**13.**
By direct computation, find the number of Sylow 3-subgroups
and the number of Sylow 5-subgroups of the symmetric group S_{5}.
Check that your calculations are consistent with the Sylow theorems.

*Solution:*
In S_{5} there are
( 5 **·** 4 **·** 3 ) / 3 = 20 three cycles.
These will split up into 10 subgroups of order 3.
This number is congruent to 1 mod 3,
and is a divisor of 5 **·** 4 **·** 2.

There are ( 5! ) / 5 = 24 five cycles.
These will split up into 6 subgroups of order 5.
This number is congruent to 1 mod 5,
and is a divisor of 4 **·** 3 **·** 2.

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**14.**
How many elements of order 7 are there in a simple group of order 168?

*Solution:*
First, 168 = 2^{3} **·** 3 **·** 7.
The number of Sylow 7-subgroups must be congruent to 1 mod 7
and must be a divisor of 24.
The only possibilities are 1 and 8.
If there is no proper normal subgroup,
then the number must be 8.
The subgroups all have the identity in common,
leaving 8 **·** 6 = 48 elements of order 7.

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**15.**
Prove that a group of order 48 must have a normal subgroup
of order 8 or 16.

*Solution:*
The number of Sylow 2-subgroups is 1 or 3.
In the first case there is a normal subgroup of order 16.
In the second case,
let G act by conjugation on the Sylow 2-subgroups.
This produces a homomorphism from G into S_{3}.
Because of the action, the image cannot consist of just 2 elements.
On the other hand, since no Sylow 2-subgroup is normal,
the kernel cannot have 16 elements.
The only possibility is that the homomorphism maps G onto S_{3},
and so the kernel is a normal subgroup of order 48 / 6 = 8.

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**16.**
Let G be a group of order 340.
Prove that G has a normal cyclic subgroup of order 85
and an abelian subgroup of order 4.

*Solution:*
First, 340 = 2^{2} **·** 5 **·** 17.
There exists a Sylow 2-subgroup of order 4, and it must be abelian.
No divisor of 68 = 2^{2} **·** 17
is congruent to 1 mod 5,
so the Sylow 5-subgroup is normal.
Similarly, then Sylow 17-subgroup is normal.
These subgroups have trivial intersection,
so their product is a direct product,
and hence must be cyclic of order 85 = 5 **·** 17.
The product of two normal subgroups is again normal,
so this produces the required normal subgroup of order 85.

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**17.**
Show that there is no simple group of order 200.

*Solution:*
Since 200 = 2^{3} **·** 5^{2},
the number of Sylow 5-subgroups is congruent to 1 mod 5
and a divisor of 8.
Thus there is only one Sylow 5-subgroup,
and it is a proper nontrivial normal subgroup.

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**18.**
Show that a group of order 108 has a normal subgroup of order 9 or 27.

*Solution:*
Let S be a Sylow 3-subgroup of G.
Then [G:S] = 4, since |G| = 2^{2} 3^{3},
so we can let G act by multiplication on the cosets of S.
This defines a homomorphism µ : G -> S_{4},
so it follows that | µ(G) | is a divisor of 12,
since it must be a common divisor of 108 and 24.
Thus | ker(µ) |
9,
and it follows from Exercise 7.3.1 of the text that
ker(µ) S,
so | ker(µ) | must be a divisor of 27.
It follows that | ker(µ) | = 9
or | ker(µ) | = 27.

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**19.**
If p is a prime number,
find all Sylow p-subgroups of the symmetric group S_{p}.

*Solution:*
Since |S_{p}| = p!, and p is a prime number,
the highest power of p that divides |S_{p}| is p.
Therefore the Sylow p-subgroups are precisely
the cyclic subgroups of order p, each generated by a p-cycle.
There are (p-1)! = p! / p ways to construct a p-cycle
(a_{1}, . . . , a_{p}).
The subgroup generated by a given p-cycle
will contain the identity and the p-1 powers of the cycle.
Two different such subgroups intersect in the indentity,
since they are of prime order,
so the total number of subgroups of order p in S_{p} is
(p-2)! = (p-1)! / (p-1).

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**20.**
Prove that if G is a group of order 56,
then G has a normal Sylow 2-subgroup
or a normal Sylow 7-subgroup.

*Solution:*
The number of Sylow 7-subgroups is either 1 or 8.
Eight Sylow 7-subgroups would yield 48 elements of order 7,
and so the remaining 8 elements would constitute
the (unique) Sylow 2-subgroup.

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**21.**
Prove that if N is a normal subgroup of G
that contains a Sylow p-subgroup of G,
then the number of Sylow p-subgroups of N
is the same as that of G.

*Solution:*
Suppose that N contains the Sylow p-subgroup P.
Then since N is normal it also contains all of the conjugates of P.
But this means that N contains all of the Sylow p-subgroups of G,
since they are all conjugate.
We conclude that N and G have the same number of Sylow p-subgroups.

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**22.**
Prove that if G is a group of order 105,
then G has a normal Sylow 5-subgroup
and a normal Sylow 7-subgroup.

*Solution:*
The notation n_{p}(G) will be used for the nubmer
of Sylow p-subgroups of G.
Since 105 = 3 **·** 5 **·** 7,
we have n_{3}(G) = 1 or 7, n_{5}(G) = 1 or 21,
and n_{7}(G) = 1 or 15 for the numbers of Sylow subgroups.
Let P be a Sylow 5-subgroup and let Q be a Sylow 7-subgroup.
At least one of these subgroups must be normal,
since otherwise we would have 21 **·** 4 elements of order 5
and 15 **·** 6 elements of order 7.
Therefore PQ is a subgroup,
and it must be normal since its index is the smallest prime divisor of |G|,
so we can apply the result in the previous problem.
Since PQ is normal and contains a Sylow 5-subgroup,
we can reduce to the number 35
when considering the number of Sylow 5-subgroups,
and thus n_{5}(G) = n_{5}(PQ) = 1.
Similarly, since PQ is normal and contains a Sylow 7-subgroup,
we have n_{7}(G) = n_{7}(PQ) = 1.

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