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    Say we are given G = {a,b,c,d} and * a binary operation on G such that (G,*) is a group and am told that a is the identity element and asked to write down all the possible cayley tables.

    Basically the way to do this is rearrange a,b,c,d however many times (e.g bacd, badc, cabd etc) until all possibilities are exhausted, right?
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    (Original post by nuodai)
    Well if a is your identity then you may as well use it for the first row and column of your tables. Then you just have to work out what happens to b, c and d. But all groups of order 3 are isomorphic to each other so you'll find that there aren't many possibilities. [Although this does seem like quite a tedious exercise.]
    Well now I am on my 8th table and continuing on, so yeah it is kind of tedious! So basically all the possibilities will be isomorphic?
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    (Original post by boromir9111)
    Well now I am on my 8th table and continuing on, so yeah it is kind of tedious! So basically all the possibilities will be isomorphic?
    Forget what I said about groups of order 3 being isomorphic, I was being an eejit. There are two types of group of order 4, one is cyclic and one is known as the 'Klein group'. For the cyclic groups, you could let c=b², d=b³ and then write out the corresponding table and then just permute b,c,d for the others. For the Klein groups, you can let d=bc and, again, permute the elements.

    (Original post by ghostwalker)
    Granted, but this is order 4.
    Indeed, my brain's not with me. For some reason I was treating {b,c,d} as a group in its own right; I think I've been spending too much time on TSR
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    (Original post by nuodai)
    Forget what I said about groups of order 3 being isomorphic, I was being an eejit. There are two types of group of order 4, one is cyclic and one is known as the 'Klein group'. For the cyclic groups, you could let c=b², d=b³ and then write out the corresponding table and then just permute b,c,d for the others. For the Klein groups, you can let d=bc and, again, permute the elements.
    Why do you have to do it that way? the way I am doing it is rearranging {a,b,c,d} into different combinations, is this the wrong way of doing it?
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    (Original post by nuodai)
    Indeed, my brain's not with me. For some reason I was treating {b,c,d} as a group in its own right; I think I've been spending too much time on TSR
    You're doing a great job though. I'd rep you for the huge proprotion of threads you've responded to, but you'll have to wait until I've repped 40 others first.
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    (Original post by boromir9111)
    Why do you have to do it that way? the way I am doing it is rearranging {a,b,c,d} into different combinations, is this the wrong way of doing it?
    I don't know what you mean by that... give me an example of what you're doing. I mean, you're told that a is the identity, so if you write out a table for some group and then just swap a and b (say) with each other then a won't be the identity in the resulting group. This is why you're best to write out two non-isomorphic groups consisting of {a,b,c,d} (where a is the identity) and then permute b, c and d accordingly.
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    (Original post by nuodai)
    I don't know what you mean by that... give me an example of what you're doing. I mean, you're told that a is the identity, so if you write out a table for some group and then just swap a and b (say) with each other then a won't be the identity in the resulting group. This is why you're best to write out two non-isomorphic groups consisting of {a,b,c,d} (where a is the identity) and then permute b, c and d accordingly.
    Oh, so you're basically saying keep where a is and then rearrange bcd like bdc, cbd,cdb, dcb etc?
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    (Original post by boromir9111)
    Oh, so you're basically saying keep where a is and then rearrange bcd like bdc, cbd,cdb, dcb etc?
    Yes, provided you have the right group structures in your tables.
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    (Original post by nuodai)
    Yes, provided you have the right group structures in your tables.
    Yes, that makes sense. Thanks, shall get on this!
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    (Original post by nuodai)
    Yes, provided you have the right group structures in your tables.
    Hey. I have 6 six cayley tables where a is the identity element {a,b,c,d}, {a,b,d,c}, {a,c,b,d}, {a,c,d,b}, {a,d,b,c} and {a,d,c,b} now these are all isomorphic to one another as far as I can tell, right?
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    (Original post by boromir9111)
    Hey. I have 6 six cayley tables where a is the identity element {a,b,c,d}, {a,b,d,c}, {a,c,b,d}, {a,c,d,b}, {a,d,b,c} and {a,d,c,b} now these are all isomorphic to one another as far as I can tell, right?
    {a,b,c,d} isn't a Cayley table. The structure of the group depends not only on how you rearrange a,b,c,d but also on how they 'multiply' together under the binary operation. If you wrote out one valid table and then permuted b, c and d (which I'm guessing is what you did), then they're all isomorphic -- the isomorphism is given by the permutation, so like "b<->c" is an isomorphism between your {a,b,c,d} and {a,c,b,d}; but there is a second set of Cayley tables you can have (which are isomorphic to each other, but not to the ones you currently have). That is, you can write two non-isomorphic Cayley tables both of which have a,b,c,d in that order along the top and down the side.
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    (Original post by nuodai)
    {a,b,c,d} isn't a Cayley table. The structure of the group depends not only on how you rearrange a,b,c,d but also on how they 'multiply' together under the binary operation. If you wrote out one valid table and then permuted b, c and d (which I'm guessing is what you did), then they're all isomorphic -- the isomorphism is given by the permutation, so like "b<->c" is an isomorphism between your {a,b,c,d} and {a,c,b,d}; but there is a second set of Cayley tables you can have (which are isomorphic to each other, but not to the ones you currently have). That is, you can write two non-isomorphic Cayley tables both of which have a,b,c,d in that order along the top and down the side.
    The one valid table is {a,b,c,d} which is G and then from there I permutated b,c and d to get 6 cayley tables? I get what you mean permutating b with c, they are identical! if you permute c and d aren't they isomorphic?
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    I have to say, I think this "permuting the elements" is a bit of a red-herring (after all, if you simply permute the elements, you haven't changed the Cayley table, at least to my mind).

    Since a is the identity, you "start off" knowing your table looks like:

    abcd
    b???
    c???
    d???

    (where ? means don't know yet).

    You know that you can't have two elements the same in any row or column.

    So, what are the possibilities for b * b? For each possibility, there are only a couple more possibilities for b * c. Keep working methodically and you'll list all the possible tables.
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    (Original post by DFranklin)
    I have to say, I think this "permuting the elements" is a bit of a red-herring (after all, if you simply permute the elements, you haven't changed the Cayley table, at least to my mind).

    Since a is the identity, you "start off" knowing your table looks like:

    abcd
    b???
    c???
    d???

    (where ? means don't know yet).

    You know that you can't have two elements the same in any row or column.

    So, what are the possibilities for b * b? For each possibility, there are only a couple more possibilities for b * c. Keep working methodically and you'll list all the possible tables.
    Ok, let me try this!
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    (Original post by DFranklin)
    I have to say, I think this "permuting the elements" is a bit of a red-herring (after all, if you simply permute the elements, you haven't changed the Cayley table, at least to my mind).

    Since a is the identity, you "start off" knowing your table looks like:

    abcd
    b???
    c???
    d???

    (where ? means don't know yet).

    You know that you can't have two elements the same in any row or column.

    So, what are the possibilities for b * b? For each possibility, there are only a couple more possibilities for b * c. Keep working methodically and you'll list all the possible tables.
    You can only get 3 cayley tables this way. Is this right?
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    Since there are 3 cayley tables, would you say they are all isomorphic right?
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    Why would you think that?
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    (Original post by DFranklin)
    Why would you think that?
    Well, they have the same number of elements, group stucture is similar.......it's bijective as well, every element maps on to another element?
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    (Original post by boromir9111)
    Well, they have the same number of elements, group stucture is similar.......it's bijective as well, every element maps on to another element?
    Some of the groups have different structures - they are NOT all isomorphic.
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    (Original post by DFranklin)
    Some of the groups have different structures - they are NOT all isomorphic.
    hhmmmm, in what way are the structures of the 3 cayley tables different?
 
 
 
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