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Homework Help: Non abelian groups of order 6 isomorphic to S_3

  1. Oct 16, 2008 #1
    How can I show that all Non Abelian Groups of order 6 are isomorphic to S_3 without using Sylow's Theorems?

    I have shown the following:
    G has a non-normal group of normal subgroup of order 2
    The elements of G look like:
    1, a, a^2, b, c, d, where a,a^2 have order 3 and others have order 2

    I want G to act on the set of left cosets of a subgroup of order 2 to get a homomorphism from G onto S3, but I'm kind of confused.

    Any ideas? Thanks
  2. jcsd
  3. Oct 16, 2008 #2


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    Did you mean to say "G has a non-normal subgroup of order 2"?

    That's a good idea. Call the subgroup G is acting on H. Look at the kernel of the induced homomorphism into S_3: it's a subgroup of H.
  4. Oct 16, 2008 #3
    Oops, yes I meant non-normal s/gp of order 2.

    So if H = (1,a) where a has order 2, and non-normal, and we let G act on the left cosets of H, then the kernel is [tex]\cap_{x\in G} xHx^{-1}[/tex], so the kernel is trivial.

    I'm not sure where I'm going with this, which makes me think I did not fully understand your advice.:confused:
  5. Oct 16, 2008 #4


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    G has six elements. S_3 has six elements. We have an injective homomorphism from G into S_3. So ...!
  6. Oct 16, 2008 #5
    That would imply isomorphism, but I'm confused as to why a trivial kernel implies that the homom. is injective. (assuming I did that correctly)
  7. Oct 16, 2008 #6


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    Try to prove it. It's very straightforward. Start with what it means for a homomorphism to be injective.
  8. Oct 17, 2008 #7
    So suppose Ker p = 1 where p is homom.

    Then let x,y be in G such that p(x)=p(y)

    Then p(xy^-1)=p(x)p(y)^-1 =1 .. so x=y

    So I guess this all proves that G is isomorphic to S3.

    Thank you very much!
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