SOLUTION: please help me .1.prove if G is a simple group of order 60,then show that G=A5

Question 875148: please help me .1.prove if G is a simple group of order 60,then show that G=A5
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
if G is a simple group of order 60, then show that G = A5 ( alternating subgroup of order 5)
Consider Cayley's Theorem: If G is a finite group then G is a permutation group i.e. G is a subgroup of the symmetric group.
First factor 60 = 2^2 * 3 * 5
Let n_2, n_3, n_5 be the number of Sylow 2-subgroups, Sylow 3-subgroups, Sylow 5-subgroups
Sylow's Theorem 3: Let p be a prime factor with multiplicity n of the order of a finite group G, so that the order of G can be written as pnm, where n > 0 and p does not divide m. Let np be the number of Sylow p-subgroups of G. Then the following hold:
np divides m, which is the index of the Sylow p-subgroup in G.
np ≡ 1 mod p.
np = |G : NG(P)|, where P is any Sylow p-subgroup of G and NG denotes the normalizer.
Using Sylow's third theorem, we know n_p is equivalent to 1 mod p and that n_p divides 60. Now we can list the possibilities for n_p where p = 2, 3, 5
n_2 = 1, 3, 5, 15
n_3 = 1, 4, 10
n_5 = 1, 6
Since G is simple it means that it cannot have a unique Sylow p-subgroup for any p. Which means the possibilities for n_p are:
n_2 = 3, 5 15
n_3 = 4, 10
n_5 = 6
now we can say G < S_n = n! and G < n_p!, we know G = 60, so we can rule out n_2 = 3 and n_p = 4 and the possibilities for n_p are
n_2 = 5, 15
n_3 = 10
n_5 = 6
now we have
In n_3 there are ( 5 � 4 � 3 ) / 3 = 20 elements at least of order 3 in G. 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.
In n_5 there are ( 5! ) / 5 = 24 elements at least of order 5 in G. . 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.
In n_2, we have 15 elements of order 2 for a total of 30 elements
(1 2)(3 4)
(1 2)(3 5)
(1 2)(4 5)
(1 3)(2 4)
(1 3)(2 5)
(1 3)(4 5)
(1 4)(2 3)
(1 4)(2 5)
(1 4)(3 5)
(1 5)(2 3)
(1 5)(2 4)
(1 5)(3 4)
(2 3)(4 5)
(2 4)(3 5)
(2 5)(3 4)
Thus, G has at least 1+24+20+30 = 75 elements - which is a contradiction. This means G must have precisely 5 Sylow 2-subgroups, namely n_2 = 5

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