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The Group A_4 as a Split Extension.

  1. Apr 15, 2007 #1

    Tom Mattson

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    1. The problem statement, all variables and given/known data
    Let [itex]H[/itex] be all the elements of the alternating group [itex]A_4[/itex] of exponent 2. Show that [itex]A_4[/itex] is a split extension of [itex]H[/itex] by [itex]\mathbb{Z}_3[/itex].


    2. Relevant equations
    None.


    3. The attempt at a solution
    I have already shown that [itex]H[/itex] is normal in [itex]A_4[/itex]. I also have the proof of the following proposition:

    Let [itex]H[/itex] and [itex]K[/itex] be groups whose orders are relatively prime and let [itex]f:G \rightarrow K[/itex] be an extension of [itex]H[/itex] by [itex]K[/itex]. Then the extension splits iff [itex]G[/itex] has a subgroup of order [itex]|K|[/itex].

    Let [itex]H=H[/itex] (as defined in the problem statement), let [itex]K=\mathbb{Z}_3[/itex], and let [itex]G=A_4[/itex]. Then certainly the orders of [itex]H[/itex] and [itex]K[/itex] are relatively prime. Also, [itex]A_4[/itex] certainly has a subgroup of order 3. Just take the identity, a product of disjoint transpositions, and its inverse and you have a subgroup. So referring back to the definition of an extension, I've reduced the problem to finding a surjective homomorphism [itex]f: A_4 \rightarrow \mathbb{Z}_3[/itex] with [itex]\ker(f)=H[/itex].

    Now I'm stuck. I know I need the homomorphism to map all 4 elements of [itex]H[/itex] to 0 in [itex]\mathbb{Z}_3[/itex]. But there doesn't seem to be enough variety in the remaining elements of [itex]A_4[/itex] for me to decide which of them should be mapped to 1 and which should be mapped to 2.

    Little help?
     
  2. jcsd
  3. Apr 15, 2007 #2

    StatusX

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    It doesn't matter, there's an automorphism of Z_3 taking 1 to 2. Remember the map is constant on cosets of H.
     
  4. Apr 15, 2007 #3

    Tom Mattson

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    Good grief, I can't believe I didn't notice that. :redface:

    Thank you.
     
  5. Apr 16, 2007 #4

    Tom Mattson

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    I finished the problem. In case anyone's interested, here's the rest of the solution. Not being a mathematical genius, I had to write out the multiplication table for the elements of [itex]A_4[/itex] that are not in [itex]H[/itex]. I noticed that there were two distinct equivalence classes of elements, that when multiplied by each other give an element of [itex]H[/itex] and when multiplied by an element of the other class given an element of [itex]A_4[/itex] not in [itex]H[/itex]. So I arbitrarily chose to send one class to 1 and the other to 2 (in [itex]\mathbb{Z}_3[/itex], that is).

    Elements going to 1:
    (1 2 3)
    (1 4 2)
    (1 3 4)
    (2 4 3)

    Elements going to 2:
    (1 3 2)
    (1 2 4)
    (1 4 3)
    (2 3 4)

    Clearly, [itex]f[/itex] defined in this way is a surjective homomorphism from [itex]A_4[/itex] to [itex]\mathbb{Z}_3[/itex] whose kernel is [itex]H[/itex]. And now that I'm done, I see that I have an even slicker proof that [itex]H[/itex] is normal in [itex]A_4[/itex], since it is the kernel of a homomorphism out of [itex]A_4[/itex]. I rule. :cool: :biggrin:
     
  6. Apr 16, 2007 #5

    Tom Mattson

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    I caught a mistake in the opening post.

    That doesn't give a subgroup at all. To get the desired subgroup of order 3 you have to take the identity, a 3 cycle, and its inverse.

    OK, I think I'm finally done with this. :zzz:
     
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