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Algebra: Non-isomorphic groups

  1. Nov 22, 2015 #1
    1. The problem statement, all variables and given/known data
    How many non-isomorphic groups of two elements are there?

    2. Relevant equations


    3. The attempt at a solution
    I don't understand exactly what we are being asked.
    If we have a group of two elements under, say, addition, then [itex]G =\{0, g\}[/itex].
    Then also [itex]g+g = 0[/itex] must be true, means [itex]g[/itex] is its own opposite. (of order 2).

    Now, how should I construct a group under some operation ##*##: [itex]G' = \{e, g'\}[/itex], where ##e## is the unit/zero element (depending on operation) such that [itex]G[/itex] and [itex]G'[/itex] are not isomorphic?

    [itex]G\cong G'[/itex] iff there exists a bijective group homomorphism [itex]f: G\to G'[/itex]
    I can define [itex]f[/itex] such that:
    [itex]f(0) = e[/itex] (satisfies one of the group homormorphism requirement)
    [itex]f(g) = g'[/itex]
    Is bijection and

    [itex]f(0+g) = f(g) = g' = e*g'[/itex]
    [itex]f(g+g) = f(0) = e = g'*g'[/itex]
    so [itex]G\cong G'[/itex]

    What must I do to generate non-isomorphic groups of two elements?

    In general, for any ##n##, how can I determine the number of non-isomorphic groups of ##n## elements?
     
    Last edited: Nov 22, 2015
  2. jcsd
  3. Nov 22, 2015 #2

    Samy_A

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    Do you think that is possible?

    I think that is not easy at all.
     
  4. Nov 22, 2015 #3

    pasmith

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    You can't. You are asked "how many non-isomorphic groups of two elements are there?" and you've shown that the answer is "all groups of order two are ismorphic to the cyclic group of order two."
     
  5. Nov 22, 2015 #4
    EDIT: nope
     
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