Normal group of order 60 isomorphic to A_5

  • #1
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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?
 

Answers and Replies

  • #2
member 587159
Let me start with proving the hint.

Note first that it is possible that ##x## has order 4. You don't necessarily know that ##x## has order ##2##. What you do know is that by Cayley's theorem that ##P\cap Q## contains some element of order ##2##. Alright, since ##P,Q## are groups of order ##4## , they are abelian so you know that ##x## commutes with every element in ## P## and ##Q## hence, since P
##P## and ##Q## are distinct you know that ##x## commutes with at least 5 distinct elements´. Hence ##|C_G(x)|\geq 5## and thus since this order also divides ##60## we have ##|C_G(x)|\geq 12## (order ##6## is not possible since then this centraliser contains a subgroup of order 3 and then we have ##5+2## distinct elements already), so the index is at most ##5##.

If ##n=[G:C_G(x)]##, then ##G## acts on the cosets of ##C_G(x)## and we have an induced group morphism ##G\to \operatorname{Sym}(G/C_G(x))\cong S_n## and since ##G## is simple the kernel is trivial so this is an injection´. Hence, ##60=|G|\leq n!##, from which it follows that ##n\geq 5##.
 
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  • #3
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Thank you! My only question is, how did we rule out ##C_G(x) \neq 10##?

Proof: Let ##P, Q## be distinct Sylow ##2## subgroups of ##G##. Suppose ##P, Q## do not intersect trivially. Then there exists non identity element ##x \in P \cap Q##. Since ##P, Q## have order ##4##, they are abelian. Hence, ##x## commutes with at least ##5## elements. We note these ##5## elements have order ##1, 2## or ##4##. So, ##\vert C_G(x) \vert \ge 5##. We know ##\vert C_G(x) \vert \vert 60##. If ##C_G(x) \ge 15##, then we can find an injective homomorphism ##\psi : G \rightarrow S_k## for some ##k \le 4##. Since ##G## has more elements than ##S_k## for ##k \le 4##, we can conclude ##C_G(x) \le 12##.

From a previous note, ##\vert C_G(x) \vert \neq 5## since it contains an element of order ##2## or ##4##. So, ##\vert C_G(x) \vert = 10## or ##12##. So, the index of ##C_G(x)## is ##5## or ##6##.

If ##n = [G : C_G(x)]##, then there is an homomorphism ##\psi : G \rightarrow S_n## with ##\ker \psi = 1##. Hence, ##G## is isomorphic to some subgroup of ##S_n##. This implies ##\vert G \vert \le n!##. So, ##n \ge 5##.

On the other hand, suppose any two distinct Sylow ##2## subgroups of ##G## intersect trivially. In the OP, we see ##n_2 = 5## or ##15##. If ##n_2 = 15##, then we get ##45## elements in addition to the ##45## we've already found, a contradiction. Hence, ##n_2 = 5##. Let ##P## be a Sylow ##2## subgroup. Then ##\vert \mathcal{O}(P) \vert = 5## where ##\mathcal{O}(P)## is the orbit of ##G## acting on ##P## by conjugation.

In either case, we've found a subgroup of ##G## with index ##5##. So there is an homomorphism ##\psi: G \rightarrow S_5## with ##\ker\psi = 1##. So, ##G## is isomorphic to some subgroup of ##S_5## of order ##60##. The only such subgroup of ##S_5## is ##A_5##. Hence, ##G \cong A_5##.[]
 
  • #5
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Response to post #4

I think I can do the element of order ##2## part, but showing ##N_G(x)## has index ##5## has me stuck, unless we use post #2.

Proof: Continuing from the OP, we have ##n_2 = 5## or ##15##. Suppose ##n_2 = 5## and let ##P## be a Sylow 2 subgroup of ##G##. Then ##N_G(P)## has index ##5##.

On the other hand, suppose ##n_2 = 15##. Then there's exactly ##15## distinct Sylow 2 subgroups. If all pairs of these intersect trivially, then we get ##45## new elements in addition to the 45 we already found (in the OP), a contradiction. So there must be distinct Sylow 2 subgroups ##P## and ##Q## such that ##P \cap Q \neq 1##. Let ##x \in P \cap Q## be a nonidentity element. If the order of ##x = 4##, then it generates both ##P## and ##Q##, a contradiction. Hence, the order of ##x## is ##2##.
 
  • #6
member 587159
Thank you! My only question is, how did we rule out ##C_G(x) \neq 10##?
##C_G(x)## contains ##P##, a group of order ##4##. Thus ##4## divides the order of ##C_G(x)## and thus this cannot be of order ##6## or ##10##.

Let me know if you need any further help!
 
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  • #7
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##C_G(x)## contains ##P##, a group of order ##4##. Thus ##4## divides the order of ##C_G(x)## and thus this cannot be of order ##6## or ##10##.

Let me know if you need any further help!
That makes sense, thank you again!
 
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