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Math Help - Isomorphism

  1. #1
    Super Member Showcase_22's Avatar
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    Isomorphism

    Let G=S_4 be the group of all permutations of \{1,2,3,4 \}.

    The subset N=\{1,(1,2)(3,4),(1,3)(2,4),(1,4)(2,3)\} of G is a normal subgroup of G.

    By proposition 3.5 of the notes, since |G-N|=|G|/ |N|=6,either G -N \cong C_6 (a cyclic group of order 6) or G-N \cong D_6.

    Decide which and explain your answer.
    G-N is abelian since every element is it's own inverse (I proved this in the last homework I had).

    D_6 is not abelian. This is the group of isometries of a triangle. If we label the vertices 1,2 and 3 respectively, the permutations are a=(1,2,3) and b=(2,3). ab sends 2 to 1 but ba sends 2 back to 2. Hence it isn't abelian.

    For two groups to be isomorphic, they must be structurally similar (ie. G-N must be isomorphic to another abelian group).

    Therefore G-N \cong C_6.

    I think this is right, but one of my friends has got it isomorphic to D_6. Which one of us is right? (I can't see a flaw in my reasoning).
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    G-N is abelian since every element is it's own inverse (I proved this in the last homework I had).

    D_6 is not abelian. This is the group of isometries of a triangle. If we label the vertices 1,2 and 3 respectively, the permutations are a=(1,2,3) and b=(2,3). ab sends 2 to 1 but ba sends 2 back to 2. Hence it isn't abelian.

    For two groups to be isomorphic, they must be structurally similar (ie. G-N must be isomorphic to another abelian group).

    Therefore G-N \cong C_6.

    I think this is right, but one of my friends has got it isomorphic to D_6. Which one of us is right? (I can't see a flaw in my reasoning).
    Let g \in S_4 such that <gN> = G/N \cong C_6, the cyclic group of order 6. Then clearly o(g) \geq 6 (specifically, 6 | o(g)).

    So, can you find an element in your group of order greater than or equal to 6?
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  3. #3
    Super Member Showcase_22's Avatar
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    Well, since every element is it's own inverse the highest order we have is 2.

    Then clearly <br /> <br />
o(g) \geq 6<br />
(specifically <br /> <br />
6 | o(g)<br />
).
    I'm afraid this isn't so clear! This looks like Lagrange's theorem, but isn't it the wrong way round?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    G-N is abelian since every element is it's own inverse (I proved this in the last homework I had).

    D_6 is not abelian. This is the group of isometries of a triangle. If we label the vertices 1,2 and 3 respectively, the permutations are a=(1,2,3) and b=(2,3). ab sends 2 to 1 but ba sends 2 back to 2. Hence it isn't abelian.

    For two groups to be isomorphic, they must be structurally similar (ie. G-N must be isomorphic to another abelian group).

    Therefore G-N \cong C_6.

    I think this is right, but one of my friends has got it isomorphic to D_6. Which one of us is right? (I can't see a flaw in my reasoning).
    If you proved that every element of G/N is of order two, how could it be a cyclic group??? You would have to have an element of order six.
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  5. #5
    Super Member Showcase_22's Avatar
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    I see!

    So it can't be isomorphic to C_6, so it must be isomorphic to D_6 instead.

    Thankyou!
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    Well, since every element is it's own inverse the highest order we have is 2.
    So...every (non-trivial) element has order 2? This is another problem. If you haven't already, you will encounter a theorem called Sylow's theorem. If you have encountered it, apply it to your comment I just quoted. If not, what can you tell me about the generating elements of D_6?
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  7. #7
    Super Member Showcase_22's Avatar
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    2 is prime.

    Therefore Given a prime number p, \ (\forall \ g \in G)( \exists n \in \mathbb{N})[g^{p^n}=1] \ .

    As I said before, n=1 and p=2.

    G/N is not a proper subgroup of any other p-subgroup of G. This is because |G|/|N|=6 so the only possible primes that divide 6 are 3 and 2.

    Let n_2 be the number of sylow-2 subgroups. Then [tex]n_2|3[/MATh] and n_2=1 \mod 2. Then we have that n_2=1 or 3 since these are the only numbers that satisfy these constraints.<--- I'm not sure where to go from here, i'm trying to show that G/N is not a proper subgroup of any other p-subgroup of G. I think want to show that n_2=3?

    As for your other question, there must be more than 1 generator for D_6. Therefore it cannot be cyclic.
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  8. #8
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    2 is prime.

    Therefore Given a prime number p, \ (\forall \ g \in G)( \exists n \in \mathbb{N})[g^{p^n}=1] \ .

    As I said before, n=1 and p=2.

    G/N is not a proper subgroup of any other p-subgroup of G. This is because |G|/|N|=6 so the only possible primes that divide 6 are 3 and 2.

    Let n_2 be the number of sylow-2 subgroups. Then n_2|3 and n_2=1 \mod 2. Then we have that n_2=1 or 3 since these are the only numbers that satisfy these constraints.<--- I'm not sure where to go from here, i'm trying to show that G/N is not a proper subgroup of any other p-subgroup of G. I think want to show that n_2=3?

    As for your other question, there must be more than 1 generator for D_6. Therefore it cannot be cyclic.
    A nice corollary of Sylow's theorem (it may be a corollary of an earlier theorem, but it very obviously falls out of Sylow's theorem) is that for every prime p dividing the order of your group you have a subgroup of order p^n.

    Looking for a contradiction to what you said about every element having order 2, apply this to your case.
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