No group of order 720 is simple:

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

So for order 720, n_{3} can be

1) If n_{3} = 1 we are done.

2) If n_{3} = 4 we're done since there's a homomorphism mapping
G into S_{4}, and its kernel is normal.

3) If n_{3} = 10, G can be mapped into S_{10}
by multiplication on the cosets of the normalizer of a Sylow-3 subgroup.

If G isn't contained in A_{10} it has a normal subgroup of
index 2, so one can assume it's contained in A_{10}.

Suppose each Sylow-3 subgroup of G has an intersection of order 3 with
another. Let
P_{1} and P_{2} be Sylow-3 subgroups with an
intersection of order 3.

From Sylow's theorem, _{1} ∩ P_{2})_{1} ∩ P_{2}

It obviously doesn't have only one.

Suppose _{1} ∩ P_{2})_{1} and
P_{2}. The order of
_{1}) ∩ N(P_{2}) _{1}) _{1}) ∩ N(P_{2})]
is divisible by 3, but not by 9. It's not more than 9, so it
could be 3 or 6.

_{1}) ∩ N(P_{2}) <
_{1} ∩ P_{2}) ∩
_{1}),_{1} in
_{1} ∩ P_{2}). _{1} ∩ P_{2}) :
N(P_{1}) ∩ N(P_{2})] _{1}) ∩ N(P_{2})_{1})] = 10_{1}) ∩ N(P_{2})_{1}) : N(P_{1}) ∩
N(P_{2})] = 3 or 6,_{1}) ∩ N(P_{2})_{1} ∩ P_{2})]_{1} ∩ P_{2})

Suppose _{1} ∩ P_{2})_{1} ∩ P_{2})_{1} ∩ P_{2})

Suppose that no Sylow-3 subgroup intersects with another. Then the
orbit of a Sylow-3 subgroup acting by conjugation on the other
_{1} and P_{2} are two Sylow-3 subgroups,
_{1}) ∩ N(P_{2})_{10} is of order 54, and in G a normalizer of a Sylow-3
subgroup is of order 72.

So, a P_{1} is of the form
_{1}a_{2}a_{3})(b_{1}b_{2}b_{3})(c_{1}c_{2}c_{3}),
(a_{1}b_{1}c_{1})(a_{2}b_{2}c_{2})(a_{3}b_{3}c_{3})
_{9} is
centralized only by itself and is isomorphic to
_{3} x C_{3}._{1}) ∩ N(P_{2})_{1}), and is
isomorphic to a subgroup of
_{3} x C_{3}).

What does _{1}) ∩ N(P_{2})_{3} x C_{3})_{9}. So _{1}) ∩ N(P_{2})

Let _{1}a_{2}a_{3})(b_{1}b_{2}b_{3})(c_{1}c_{2}c_{3}),_{1}b_{1}c_{1})(a_{2}b_{2}c_{2})(a_{3}b_{3}c_{3})._{1}) ∩ N(P_{2})

^{2},

^{2}, β(y) =
x^{2}y^{2}, and powers of β;

^{2}, and
powers of γ.

By finding square roots I divined that

_{1}c_{3}b_{1}c_{2})(a_{2}a_{3}b_{3}b_{2})

_{1}b_{3}b_{1}a_{2})(b_{2}c_{3}a_{3}c_{2})

_{1}b_{2}b_{1}a_{3})(a_{2}c_{3}b_{3}c_{2}).

If another permutation in A_{9} had the same action as α,
say, on P_{1}, it would be equal to α multiplied by an
element of _{1}) = P_{1}._{2}. So, α, β and γ are
uniquely determined.

No non-identity element of
_{1}) ∩ N(P_{2})_{1} or
P_{2}.
So G has 45 conjugates of _{1}) ∩
N(P_{2}),_{1}) ∩ N(P_{2})_{1}) ∩ N(P_{2}).

Time now to look at elements of order 5! They are realized in
A_{10} as the product
of two disjoint

The normalizer of a product of two disjoint _{10} is order 100, so the order of a
normalizer of a _{5} = 36,_{10} is of order 25,
a

__Lemma:__ Elements from different

__Proof:__ If they did, they would generate a
subgroup of G which
could be regarded as contained in _{5} x A_{5}._{1},x_{2}) and
y = (y_{1},y_{2}) _{5} x A_{5}._{1} and y_{1}_{5}, then they generate
A_{5}. (The subgroup generated by two
_{5} can't be of order 10,
because the _{5} would act by multiplication on the cosets of the
group with a nontrivial kernel.)
So _{1} and y_{1}_{10}, and some power of this element has order 3. This isn't
possible, because the _{10} as permutations that fix only one number.
So x_{1} and y_{1} must be in the same
_{5}, similarly for
x_{2} and y_{2}. And y must be a power of x, because
otherwise some element of G would be realized as a single _{10}.

Since all the elements of order 4 in G are in some conjugate of
_{1}) ∩ N(P_{2}),_{1}a_{2}a_{3}a_{4})(b_{1}b_{2}b_{3}b_{4}), _{1} and c_{2}. If
a_{1} is in the same _{1} then
a_{2},a_{3},a_{4} are also, and
b_{1},b_{2},b_{3},b_{4} are in the other
_{1}a_{2}a_{3}a_{4})(b_{1}b_{2}b_{3}b_{4})

However, its square δ^{2} normalizes 4 _{5} has 5 conjugates of
<δ>, because if δ^{2} were normal in N(Q_{5}),
it would commute with Q_{5}, which is centralized only by itself.
So there are 36x5 pairings of cyclic groups of order 4 with

One can also show that the number of Sylow-5 subgroups that
δ^{2} normalizes is
^{2}) : N(δ^{2})
∩ N(Q_{5})], which is 4 or 2.

So when G is mapped into S_{36}
by conjugation on the 36 _{36}.

4) Suppose _{3} = 16._{1} and P_{2}.

Then, _{1} ∩ P_{2}_{1} and P_{2}.
Also, since a 5-cycle normalizes P_{1}, the 5-cycle must
centralize P_{1},
since the automorphism group of a group of order 9 doesn't have order
divisible by 5.

So _{1} ∩ P_{2}),_{1} ∩ P_{2}_{1}) and _{2}) . It
properly contains _{1}) since
it has more than one _{1} ∩ P_{2}_{1}) is a normal subgroup of
index 2 in _{1} ∩ P_{2}_{1} ∩ P_{2}_{4}
or S_{2}. And it can't be order 720 because
_{1} ∩ P_{2}

5) Suppose _{3} = 40._{1} and P_{2} intersect in a
group of order 3.
Consider again _{1} ∩ P_{2}_{1} ∩ P_{2}

It can't have order 9x2 or 9x5 because its

Suppose _{1} ∩ P_{2})_{1} ∩ P_{2})_{1} ∩ P_{2})/
(P_{1} ∩ P_{2})_{1} ∩ P_{2})_{1} ∩
P_{2}_{1} ∩ P_{2})_{1} ∩ P_{2})._{1} ∩ P_{2}),_{1} ∩ P_{2})_{1} ∩
P_{2}_{1} ∩ P_{2})_{1} ∩ P_{2}_{1} ∩ P_{2})_{1} ∩ P_{2})

Mapping G into S_{10} by multiplication on the cosets of
_{1} ∩ P_{2})_{1} ∩ P_{2}_{1} ∩ P_{2}_{1} ∩ P_{2}_{1} ∩ P_{2})_{1}a_{2}a_{3})(b_{1}b_{2}b_{3})(c_{1}c_{2}c_{3}).
However, the normalizer of such a permutation in S_{10} has
order 81x4, which is not divisible by 72, which is the order of
_{1} ∩ P_{2})

Suppose _{1} ∩ P_{2})_{1} ∩ P_{2}) = C
(P_{1} ∩ P_{2})_{10} by multiplication on the cosets of
_{1} ∩ P_{2})_{1} ∩ P_{2}_{10}.

What if _{1} ∩ P_{2}_{1} ∩ P_{2})/ (P_{1}
∩ P_{2}_{1} ∩ P_{2})/ (P_{1}
∩ P_{2}_{1} ∩ P_{2})_{1} ∩ P_{2})

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