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Math Help - Groups - Alternating group question

  1. #1
    Member Jason Bourne's Avatar
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    Groups - Alternating group question

    Define A_n as the alternating group. How do I figure out how many 5-cycles A_5 has?
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    MHF Contributor

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    Quote Originally Posted by Jason Bourne View Post

    Define A_n as the alternating group. How do I figure out how many 5-cycles A_5 has?
    24. suppose \sigma=(a \ b \ c \ d \ e). clearly we have 5! = 120 of these cycles but \sigma=(b \ c \ d \ e \ a)=(c \ d \ e \ a \ b)=(d \ e \ a \ b \ c)=(e \ a \ b \ c \ d). so the number of distinct 5-cycles is 120/5 = 24.
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    Member Jason Bourne's Avatar
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    Thank you for your post. There are a few more questions concerning this and Sylow theorem that maybe you can help me with.

    (i) What is the order of every Sylow 5-subgroup of A_5 ?
    (ii) How many Sylow 5-subgroups does A_5 have?

    For (i) I think the answer is 5 because 60=2^{2}.3.5^{1}
    for (ii) I think using one of the Sylow theorems, that the number of Sylow 5-subgroups is of the form 5k+1 and divides the order of A_5, so I guess from this there is either 1 or 6 Sylow 5-subgroups in A_5, so the answer is 6?? Can anyone verify any of this?

    (iii) If P is a Sylow p-subgroup of a finite group G, and if Q is a Sylow p-subgroup of a finite group H, prove that the direct product P \times Q is a Sylow p-subgroup of the direct product G \times H.

    For (iii) i attempted to prove it like follows: Take (p_1,q_1),(p_2,q_2) \in P \times Q (p_1,p_2 \in P, q_1,q_2 \in Q)

    (p_1,q_1)(p_2^{-1},q_2^{-1}) = (p_1p_2^{-1},q_1q_2^{-1}) \in P \times Q

    Is this the right method? Thanks for any help anyone can provide.
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    Quote Originally Posted by Jason Bourne View Post
    Thank you for your post. There are a few more questions concerning this and Sylow theorem that maybe you can help me with.

    (i) What is the order of every Sylow 5-subgroup of A_5 ?
    (ii) How many Sylow 5-subgroups does A_5 have?

    For (i) I think the answer is 5 because 60=2^{2}.3.5^{1}
    for (ii) I think using one of the Sylow theorems, that the number of Sylow 5-subgroups is of the form 5k+1 and divides the order of A_5, so I guess from this there is either 1 or 6 Sylow 5-subgroups in A_5, so the answer is 6?? Can anyone verify any of this?

    (iii) If P is a Sylow p-subgroup of a finite group G, and if Q is a Sylow p-subgroup of a finite group H, prove that the direct product P \times Q is a Sylow p-subgroup of the direct product G \times H.

    For (iii) i attempted to prove it like follows: Take (p_1,q_1),(p_2,q_2) \in P \times Q (p_1,p_2 \in P, q_1,q_2 \in Q)

    (p_1,q_1)(p_2^{-1},q_2^{-1}) = (p_1p_2^{-1},q_1q_2^{-1}) \in P \times Q

    Is this the right method? Thanks for any help anyone can provide.
    your answers to (i) and (ii) are correct, although you need to explain in (ii) why the number of Sylow 5-subgroups cannot be 1. for that you need to look at your first question about the number

    of 5 cycles. for (iii) look at the orders: since P and Q are Sylow p-subgroups we have |P|=p^m, \ |Q|=p^n, \ |G|=p^m r, \ |H|=p^n s, where \gcd(p,r)=\gcd(p,s)=1. then |P \times Q|=p^{m+n} and

    |G \times H|=p^{m+n}rs and clearly \gcd(rs,p)=1. this completes the proof.
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  5. #5
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by Jason Bourne View Post
    (ii) I think using one of the Sylow theorems, that the number of Sylow 5-subgroups is of the form 5k+1 and divides the order of A_5, so I guess from this there is either 1 or 6 Sylow 5-subgroups in A_5, so the answer is 6?? Can anyone verify any of this?
    Hi Jason Bourne.

    If you are allowed to assume that A_5 is simple, then the answer is immediately obvious. If there were just 1 Sylow 5-subgroup, this would be a proper nontrivial normal subgroup of A_5; since A_5 is simple, there must therefore be more than 1 Sylow 5-subgroup (indeed more than 1 Sylow subgroup of any order).
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    Member Jason Bourne's Avatar
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    Quote Originally Posted by TheAbstractionist View Post
    If there were just 1 Sylow 5-subgroup, this would be a proper nontrivial normal subgroup of A_5
    Thanks. How do you know that if there is a unique Sylow 5-subgroup then this is Normal in A_5?
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    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by Jason Bourne View Post
    Thanks. How do you know that if there is a unique Sylow 5-subgroup then this is Normal in A_5?
    Hi Jason Bourne.

    Are you aware of the result that for each prime p dividing the order of a finite group G, all Sylow p-subgroups of G are conjugate to each other? It follows that if G has a unique Sylow p-subgroup, then that Sylow p-subgroup is normal in G.

    Think about it.
    Last edited by TheAbstractionist; May 8th 2009 at 09:09 PM.
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  8. #8
    Member Jason Bourne's Avatar
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    Quote Originally Posted by TheAbstractionist View Post
    Hi Jason Bourne.

    Are you aware of the result that for each prime p dividing the order of a finite group G, all Sylow p-subgroups of G are conjugate to each other? It follows that if G has a unique Sylow p-subgroup, then that Sylow p-subgroup is normal in G.

    Think about it.
    Oh yes I just got it! Thank you ,NonCommAlg, TheAbstractionist,thank you Mr Sylow thank you MHF! I don't know why the penny didn't drop before.
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