1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    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

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    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

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Non abelian groups of order 6 isomorphic to S_3
Loading...