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    (Original post by boromir9111)
    hhmmmm, in what way are the structures of the 3 cayley tables different?
    Where have you got the number 3 from? Write out what you've done; I don't think me or DFranklin are following your reasoning at all here.
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    (Original post by nuodai)
    Where have you got the number 3 from? Write out what you've done; I don't think me or DFranklin are following your reasoning at all here.
    hhhmmm, since a is the identity element the table will first be

    abcd
    badc
    cdab
    dcba

    abcd
    bcda
    cdab
    dabc

    abcd
    bdac
    cadb
    dcba

    This is the only "possibilities" I can think of?
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    You sure that the first and third are isomorphic? I think it's 2nd and 3rd.
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    (Original post by boromir9111)
    hhhmmm, since a is the identity element the table will first be

    abcd
    badc
    cdab
    dcba

    abcd
    bcda
    cdab
    dabc

    abcd
    bdac
    cadb
    dcba

    This is the only "possibilities" I can think of?
    The second and third tables are both cyclic, so they're isomorphic; there are only two non-isomorphic groups of order 4.

    (Original post by DFranklin)
    You sure that the first and third are isomorphic? I think it's 2nd and 3rd.
    After staring at it for more than 5 seconds I changed my mind :p:
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    (Original post by nuodai)
    The second and third tables are both cyclic, so they're isomorphic; there are only two non-isomorphic groups of order 4.
    ahhh yes, that makes sense but am I correct in saying there are only 3 cayley tables possible?
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    (Original post by boromir9111)
    ahhh yes, that makes sense but am I correct in saying there are only 3 cayley tables possible?
    No; for example:

    abcd
    badc
    cdba
    dcab

    is another.
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    (Original post by nuodai)
    No; for example:

    abcd
    badc
    cdba
    dcab

    is another.
    Oh yeah. So there are only 4 cayley table possibilties and you was saying there are only 2 non-isomorphic groups, I take it's 1 and 4?
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    (Original post by boromir9111)
    Oh yeah. So there are only 4 cayley table possibilties and you was saying there are only 2 non-isomorphic groups, I take it's 1 and 4?
    Well 2,3,4 are all isomorphic, i.e. they all represent the same group, but with different elements labelled differently. [That's essentially all an isomorphism is: a relabelling.] So #1 and #2 (which is 'the same' as #3 and #4) are distinct groups. Or you could say #1 and #3 are distinct groups, or #1 and #4 are distinct groups... it doesn't matter.
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    (Original post by nuodai)
    Well 2,3,4 are all isomorphic, i.e. they all represent the same group, but with different elements labelled differently. [That's essentially all an isomorphism is: a relabelling.] So #1 and #2 (which is 'the same' as #3 and #4) are distinct groups. Or you could say #1 and #3 are distinct groups, or #1 and #4 are distinct groups... it doesn't matter.
    That's why I thought all of them were isomorphic cause they all was the same but just relabelled which is why I don't understand why 1 isn't isomorphic? all I can think of is that the diagonal starting from the top left going to bottom right all consist of a where as for groups 2,3 and 4 they don't?
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    (Original post by boromir9111)
    That's why I thought all of them were isomorphic cause they all was the same but just relabelled which is why I don't understand why 1 isn't isomorphic? all I can think of is that the diagonal starting from the top left going to bottom right all consist of a where as for groups 2,3 and 4 they don't?
    In #2,3,4 you have one element which squares to the identity, and two elements which square to that element. That is, your group is of the form \{ e,x,x^2,x^3 \}, where x^2 is the element that squares to the identity and x,x^3 both square to x^2. Since you're forced to have a=e, it is left open to relabel x,x^2,x^3 as a,b,c. But since we could let y=x^3 and the group would be \{e,y^3,y^2,y\} = \{e,y,y^2,y^3 \} it makes no difference whether an element is x or x³, so the only element where the choice "matters" is x^2, hence three different-looking tables (if you put a,b,c,d in that order).

    In #1 every element of your group squares to the identity, and so it's of the form \{ e, x, y, xy \}. But notice that you could let, say, p=x, q=xy, and then you'd have \{ e, p, pq, q \} = \{ e, p, q, pq \}, so nothing changes no matter what you call any of the elements, hence only one different-looking table (if you put a,b,c,d in that order).
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    (Original post by nuodai)
    In #2,3,4 you have one element which squares to the identity, and two elements which square to that element. That is, your group is of the form \{ e,x,x^2,x^3 \}, where x^2 is the element that squares to the identity and x,x^3 both square to x^2. Since you're forced to have a=e, it is left open to relabel x,x^2,x^3 as a,b,c. But since we could let y=x^3 and the group would be \{e,y^3,y^2,y\} = \{e,y,y^2,y^3 \} it makes no difference whether an element is x or x³, so the only element where the choice "matters" is x^2, hence three different-looking tables (if you put a,b,c,d in that order).

    In #1 every element of your group squares to the identity, and so it's of the form \{ e, x, y, xy \}. But notice that you could let, say, p=x, q=xy, and then you'd have \{ e, p, pq, q \} = \{ e, p, q, pq \}, so nothing changes no matter what you call any of the elements, hence only one different-looking table (if you put a,b,c,d in that order).
    Perfect explanation. I was hinting at that but couldn't explain it in a mathematical way which you just did. Thanks mate!
 
 
 
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