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Math Help - Groups of Order 6

  1. #1
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    Groups of Order 6

    Can anyone prove that there are only 2 non-isomorphic groups of order 6.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by pleasehelpme1 View Post
    Can anyone prove that there are only 2 non-isomorphic groups of order 6.
    What knowledge do you have? What theorems do you know that can be used? I mean, if you are a masochist you can draw up the Cayley tables.
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  3. #3
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    it's a nicer problem to prove that for any prime number p there are only two groups (up to isomorphism) of order 2p.
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    ...

    can your prove that then.

    that for any prime number there are only two groups (up to isomorphism) of order
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  5. #5
    Senior Member Dinkydoe's Avatar
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    Every group has generators:

    You can start by showing that G is either
    1.generated by an element of order 6
    2.an element of order 3 and an element of order 2.

    Since all orders of the generators divide the group-order
    (This follows from the Lagrange Theorem: #[G/H].#[H] = #G
    For any subgroup H)


    Once you figured that out you show that (1) and (2) are not isomorphic.
    Secondly you show that 2 groups G and G' of type (1) are isomorhpic
    and 2 groups of type (2) are isomorphic.

    (This isomorphism is made by sending generators to generators)
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Dinkydoe View Post
    Every group has generators:

    You can start by showing that G is either
    1.generated by an element of order 6
    2.an element of order 3 and an element of order 2.

    Since all orders of the generators divide the group-order
    (This follows from the Lagrange Theorem: #[G/H].#[H] = #G
    For any subgroup H)


    Once you figured that out you show that (1) and (2) are not isomorphic.
    Secondly you show that 2 groups G and G' of type (1) are isomorhpic
    and 2 groups of type (2) are isomorphic.

    (This isomorphism is made by sending generators to generators)
    In (2) you also want the generators to not commute with one another, as otherwise you get your cyclic group again.
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  7. #7
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    Quote Originally Posted by Swlabr View Post
    In (2) you also want the generators to not commute with one another, as otherwise you get your cyclic group again.
    suppose G is a group of order 6 and let x,y \in G be with o(x)=3, \ o(y)=2. since [G:<x>]=2, we have <x> \lhd G. now consider two cases:

    1) xy=yx: in this case o(xy)=6=|G| and thus G = <xy> \cong C_6. where C_6 is the cyclic group of order 6.

    2) xy \neq yx: since <x> \lhd G, we have yxy^{-1}=x^k, for some k=0,1,2. if k=0, then x=1_G, which is not true. if k=1, then xy=yx, which is

    the first case. so k=2 and yxy^{-1}=x^2=x^{-1}. therefore G \cong D_6, the dihedral group of order 6.
    Last edited by NonCommAlg; December 18th 2009 at 11:41 PM.
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