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Math Help - Proof for (Abelian) Group - or not?

  1. #1
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    Proof for (Abelian) Group - or not?

    Hello together!

    I need to show whether the following is a(n abelian) group or not:

    G = \{w, f\}; a*b := (a \leftrightarrow b)

    I try to show the axioms.
    1) Closure
    Is given, since
    w*f = f
    f*w = f
    w*w = w
    f*f = f

    2.) Associativity
    ???

    3.) Identity
    I need an element e with e*a = a*e = a. (for all elements of G)
    So..

    a*e = a <-> e = w.
    because w*w = w and f*f = f.

    4.) Inverse element;
    I need an element i with
    a*i = i*a = e.

    the inverse for w is w, since w*w=w=e.
    the inverse for f is f, since f*f=w=e.

    5.) Commutativity:

    Is a*b=b*a ?
    This can be shown be the few examples:
    w*f = f
    f*w = f
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  2. #2
    MHF Contributor
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    Re: Proof for (Abelian) Group - or not?

    Ha-ha, you forgot to translate w into English!

    Quote Originally Posted by MageKnight View Post
    2.) Associativity
    ???
    Wikipedia says that associativity holds. In the worst case, you have 8 cases to check.

    Quote Originally Posted by MageKnight View Post
    3.) Identity
    I need an element e with e*a = a*e = a. (for all elements of G)
    So..

    a*e = a <-> e = w.
    because w*w = w and f*f = f.
    You mean f * w = f.

    The rest seems OK.
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  3. #3
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    Re: Proof for (Abelian) Group - or not?

    I am not sure I understand your definition for a \leftrightarrow b. To me, that reads a if and only if b. But, the multiplication you give in part (1) suggests otherwise. Then, for part (4), your multiplication changes: f*f = w in part (4) but f*f = f in part (1) and part (3). I think you are correct in part (4) since that multiplication is correct for the biconditional.

    For part (3), you say w*w = w and f*f=f. I think you want to show w*w = w and f*w = w*f = f. So, w is the identity element.

    For part (4), I agree with you.

    For part (5), you have f*f = f*f = w, f*w = w*f = f, and w*w = w. Those are the three cases.

    Anyway, for associativity, you need to show that a*(b*c) = (a*b)*c for all a,b,c\in G. So, try it out. There are only eight possible combinations for a,b,c \in G.
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