Normal group of order 60 isomorphic to A_5

In summary, Proof shows that G contains a 1 element of order 1, 24 elements of order 5, and 20 elements of order 3.
  • #1
fishturtle1
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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?
 
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  • #2
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
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
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
fishturtle1 said:
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
Math_QED said:
##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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FAQ: Normal group of order 60 isomorphic to A_5

What is a normal group of order 60?

A normal group of order 60 is a group that has 60 elements and satisfies the condition that all its subgroups are normal. This means that for any subgroup of the group, the conjugate of that subgroup by any element in the group is also a subgroup of the group.

What does it mean for a group to be isomorphic?

Two groups are said to be isomorphic if there exists a bijective homomorphism between them. This means that the two groups have the same structure and properties, but may have different elements and operations.

What is A5?

A5 is the alternating group of degree 5, which is a subgroup of the symmetric group S5. It consists of all even permutations of 5 elements, and has order 60.

Why is it significant that a normal group of order 60 is isomorphic to A5?

This is significant because it shows that A5 is the only simple group of order 60, meaning it has no non-trivial normal subgroups. This is a key result in group theory and has many applications in other areas of mathematics.

What are some real-world applications of this result?

This result has applications in various areas of mathematics, such as algebraic geometry, number theory, and topology. It also has applications in physics, particularly in the study of crystal structures and quantum mechanics. Additionally, this result has implications in cryptography and coding theory.

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