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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