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Subgroups of Z6

  1. Mar 14, 2010 #1
    Hi guys

    I don't really understand how exactly to FIND subgroups of a given group..... Is there any specific process to do so?
    1. The problem statement, all variables and given/known data

    Find all subgroups of Z6.

    2. Relevant equations



    3. The attempt at a solution

    How does one find subgroups??
    Z6 = Z2 x Z3
    Am I right in saying this?

    I think Z6 = {0,1,2,3,4,5}...

    Then do I just look at each element or something? I'm really bad at abstract algebra.. Can someone attempt and explain the steps to take to determine a subgroup?
    Thanks
     
  2. jcsd
  3. Mar 14, 2010 #2

    jbunniii

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    In general it's a hard problem to find all subgroups of a given group.

    In your case it's much easier, because [itex]Z_6[/itex] is cyclic. Therefore all of its subgroups must also be cyclic. (Why?)

    A cyclic subgroup is generated by a single element. You only have six elements to work with, so there are at MOST six subgroups. Work out what subgroup each element generates, and then remove the duplicates and you're done.

    By the way,

    [tex]Z_6 = Z_2 \times Z_3[/tex]

    is not correct. Instead write

    [tex]Z_6 \cong Z_2 \times Z_3[/tex]

    That is, [itex]Z_6[/itex] is isomorphic to [itex]Z_2 \times Z_3[/itex], but they aren't EQUAL. The elements of [itex]Z_6[/itex] are [itex]\{0, 1, 2, 3, 4, 5\}[/itex] whereas the elements of [itex]Z_2 \times Z_3[/itex] are [itex]\{(0,0), (0,1), (0,2), (1,0), (1,1), (1,2)\}[/itex]. But both are cyclic groups with order 6, and therefore they are isomorphic.

    [edit]: P.S. The fact that

    [tex]Z_6 \cong Z_2 \times Z_3[/tex]

    is irrelevant to this problem. You don't need to use that fact to find the subgroups of [itex]Z_6[/itex].
     
    Last edited: Mar 14, 2010
  4. Mar 14, 2010 #3
    One place to start would be to use Lagrange's theorem...Since Z6 has order 6, you know that any subgroup can only have order 1, 2 or 3 (and 6, but then this is Z6 itself). You also know that any subgroup must have the identity in it so that narrows your search down as well. Since (I believe) Z6 is cyclic, you also know that the order of any element of a group must divide the order of the group, so that will help too. I think it's just a matter of taking what you know about orders and using them to whittle down the possibilities. Just to be sure...you might want to check that all my assumptions are correct...I'm still just learning this stuff, as well.

    Hope that helps.

    Cheers,
    Lauren. =)
     
  5. Mar 14, 2010 #4
    oh okay
    so i have
    <0> = {0}
    <1> = {0,1,2,3,4,5} = Z6
    <2> = {0,2,4}
    <3> = {0,3}

    so there are 4 subgroups?

    thanks for your replies!
     
  6. Mar 14, 2010 #5

    jbunniii

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    Correct!
     
  7. Feb 1, 2012 #6
    didn't you forget
    [tex] <|4|>={[0],[4],[2]}[/tex]
    and
    [tex] <|5|>=<[1]>[/tex]??
     
    Last edited: Feb 1, 2012
  8. Feb 1, 2012 #7

    jbunniii

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    But <4> = <2> and <5> = <1>, so they were not omitted.
     
  9. Feb 1, 2012 #8

    Deveno

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    true enough, but this is worth verifying (at least once), to illustrate that generators of cyclic subgroups need not be unique (if a cyclic subgroup is of prime order, there are LOTS of choices for a generator).

    <3> is of order 2, so we get lucky, there is only one generator.
    <2> is of order 3, so we have 2 generators (it might seem, naively at first, that <2> and <4> might be different, they certainly are in Z).

    <1> is of order 6, and 6 isn't prime, so we would expect to find fewer than 5 generators. in fact, we have φ(6) = φ(2)φ(3) = (1)(2) = 2 generators (where φ is the euler totient function).

    gcd(k,6) = 1 ---> leads to a subgroup of order 6 (obviously the whole group Z6).
    gcd(k,6) = 3 ---> leads to a subgroup of order 6/3 = 2 (and this subgroup is, surprisingly, unique).
    gcd(k,6) = 2 ---> leads to a subgroup of order 3 (also unique. it's not immediately obvious that a cyclic group has JUST ONE subgroup of order a given divisor of the order of the whole group, but this is indeed true, and worth proving!)
    gcd(k,6) = 6 ---> leads to the trivial subgroup {0}.

    all that "greatest common denominator" stuff one learns in high school finally pays off! :)
     
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