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Find all groups of order 9, order 10, and order 11

  1. Nov 1, 2010 #1
    1. The problem statement, all variables and given/known data
    Find all groups of order 9, order 10, order 11.


    2. Relevant equations
    None


    3. The attempt at a solution
    We have already done an example in class of groups of order 4 and of order 2,3,5, or 7.
    So i'm going to base my proofs on the example of groups of order 4 except for the group of order 11 which I suspect is acting like the groups of order 2,3,5 or 7 since it is also a prime number.

    Here is my attempt at groups of order 9, i'm a little unsure about the final part.

    Let G be a group of order 9, every element has order 1, 3, or 9. If there is an element g of order 9, then <g> = G. G is cyclic and isomorphic to (Z/9, +).
    If there is no element of order 9, the (non-identity) elements must all have order 3.
    G = {e, a, a2, b, b2, c, c2, d, d2}
    G is isomorphic to Z/3 x Z/3
    a3 = e
    b3 = e
    c3 = e
    d3 = e

    Now i'll show the mappings of G onto Z/3 x Z/3:
    e -> (0,0)
    a -> (1,0)
    a2 -> (2,0)
    b -> (0,1)
    b2 -> (0,2)
    c -> (1,1)
    c2 -> (2,2)
    d -> (1,2)
    d2 -> (2,1)

    Did I do everything correctly here, and is this sufficient to find all groups of order 9 as the problem is asking?
     
  2. jcsd
  3. Nov 1, 2010 #2
    Uh, yes, this is sufficient.

    but you constructed a map from G to Z3 x Z3. I don't find it very obvious that it is an homomorfism. For example, is ab being sent to (1,1). (i.e. do you know for certain that ab=c?)
     
  4. Nov 1, 2010 #3
    Maybe it would be better to describe the group as G = {e, a, a2, b, b2, ab, (ab)2, ab2, a2b} ?

    Or am I getting myself into more trouble here?
     
  5. Nov 1, 2010 #4
    Yes, you could do that, but the problem remains. Like, what does (ab)a equal? (I know it should equal a²b, but you need to prove it)
     
  6. Nov 1, 2010 #5
    Yeah I was just thinking about that, I'm a little stuck on it. I can't assume this group is abelian can I? That would make for a pretty simple proof. Otherwise I guess I can try to use some sort of associativity proof.

    Suppose (ab)a ≠ a2b
    = (ab)a ≠ a(ab)
    = a ≠ a (this is actually true a = a) which contradicts our original supposition which means (ab)a = a2b
    (for some reason I have a feeling I can't just divide each side by (ab) like I did. hah). I'll keep trying other methods.
     
  7. Nov 1, 2010 #6
    No, you cant just divide by ab sorry.

    Maybe you should look back to the proof of the group of order 4. How did you show there that the map was a homomorphism. Maybe you could copy that...

    Honestly, I don't know any elementary methods for finding the groups of order 9. I guess only a bit of trial-and-error could do the job...
     
  8. Nov 1, 2010 #7
    The group of order 4 uses the theorem that if the square of every element in the group = e then the group is Abelian which I can't use for groups of order 9. I did find an example in my book for groups of order 6 which includes an element of order 3. I'll take a look at this
     
  9. Nov 1, 2010 #8

    Dick

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    Homework Helper

    Do you know p-groups have a nontrivial center? If not, use the class equation. It's a pretty common group theory exercise to show all groups of order p^2 are abelian.
     
    Last edited: Nov 1, 2010
  10. Nov 3, 2010 #9
    Ok my professor did an example of groups of order 8 in my last lecture which helped a lot so I think I have it figured out now:

    Groups of order 9:
    Let G be a group of order 9, every element has order 1, 3, or 9. If there is an element g of order 9, then <g> = G. G is isomorphic to Z/9.

    If there is no element of order 9, the (non-identity) elements must all have order 3.
    G = {e, a, a2, b, b2, ab, (ab)2, ab2, a2b}
    Now lets assume the following relationships:
    a3 = e
    b3 = e
    ab = ba (so we are abelian)
    Lets check that our assumptions hold true:
    aoddbodd = ab OR a OR b OR e
    aoddbeven = ab2 OR b2
    aevenbodd = a2b OR a2
    aevenbeven = a2b2 = (ab)2
    Our assumptions hold and I believe this is isomorphic to Z/3 x Z/3. Let's check the mappings:
    e --> (0,0)
    a --> (1,0)
    a2 --> (2,0)
    b --> (0,1)
    b2 --> (0,2)
    ab --> (1,1)
    (ab)2 --> (2,2)
    ab2 --> (1,2)
    a2b --> (2,1)

    I think this should be enough to prove that these are the 2 groups of order 9. What do you guys think?
     
  11. Nov 3, 2010 #10
    When checking if my assumptions hold true I should also check them with the a and b values flipped to check the ab = ba assumption, I won't write them out here but I know they hold, just assume I wrote them up there.
     
  12. Nov 3, 2010 #11

    Dick

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    Science Advisor
    Homework Helper

    You already assumed ab=ba in deriving the group structure. There's not much use in 'checking' it now. As I said before, you can prove a group of order 9 is abelian before you start.
     
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