. The elements of the Sylow-2 subgroup that are not in N(P1) ∩ N(P2)  are not of order 4, because the various possibilities for elements of order 4 are either odd permutations, or their square is not in N(P1) ∩ N(P2).

Time now to look at elements of order 5!   They are realized in A10 as the product of two disjoint 5-cycles, since no element of order 5 is contained in the normalizer of a Sylow-3 group.

The normalizer of a product of two disjoint 5-cycles in A10 is order 100, so the order of a normalizer of a Sylow-5 subgroup in G divides 20. From Sylow's theorem, the only possibility is for n5 = 36, so that the normalizer of the 5-cycle is order 20. Since the centralizer of the product of two disjoint 5-cycles in A10 is of order 25, a Sylow-5 subgroup of G is centralized only by itself. So there is an element of order 4 in the normalizer.   Let's see what this element looks like as a permutation on the Sylow-5 subgroups.

Lemma: Elements from different Sylow-5 subgroups can't share the same division of {1, ..., 10} between the two 5-cycles.
Proof:   If they did, they would generate a subgroup of G which could be regarded as contained in A5 x A5. Let x = (x1,x2) and y = (y1,y2) be two elements of order 5 in A5 x A5.  If x1 and y1 are in different Sylow-5 subgroups of A5, then they generate A5. (The subgroup generated by two Sylow-5 subgroups of A5 can't be of order 10, because the Sylow-5 subgroup would be normal. If it's of larger order, A5 would act by multiplication on the cosets of the group with a nontrivial kernel.)  So x1 and y1 would generate a 3-cycle.   So there's an element of order divisible by 3 in G with at least two fixed points as a permutation in A10, and some power of this element has order 3. This isn't possible, because the Sylow-3 subgroups are realized in A10 as permutations that fix only one number. So x1 and y1 must be in the same Sylow-5 subgroup of A5, similarly for x2 and y2. And y must be a power of x, because otherwise some element of G would be realized as a single 5-cycle in A10.

Since all the elements of order 4 in G are in some conjugate of N(P1) ∩ N(P2), a permutation that normalizes a Sylow-5 group is of the form (a1a2a3a4)(b1b2b3b4),   a conjugate of α, β or γ.  Call it δ.   δ doesn't move two elements, c1 and c2.  If a1 is in the same 5-cycle as c1 then a2,a3,a4 are also, and b1,b2,b3,b4 are in the other 5-cycle.   So (a1a2a3a4)(b1b2b3b4) can only normalize two Sylow-5 subgroups of G.

However, its square δ2 normalizes 4 Sylow-5 subgroups.   This is easiest to see by counting. The normalizer of a Sylow-5 subgroup Q5 has 5 conjugates of <δ>, because if δ2 were normal in N(Q5), it would commute with Q5, which is centralized only by itself. So there are 36x5 pairings of cyclic groups of order 4 with Sylow-5 subgroups that they normalize. There are 45 conjugates of the cyclic groups of order 4, so each conjugate normalizes 4 Sylow-5 subgroups.

One can also show that the number of Sylow-5 subgroups that δ2 normalizes is [N(δ2) : N(δ2) ∩ N(Q5)], which is 4 or 2.

So when G is mapped into S36 by conjugation on the 36 Sylow-5 subgroups, δ is the product of 8 4-cycles and a 2-cycle, so it is an odd permutation in S36.

4) Suppose n3 = 16. Since 16 is not 1 mod 9, one of the orbits of a Sylow-3 acting by conjugation on the other Sylow-3 subgroups has size 3. So some two Sylow-3 subgroups must have an intersection of order 3. Call two such Sylow-3 subgroups P1 and P2.

Then, P1 ∩ P2 is centralized by P1 and P2. Also, since a 5-cycle normalizes P1, the 5-cycle must centralize P1, since the automorphism group of a group of order 9 doesn't have order divisible by 5.

So C(P1 ∩ P2), the centralizer of P1 ∩ P2 in G, contains N(P1) and N(P2) .  It properly contains N(P1) since it has more than one Sylow-3 subgroup. So C(P1 ∩ P2) is of order 90, 180, 360 or 720.  If it's of order 90, then N(P1) is a normal subgroup of index 2 in C(P1 ∩ P2), which can't happen.   C(P1 ∩ P2) can't be order 180 or 360 because there would be homomorphisms of G into S4 or S2.  And it can't be order 720 because P1 ∩ P2 would be normal in G.

5) Suppose n3 = 40. Since 40 is not 1 mod 9, again some two Sylow-3 subgroups P1 and P2 intersect in a group of order 3. Consider again C(P1 ∩ P2).   C(P1 ∩ P2) might be of order 9x2, 9x4, 9x8, 9x16, 9x5, 9x10, 9x20, 9x40, 9x80.

It can't have order 9x2 or 9x5 because its Sylow-3 subgroup would be normal. It's also easy to rule out 9x16, 9x20, 9x40 and 9x80.

Suppose C(P1 ∩ P2) has order 9x4. Then C(P1 ∩ P2) has 4 Sylow-3 subgroups. C(P1 ∩ P2)/ (P1 ∩ P2) has 8 elements of order 3, so it has 12-8 = 4 elements not of order 3.   So its Sylow-2 subgroup is normal. The inverse image of this subgroup is a normal subgroup of C(P1 ∩ P2) of order 12, which is abelian since it's the product of P1 ∩ P2 and an (abelian) group of order 4. So the Sylow-2 subgroup of C(P1 ∩ P2) (call it C4) is normal in C(P1 ∩ P2).   C4 is contained in a Sylow-2 subgroup of order 16 in G, so it is a subgroup of a group of order 8, and normal in this group. So N(C4) properly contains C(P1 ∩ P2), so N(C4) is of index 10, 5, 2 or 1 in G. If it's of index 5, 2 or 1, we're done. So suppose it has index 10 in G.  Then C(P1 ∩ P2) is normal in N(C4).   But P1 ∩ P2 is characteristic in C(P1 ∩ P2) since it contains all the elements of order 3 in the center. So P1 ∩ P2 is normal in N(C4). We're done if N(P1 ∩ P2) properly contains N(C4); so assume N(P1 ∩ P2) = N(C4).

Mapping G into S10 by multiplication on the cosets of N(P1 ∩ P2), what is an element h of P1 ∩ P2 mapped into?   P1 ∩ P2 is contained in 4 Sylow-3 subgroups and has 10 conjugates in G. Since there are 40 Sylow-3 subgroups in G, two conjugates of P1 ∩ P2 cannot be contained in the same Sylow-3 subgroup.   So, h fixes no other coset of N(P1 ∩ P2). So, h is mapped into a permutation of the form (a1a2a3)(b1b2b3)(c1c2c3). However, the normalizer of such a permutation in S10 has order 81x4, which is not divisible by 72, which is the order of N(P1 ∩ P2).

Suppose C(P1 ∩ P2) has order 9x8. Then it has 4 Sylow-3 subgroups. N(P1 ∩ P2) = C (P1 ∩ P2) since otherwise G would have a subgroup of index 5, 2 or 1, which implies a normal subgroup. As above, G maps into S10 by multiplication on the cosets of N(P1 ∩ P2), which leads to a contradiction because an element of P1 ∩ P2 has a normalizer of the wrong order in S10.

What if C(P1 ∩ P2) has order 90?   Then C(P1 ∩ P2)/ (P1 ∩ P2) is of order 30,and it has 10 Sylow-3 subgroups of order 3. So it has 10 elements not of order 3. By Sylow's theorem, C(P1 ∩ P2)/ (P1 ∩ P2) could have 1 or 6 Sylow-5 subgroups. But it doesn't have room for 6 of them. So it has a normal Sylow-5 subgroup, and C(P1 ∩ P2) has a normal subgroup of order 15, which is abelian since the 5-cycle centralizes the 3-cycle. So, C(P1 ∩ P2) has a normal Sylow-5 subgroup. So the index in G of the normalizer of a Sylow-5 subgroup is a factor of 8;   applying the Sylow theorem, a Sylow-5 subgroup is normal in G.

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