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# Cayley table watch

1. (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.
2. (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
cdab
dcba

abcd
bcda
cdab
dabc

abcd
bdac
dcba

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

abcd
cdab
dcba

abcd
bcda
cdab
dabc

abcd
bdac
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
5. (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?
6. (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
cdba
dcab

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

abcd
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?
8. (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.
9. (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?
10. (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 , where is the element that squares to the identity and both square to . Since you're forced to have , it is left open to relabel as . But since we could let and the group would be it makes no difference whether an element is x or x³, so the only element where the choice "matters" is , 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 . But notice that you could let, say, , and then you'd have , 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).
11. (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 , where is the element that squares to the identity and both square to . Since you're forced to have , it is left open to relabel as . But since we could let and the group would be it makes no difference whether an element is x or x³, so the only element where the choice "matters" is , 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 . But notice that you could let, say, , and then you'd have , 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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