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Classification of Groups of Order 12.

  1. May 2, 2007 #1

    Tom Mattson

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    1. The problem statement, all variables and given/known data
    Classify the groups of order 12.


    2. Relevant equations
    None.


    3. The attempt at a solution
    The professor has worked this out up to a point. He proved a corollary that states:

    "Let G be a group of order 12 whose 3-Sylow subgroups are not normal. Then G is isomorphic to A_4."

    After the proof he states:

    "Thus, the classification of groups of order 12 depends only on classifying the split extensions of Z_3 by groups of order 4."

    OK, fine. So I know that split extensions are semidirect products, and that there are only 2 groups of order 4. So I need to compute the following:

    [itex]D_4 \times_{\alpha} \mathbb{Z}_3[/itex]
    [itex]\mathbb{Z}_4 \times_{\beta} \mathbb{Z}_3[/itex]

    (sorry, don't know how to make the symbol for semidirect products)

    Here's where the confusion begins. If I compare the semidirect products above with the definition of the same, then I see that I have to find the homomorphisms [itex]\alpha: \mathbb{Z}_3 \rightarrow Aut(D_4)[/itex] and [itex]\beta: \mathbb{Z}_3 \rightarrow Aut(\mathbb{Z}_4)[/itex].

    The second one isn't so bad, but I would really like to turn the first one around so that the homomorphism comes out of [itex]D_4[/itex]. That's because I've already done a homework exercise that gives me all of the homomorphisms out of [itex]D_{2n}[/itex].

    So, first question: Is [itex]D_4 \times_{\alpha} \mathbb{Z}_3[/itex] for some [itex]\alpha[/itex] isomorphic to [itex]\mathbb{Z}_3 \times_{\gamma} D_4[/itex] for some [itex]\gamma[/itex]? In other words, can I arrange it so that I'm looking for homomorphisms from [itex]D_4[/itex] to [itex]Aut(\mathbb{Z}_3)[/itex]?

    Hope the question is clear.
     
    Last edited: May 2, 2007
  2. jcsd
  3. May 2, 2007 #2

    StatusX

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    In general, no. In a non-trivial semi-direct product one of the groups in the product is normal and the other isn't (where we think of the groups making up the product as embedded in semi-direct product in the natural way). So it isn't symmetric. But Aut(D_4) is a pretty simple group, and the homomorphisms from Z_3 into it are really easy to classify (ie, find the automorphisms of order dividing 3).
     
  4. May 3, 2007 #3

    Tom Mattson

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    That's what I suspected.

    OK thanks, I'll try it.

    Here's a stupid question. The definition of semidirect product of H and K with respect to [itex]\alpha[/itex] is as follows.

    "Let [itex]\alpha: K \rightarrow Aut(H)[/itex] be a homomorphism. By the semidirect product of H and K with respect to [itex]\alpha[/itex], written [itex]H \times_{\alpha} K[/itex], we mean the set [itex]H \times K[/itex] with the binary operation given by setting

    [itex](h_1,k_1) \cdot (h_2,k_2) = (h_1 \cdot \alpha(k_1)(h_2),k_1k_2)[/itex]"

    I'm a little unsure of what the right hand side of that last equation means. Since here [itex]H=D_4[/itex] and [itex]K=\mathbb{Z}_3[/itex], I suppose that in the first coordinate of the ordered pair I'll be multiplying in [itex]D_4[/itex], and in the second coordinate "[itex]k_1k_2[/itex]" means "addition of [itex]k_1[/itex] and [itex]k_2[/itex] mod 3". Is that right? Also, I am supposing that the object "[itex]\alpha(k_1)(h_2)[/itex]" is to be read as the product of [itex]\alpha(k_1)[/itex] and [itex]h_2[/itex] in [itex]D_4[/itex]. Is that also right?
     
  5. May 3, 2007 #4

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    [itex]\alpha[/itex] is a homomorphism from K to Aut(H), so [itex]\alpha(k_1)[/itex] is an automorphism, and [itex]\alpha(k_1)(h_2)[/itex] is the element in H that this automorphism sends h_2 to.
     
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