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Sylow's Counting Argument

  1. Feb 7, 2008 #1
    Hey there guys.

    Let G be a group of order 12. Show by a Sylow counting argument that if G does not have a normal subgroup of order 3 then it must have a normal subgroup of order 4.

    Deduce that G has one of the following forms:
    (i) [itex] C_3 \rtimes C_4[/itex]
    (ii) [itex] C_3 \rtimes (C_2 \times C_2)[/itex]
    (iii) [itex] C_4 \rtimes C_3[/itex] or
    (iv) [itex] (C_2 \times C_2) \rtimes C_3[/itex]

    Hence, classify all groups of order 12 up to isomorphism.

    Any suggestions on how to go about doing this one please?
    I will attempt myself once I have a good idea of what to do . Thnx :)
     
    Last edited: Feb 7, 2008
  2. jcsd
  3. Feb 7, 2008 #2

    NateTG

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  4. Feb 7, 2008 #3
    ok here is what i have for the first part.

    Please verify whether this is correct or not.

    suppose G has sylow 3-subgroup: order 3 and sylow 2-subgroup: order 4.

    Let x be number of sylow 3-subgroups then x|3 and x=1(mod3). so x = 1 or 4.
    Let y be number of sylow 2-subgroups then y|2 and y=1(mod2). so y= 1 or 2.

    We have to show either x=1 or y=1 as that would imply it has a sylow subgroup which is fixed by conjugation and is therefore a normal subgroup.

    Let's contradict:
    ------------------
    let x=4, y=2. If x has 4 distinct subgroups of order 3 then the intersection of any 2 must be a subgroup and hence a proper divisor of 3. i.e. it's 1.
    Together, the 4 subgroups have 9 distinct elements.

    On the other hand, the sylow 2-subgroup has a trivial intersection with any of the sylow 3-subgroups and hence has 3 additional distinct elements.
    So alltogether so far, we have 12 elements.

    However, we still have a remaining order 4 subgroup that adds atleast 1 more element. So in total, we have atleast 13 distinct elements. This is a contradiction because the group has only 12 elements. This means either x=1 or y=1. So there is a normal subgroup of order 4.

    any ideas on how to solve the rest of the question. i.e. as below?????.
     
  5. Feb 7, 2008 #4

    NateTG

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    [tex]2 \equiv 0 (\rm{mod} 2)[/tex]
    (The choices for [itex]y[/itex] are 1 and 3.)

    The rest of the proof for the first part is OK.

    You can do the second part as cases again.
    (What are the possibilities for [itex]G/H[/itex] for each of the normal subgroups.)
     
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