**15.**
Prove that there are no simple groups of order 200.

*Solution:*
Suppose that |G| = 200 = 2^{3} **·** 5^{2}.
The number of Sylow 5 subgroups must be a divisor of 8
and congruent to 1 modulo 5,
so it can only be 1, and this gives us a proper nontrivial normal subgroup.

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**16.**
Sharpen Exercise 7.7.3 (b) of the text by showing that if G
is a simple group that contains a subgroup of index n,
where n > 2,
then G can be embedded in the alternating group A_{n}.

*Solution:*
Assume that H is a subgroup with [G:H] = n,
and let G act by multiplication on the left cosets of H.
This action is nontrivial,
so the corresponding homomorphism
µ : G -> S_{n} is nontrivial.
Therefore ker(µ) is trivial, since G is simple.
Thus G can be embedded in S_{n}. Then
A_{n} µ(G)
is a normal subgroup of µ(G), so since G is simple, either
µ(G) A_{n}, or
A_{n} µ(G) = {e}.
The second case implies |G| = 2,
since the square of any odd permutation is even,
and this cannot happen since n > 2.

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**17.**
Prove that if G contains a nontrivial subgroup of index 3,
then G is not simple.

*Solution:*
If G is simple and contains a subgroup of index 3,
then G can be embedded in A_{3}.
If the subgroup of index 3 is nontrivial,
then |G| > 3 = | A_{3} |, a contradiction.

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**18**
Prove that there are no simple groups of order 96.

*Solution:*
Suppose that |G| = 96 = 2^{5} **·** 3.
Then the Sylow 2-subgroup of G has index 3,
and so the previous problem shows that G cannot be simple.

An alternate proof is to observe that |G| is not a divisor of 3!.

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**19.**
Prove that there are no simple groups of order 132.

*Solution:*
Let G be a group of order
132 = 2^{2} **·** 3 **·** 11.
For the number of Sylow subgroups we have
n_{2}(G) = 1, 3, 11, or 33;
n_{3}(G) = 1, 4, or 22; and
n_{11}(G) = 1 or 12.
We will focus on n_{3}(G) and n_{11}(G).
If n_{3}(G) = 4 we can let the group act on the Sylow 3-subgroups
to produce a homomorphism into S_{4}.
Because 132 is not a divisor of 24 = | S_{4} |
this cannot be one-to-one and therefore has a nontrivial kernel.
If n_{3}(G) = 22 and n_{11}(G) = 12
we get too many elements: 44 of order 3 and 120 of order 11.
Thus either n_{3}(G) = 1 or n_{11}(G) = 1,
and the group has a proper nontrivial normal subgroup.

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**20.**
Prove that there are no simple groups of order 160.

*Solution:*
Suppose that |G| = 160 = 2^{5} **·** 5.
Then the Sylow 2-subgroup of G has index 5,
and 2^{5} **·** 5 is not a divisor of 5! = 120,
so G must have a proper nontrial normal subgroup.

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**21.**
Prove that there are no simple groups of order 280.

*Solution:*
Let G be a group of order
280 = 2^{3} **·** 5 **·** 7.
In this case for the number of Sylow subgroups we have
n_{2}(G) = 1, 5, 7, or 35;
n_{5}(G) = 1 or 56; and
n_{7}(G) = 1 or 8.
Suppose that n_{5}(G) = 56 *and* n_{7}(G) = 8,
since otherwise either there is 1 Sylow 5-subgroup
or 1 Sylow 7-subgroup, showing that the group is not simple.
Since the corresponding Sylow subgroups are cyclic of prime order,
their intersections are always trivial.
Thus we have 8 **·** 6 elements of order 7
and 56 **·** 4 elements of order 5,
leaving a total of 8 elements to construct
all of the Sylow 2-subgroups.
It follows that there can be only one Sylow 2-subgroup, so it is normal,
and the group is not simple in this case.

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**22.**
Prove that there are no simple groups of order 1452.

*Solution:*
Let G be a group of order
1452 = 2^{2} **·** 3 **·** 11^{2}.
We must have n_{11}(G) = 1 or 12.
In the second case,
we can let the group act by conjugation on the set of Sylow 11-subgroups,
producing a nontrivial homomorphism from the group into S_{12}.
But 1452 is not a divisor of | S_{12} | = 12!
since it has 11^{2} as a factor, while 12! does not.
Therefore the kernel of the homomorphism
is a proper nontrivial normal subgroup,
so the group cannot be simple.

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