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Partition groups into subcollection

  1. Mar 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Partition the following collection of groups into subcollections of isomorphic groups. a * superscript means all nonzero elements of the set.
    integers under addition
    [tex]S_{2}[/tex]
    [tex]S_{6}[/tex]
    [tex]integer_{6}[/tex]
    [tex]integer_{2}[/tex]
    [tex]real^{*}[/tex] under multiplication
    [tex]real^{+}[/tex] under multiplication
    [tex]rational^{*}[/tex] under multiplication
    [tex]complex^{*}[/tex] under multiplication
    17(integer) under addition
    rational under addition
    The subgroup (pi) of [tex]real^{*}[/tex] under multiplication
    3(integer) under multiplication
    real under addition
    The subgroup G of [tex]S_{5}[/tex] generated by :
    top row:(1 2 3 4 5)
    bottom row (3 5 4 1 2)





    2. Relevant equations



    3. The attempt at a solution
    i don't understand the question and what this has to do with the permutation chapter in the book.
     
  2. jcsd
  3. Mar 9, 2009 #2
    Does this make any sense with the permutation chapter
     
  4. Mar 17, 2009 #3
    I'm still clueless on this one. If anyone has any ideas, it would really help. I'm racking my brain and not coming up with anything.
     
  5. Mar 17, 2009 #4
    "Partition the following collection of groups into subcollections of isomorphic groups" is just a fancy way of asking which of these groups are isomorphic.

    To determine if two groups are isomorphic you can use cardinality (number of elements) as a first hint, if they don't have the same cardinality, they can't be isomorphic. To prove that two groups actually are isomorphic, you need to construct an isomorphism (the exponential function should be helpful in some cases).
     
  6. Mar 17, 2009 #5
    so, like s6 and integer6 would have same cardinality?
     
  7. Mar 18, 2009 #6
    No, [tex]S_6[/tex] is a http://en.wikipedia.org/wiki/Symmetric_group" [Broken] and has 6!=720 elements, the cyclic group [tex]\mathbb{Z}_6[/tex] on the other hand has only 6 elements.

    But try [tex]S_2[/tex] and [tex]\mathbb{Z}_2[/tex].
     
    Last edited by a moderator: May 4, 2017
  8. Mar 18, 2009 #7
    Oh, I see since S2 has 2! elements and Z2 has 2 elements
     
  9. Mar 30, 2009 #8
    Ok, I have an idea.
    I have s2 and z2 as one collection
    I have z,3z,17z as one collection
    My problem is what to do with the reals, rationals and complex?
     
  10. Mar 30, 2009 #9
    What is 3z? Is it [tex]\mathbb{Z}_3[/tex]?

    Think again about cardinality: The rationals are countable, the reals are uncountable. Also consider my hint about the exponential function (Post #4), it relates addition and multiplication.
     
  11. Mar 30, 2009 #10
    3z is the set of integers Z times 3.
    Ok reals are uncountable. is this true for complex, too?
    I'm just having trouble with the exponential; function and how to use it.
    Are real* and complex* under multiplication related?
     
  12. Mar 30, 2009 #11
    The reals are a subset of the complexes and if the reals are uncountable, then ...

    Of course, that doesn't mean they both have the same cardinality.

    A further hint about the exponential function: exp(x + y) = exp(x) * exp(y)
     
  13. Mar 30, 2009 #12
    then complex are uncountable
    ok, ao does that hint say something like reals under addition and real under multiplications are a collection?
     
  14. Mar 30, 2009 #13
    Good.

    You meant to say isomorphic instead of collection, I hope. Can you prove, use exp, that they are isomorphic?
     
  15. Mar 30, 2009 #14
    yeah, I meant to say isomorphic.
    let x and y be 2 real numbers under addition
    Then exp(x+y)=exp(x)*exp(y)
     
  16. Apr 13, 2009 #15
    Ok still having trouble.
    I'm saying R^+, R is one isomorphism
    C*,R* is one isomorphism
    Q,Q* is one isomorphism
     
  17. Apr 27, 2009 #16
    Is this correct?
     
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