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 Homework Statement:
 Prove that any simple group ##G## of order ##60## is isomorphic to ##A_5##. (Hint. If ##P## and ##Q## are distinct Sylow ##2## subgroups having a nontrivial element ##x## in their intersection, then ##C_G(x)## has index ##5##; otherwise, every two Sylow ##2## subgroups intersect trivially and ##N_G(P)## has index ##5##.
 Relevant Equations:

Theorem: Let ##H## be a subgroup of ##G## with index ##n##. Then there exists homomorphism ##\psi : G \rightarrow S_n##.
Sylow's Theorem: Let ##p## be a prime divisor of ##G## and ##n_p## be the number of Sylow p subgroups of ##G##. Then ##n_p \vert \vert G \vert## and ##n_p = 1 (mod p)##. Also, the Sylow ##p## subgroups are conjugate to one another.
Proof: We note ##60 = 2^2\cdot3\cdot5##. By Sylow's theorem, ##n_5 = 1## or ##6##. Since ##G## is simple, we have ##n_5 = 6##. By Sylow's theorem, ##n_3 = 1, 4, ## or ##10##. Since ##G## is simple, ##n_3 \neq 1##. Let ##H## be a Sylow ##3## subgroup and suppose ##n_3 = 4##. Then ##[N_G(H) : G] = 4## which implies there is homomorphism ##\psi : G \rightarrow S_4##. But ##\ker \psi = 1##. By 1st iso, ##S_4## contains a subgroup isomorphic to ##G##. But ##S_4## contains strictly less elements than ##G##, contradiction. Hence, ##n_3 \neq 4##. We may conclude ##n_3 = 10##. Hence ##G## contains ##1## element of order ##1##, ##24## elements of order ##5##, and ##20## elements of order ##3##.
By Sylow's theorem, ##n_2 = 1, 3, 5,## or ##15##. By a similar argument as above, we see ##n_2 \neq 1, 3##. I think we need to show ##n_2 = 5## but i'm not sure how.
I'm also having trouble with the first part of the hint: Suppose there are distinct Sylow ##2## subgroups ##P \cap Q## that don't interact trivially. Let ##x \in P \cap Q## be a nonidentity element. Then ##x## has order ##2##. I see that if I can prove the hint, then there is an homorphism ##\psi : G\rightarrow S_5## with ##ker \psi = 1##. And therefore ##G## iso. to ##A_5##(i think...) But I'm having trouble with the first part of the hint. How can I proceed, please?
By Sylow's theorem, ##n_2 = 1, 3, 5,## or ##15##. By a similar argument as above, we see ##n_2 \neq 1, 3##. I think we need to show ##n_2 = 5## but i'm not sure how.
I'm also having trouble with the first part of the hint: Suppose there are distinct Sylow ##2## subgroups ##P \cap Q## that don't interact trivially. Let ##x \in P \cap Q## be a nonidentity element. Then ##x## has order ##2##. I see that if I can prove the hint, then there is an homorphism ##\psi : G\rightarrow S_5## with ##ker \psi = 1##. And therefore ##G## iso. to ##A_5##(i think...) But I'm having trouble with the first part of the hint. How can I proceed, please?