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How to find group types for a particular order?

  1. Sep 22, 2015 #1
    Hi,

    How do we determine the group types for a particular order? I know the example of using the Cayley table to show there are only two types of groups of order 4 but do not know how to determine this for other groups. For example, suppose I wanted to show there is a group of order 10 in the alternating group ##A_5##. Now, according to Wikipedia, this group must be cyclic or dihedral. Wikipedia lists a table of small groups:

    https://en.wikipedia.org/wiki/List_of_small_groups

    How do we determine this?

    Here's what I have so far:

    (1) If the group order is prime then it must be a cyclic group. Therefore, there are no non-abelian groups with prime order.

    (2) Let G be a finite Abelian group. Then G is isomorphic to a product of groups of the form ##H_p = Z/p^{e1}Z×···×Z/p^{en}Z##, in which ##p## is a prime number and ##1\leq e1 \leq···\leq en## are positive integers

    not real clear on #2 but let's assume I got that. So that leaves non-abelian groups. Take for example a group of size 6. How do I conclude that the only non-abelian group of order 6 is ##D_6## and all other non-abelian groups with that size are isomorphic to ##D_6##?

    Then take for example a non-abelian group of size ##8##. Well ##D_8## and ##Q## (quaternions) are of that size. So then any other are isomorphic to those two. How would I prove this? That no matter what kind of non-abelian group of size 8 I could invent, it must be isomorphic to these two.

    (3) Let G by a group of order 2p with p a prime number, then G is either cyclic or isomorphic to ##D_p##. (need to review this proof I think). Ok, then that answers my question for the case of ##|G|=10##.

    Guess I should know these already.

    What about when it's not 2p though, say for 12? Well, one of them is dihedral, ##D_{12}##. Is that the only non-abelian group of size 12?. I think there is a dicyclic also however. Certainly exists a cyclic group ##C_{12}## also. But there is also a product one ##Z_6\times Z_2##. But then why isn't there a ##Z_3\times Z_4## then? Maybe they're isomorphic. Need to check.

    There is a ##Z_3\times Z_4##. Also not hard to show ##Z_{10}\cong Z_5\times Z_2##. Same dif for ##Z_{12}\cong Z_3\times Z_4##. However, ##Z_6\times Z_2 \ncong Z_3\times Z_4## (try coming up with an isomorphism) and so therefore ##Z_{12}\ncong Z_6\times Z_2##.

    I think I'm getting it guys.




    Thanks,
    Jack
     
    Last edited: Sep 22, 2015
  2. jcsd
  3. Sep 25, 2015 #2

    mathwonk

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  4. Sep 25, 2015 #3
    Thanks mathwonk. That's very informative and I'm sure I'll use it while I'm studying abstract algebra. For example, just briefly running though it I noticed it had a very good summary of the various generators for the symmetric group including the one I had worked on recently here and elsewhere about when ##n## is a prime ##p## than any transposition and any ##p##-cycle generates the group. I'm also finding GAP very useful in that regards (thanks to micromass).

    Jack
     
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