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Isomorphic groups

  1. Jul 4, 2014 #1
    Two groups are isomorphic if they has same number of elements and if they has same number of elements of same order? Is it true? Where can I find the prove of this theorem?
     
    Last edited: Jul 5, 2014
  2. jcsd
  3. Jul 4, 2014 #2

    micromass

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    I really don't understand your first sentence.
     
  4. Jul 5, 2014 #3
    Sorry Sir. I try to say. If i had to groups with same number of elements ##|G_1|=|G_2|## and groups ##(G_1,\cdot)## and ##(G_2,*)## has the same number of element with the same order are then the groups are isomorphic? Is there such theorem?
     
  5. Jul 5, 2014 #4

    micromass

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    No, there is no such theorem because it is false. I think the two groups ##\mathbb{Z}_p\times\mathbb{Z}_p\times\mathbb{Z}_p## and ##(\mathbb{Z}_p\times\mathbb{Z}_p)\rtimes\mathbb{Z}_p## constitute a counterexample for ##p## prime.
     
  6. Jul 5, 2014 #5
    Tnx a lot and what are that group. What is ##\mathbb{Z}_p##? And what is difference between
    ##\mathbb{Z}_p\times\mathbb{Z}_p\times\mathbb{Z}_p##
    and
    ##(\mathbb{Z}_p\times\mathbb{Z}_p)\rtimes\mathbb{Z}_p##?
     
  7. Jul 5, 2014 #6
    ##\mathbb{Z}_p## (also denoted as the quotient group ##\mathbb{Z}/p\mathbb{Z})## is the set of residue classes modulo ##p##. So for example, ##\mathbb{Z}_5=\{0,1,2,3,4\}##. In the above example, ##\rtimes## denotes the semi direct product, as opposed to ##\times## which is the direct product. See http://en.wikipedia.org/wiki/Direct_product and http://en.wikipedia.org/wiki/Semidirect_product for good explanations of these.

    The answer above is not likely to be of help to you, since your question implies that you are a novice in group theory. A more fitting answer for a counterexample is to consider ##\mathbb{Z}/6\mathbb{Z}## and ##S_3##, the group of permutations of ##3## elements. They both have order ##6##, and they certainly are not isomorphic.
     
  8. Jul 5, 2014 #7

    micromass

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    I think the OP wants two groups ##G## and ##G^\prime## that have the same order, and such that ##G## and ##G^\prime## have the same number of elements with a given order. For example, ##G## and ##G^\prime## both have the same number of elements of order ##2##.
     
  9. Jul 5, 2014 #8
    Yes. Exactly!
     
  10. Jul 5, 2014 #9
    Tnx. For this definitions. I will read that.
     
  11. Jul 5, 2014 #10

    micromass

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    Well, it's not true. The simplest possible example involve the group ##\mathbb{Z}_2\times\mathbb{Z}_8## and the modular group. These are groups of order 16.

    http://en.wikipedia.org/wiki/Cycle_graph_(algebra [Broken])
    http://en.wikipedia.org/wiki/Modular_group
     
    Last edited by a moderator: May 6, 2017
  12. Jul 5, 2014 #11
    Ah, the question was not very eloquently phrased.

    To answer the question, mircomass is correct in saying that the smallest example comes with order ##16##. Consider ##\mathbb{Z}/8\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}\not\simeq\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/4\mathbb{Z}\not\simeq H## where ##H=<a,b>## such that ##|a|=|b|=4##. It's quite clear that these three groups have ##12## elements of order ##4## and ##3## elements of order ##2## (and obviously the identity being the sixteenth). Is this perfect?

    More complicated examples can be arrived at by the notion of ##p##-groups, which I imagine OP has not studied yet.
     
  13. Jul 5, 2014 #12

    micromass

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    I think ##\mathbb{Z}/8\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}## has an element of order 8 though.
     
  14. Jul 5, 2014 #13
    You're correct. Small oversight, but I think I meant ##Q\times \mathbb{Z}/2\mathbb{Z}## with ##Q## being the quaternion group. Surely that doesn't have an element of order ##8##?
     
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