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    Say we have a set, S, with two elements in it and I wanted to find out how many different binary operations * can I define on S, how many of these are a group and is isomorphic?

    Searching around the method for finding the number of binary operations is S^(S^2) where S is the number of elements in the set. So would it be 2^4 = 16?

    Now to find how many of these are a group. It leads me to believe that there will be 2 but I don't see why this is the case? What are you exactly looking for here?

    Wouldn't both of the groups be isomorphic?
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    The groups will be isomorphic, but don't forget it's asking you to count the number of binary operations that form a group, not the groups themselves.
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    (Original post by DFranklin)
    The groups will be isomorphic, but don't forget it's asking you to count the number of binary operations that form a group, not the groups themselves.
    2 out of the 16 binary operations will be a group, right? therefore both are isomorphic?
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    I don't understand why you think your 2nd sentence follows ("therefore...") from the first?
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    So both are Isomorphic is what I meant. Thanks.
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    What makes you think you can define two binary operations on a group of order 2? What are they? [There is only one binary operation you can define.]

    To define a binary operation you need to draw out a multiplication table. Say your group is \{e, x \}, and * is some binary operation. Then you have no choice over what e*e, e*x, x*e are equal to because (G, *) has to satisfy group axioms. But you have two "choices" over what x*x can be: it can either be e or x. However, one of these leads to a contradiction, so it turns out you only have one choice over what x*x can be. And hence there is only one binary operation. [Or at least, if \circ is another binary operation then the map g \mapsto g from (G, *) to (G, \circ) is an isomorphism.]
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    (Original post by nuodai)
    What makes you think you can define two binary operations on a group of order 2? What are they? [There is only one binary operation you can define.]

    To define a binary operation you need to draw out a multiplication table. Say your group is \{e, x \}, and * is some binary operation. Then you have no choice over what e*e, e*x, x*e are equal to because (G, *) has to satisfy group axioms. But you have two "choices" over what x*x can be: it can either be e or x. However, one of these leads to a contradiction, so it turns out you only have one choice over what x*x can be. And hence there is only one binary operation. [Or at least, if \circ is another binary operation then the map g \mapsto g from (G, *) to (G, \circ) is an isomorphism.]
    http://answers.yahoo.com/question/in...3102041AAVgcah

    This link is confusing then to what you have said!
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    (Original post by boromir9111)
    http://answers.yahoo.com/question/in...3102041AAVgcah

    This link is confusing then to what you have said!
    D'oh! I never considered treating x as the identity.
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    (Original post by nuodai)
    D'oh! I never considered treating x as the identity.
    So the link is correct?
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    (Original post by boromir9111)
    So the link is correct?
    Yup. I fell into the trap of naming the elements of the group suggestively; i.e. calling one of them 'e'. Instead what you should do is let G={x,y}, and then for each choice of identity (either e=x or e=y) there is just one binary operation (as in my post), but since you have two choices of identity and there is one binary operation for each, there are two choices of binary operation. The multiplication tables make it fairly obvious whether or not the two are isomorphic.
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    (Original post by nuodai)
    Yup. I fell into the trap of naming the elements of the group suggestively; i.e. calling one of them 'e'. Instead what you should do is let G={x,y}, and then for each choice of identity (either e=x or e=y) there is just one binary operation (as in my post), but since you have two choices of identity and there is one binary operation for each, there are two choices of binary operation. The multiplication tables make it fairly obvious whether or not the two are isomorphic.
    Cheers!
 
 
 
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