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Math Help - Problem re: Sylow's Theorems

  1. #1
    r45
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    Problem re: Sylow's Theorems

    Hi there,

    I had this question on an exam today and I'm not sure how to do it.

    "Let G = \mathbb{Z}_3 x A_4. List all the Sylow 2-subgroups of G, and all the Sylow 3-subgroups of G, giving reasons that your lists are complete"

    I found that G has order 36, and so I determined that n_2 = 1, 3 or 9 and that n_3 = 1 or 4 (where n_i denotes the number of Sylow i-subgroups) but from there I couldn't do much. By brute force I managed to find one subgroup of order 4 and one of order 9, so in the end I just put that n_2 = 1 and n_3= 1.

    Can anyone verify, or correct this?

    Many thanks
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by r45 View Post
    Hi there,

    I had this question on an exam today and I'm not sure how to do it.

    "Let G = \mathbb{Z}_3 x A_4. List all the Sylow 2-subgroups of G, and all the Sylow 3-subgroups of G, giving reasons that your lists are complete"

    I found that G has order 36, and so I determined that n_2 = 1, 3 or 9 and that n_3 = 1 or 4 (where n_i denotes the number of Sylow i-subgroups) but from there I couldn't do much. By brute force I managed to find one subgroup of order 4 and one of order 9, so in the end I just put that n_2 = 1 and n_3= 1.

    Can anyone verify, or correct this?

    Many thanks
    A_4 has four subgroups of order 3. For example, \left|\langle (1,2,3)\rangle\right|=3. It follows then that n_3=4 (these can actually be all found by using the fact that all Sylow p-subgroups are conjugate). Similarly, there is one subgroup of order 4 of A_4 namely \{e,(1,2)(3,4),(1,3)(2,4),(1,4)(2,3)\} from where it pretty easily follows that n_2=1.
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  3. #3
    r45
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    I thought that when looking for n_3 I was trying to find subgroups of order 9? Because 36 = 3^{2}.4, so in this case Sylow 3-subgroups have order 3^{2}=9?
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  4. #4
    r45
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    So to clarify on our Sylow 2-subgroup in the context of G as I defined above, would it be:

    \{(1, (e)), (1, (13)(24)), (1, (12)(34)), (1, (14)(23))\}?

    By the way thankyou very much for the response.
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  5. #5
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    Quote Originally Posted by r45 View Post
    I thought that when looking for n_3 I was trying to find subgroups of order 9? Because 36 = 3^{2}.4, so in this case Sylow 3-subgroups have order 3^{2}=9?

    You're right. Drexel apparently misread and thought you were talking of A_4 and not of G:=\mathbb{Z}_3\times A_4...or perhaps

    he thought to mention the 3-Sylow sbgps. of A_4 as a hint for you.

    Anyway, his description gives you some hints: any 3-Sylow sbgp. of G will be the product of \mathbb{Z}_3 with a sbgp. of order 3 of A_4 ...

    Tonio
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tonio View Post
    or perhaps

    he thought to mention the 3-Sylow sbgps. of A_4 as a hint for you.

    Anyway, his description gives you some hints: any 3-Sylow sbgp. of G will be the product of \mathbb{Z}_3 with a sbgp. of order 3 of A_4 ...

    Tonio
    This is what I was doing, sorry if there was any confusion!
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  7. #7
    MHF Contributor Swlabr's Avatar
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    You can prove there is one Sylow 2-subgroup by proving that your subgroup of order 4 is normal; in general, once you have a maximal subgroup of order p^i, conjugate it with everything you can find to find the other Sylow p-subgroups...
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