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Abstract Alg- Group theory and isomorphic sets.

  1. Apr 24, 2010 #1
    1. The problem statement, all variables and given/known data

    I am suppose to determine if the following list of groups are isomorphic and if they are define an isomorphic function for them.
    a. [5[tex]Z[/tex], +],[12[tex]Z[/tex], +] where n[tex]Z[/tex] = {nz | z[tex]\in[/tex][tex]Z[/tex]}

    b. [[tex]Z[/tex]6, +6]], [S6, [tex]\circ[/tex]]

    c. [[tex]Z[/tex]2, +2]], [S2, [tex]\circ[/tex]]

    2. Relevant equations
    +6 means x +6] y = the remainder of (x+y)/6

    To prove not isomorphic we are suppose to show that the two sets are not one-to-one, or one is commutative while the other is not, etc.


    3. The attempt at a solution
    For a, I am fairly certain they are isomorphic and that the function should be f(x) = (12/5)x since it is a bijective function and f(x+y) = f(x) + f(y).

    For b, My gut feeling is that it is not isomorphic however I can't find a good reason why. Perhaps because the second group is not commutative. However that answer just doesn't sit well with me.

    Finally, for c I am confused because S2 = {(1,2), (2,1)} while [tex]Z[/tex]2 ={0, 1,2} so it seems like there could be an isomorphic function but I'm uncertain what that function could be without it being piecewise for each element 0, 1, and 2.
     
    Last edited: Apr 24, 2010
  2. jcsd
  3. Apr 24, 2010 #2
    If you're gut feeling is that one of the groups is not commutative, then try showing it by finding two elements that don't commute with each other.

    Also remember that the cardinality of the groups must be the same for a bijective correspondence
     
  4. Apr 24, 2010 #3
    A is correct.

    [tex]Z_2=\left\{0,1\right\}[/tex], not [tex]\left\{0,1,2\right\}[/tex].

    [tex]S_2=\left\{id,(1\;2)\right\}[/tex], and it's easy to see C is isomorphic. The function is [tex](0\mapsto id),(1\mapsto (1\;2))[/tex] (this notation is probably not right, though), and one can verify it by enumerating all possibilities.

    It's also easy to see that B is not isomorphic, since the cardinality of [tex]S_6[/tex] is 720, while the cardinality of [tex]Z_6[/tex] is 6.
     
    Last edited: Apr 24, 2010
  5. Apr 24, 2010 #4
    Thank you to both of you for your help.

    [tex]Z_2=\left\{0,1\right\}[/tex], not [tex]\left\{0,1,2\right\}[/tex].

    Thank you for point that out! I was going from my lecture notes and I had written it down wrong. This makes the problem make a lot more sense.
     
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