Cayley Table and Groups

  • Thread starter Bleys
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If the Cayley table, of a finite set G, is a latin square (that is, any element g appears once and only once in a given row or column) does it follow G is a group?
I know the converse is true, and it seems reasonable that this is true. Since the array will be of size |G|x|G|, inverses exist and are unique. There will be one row (and one column) that will represent the multiplication by the identity (since the rows and columns are permutations of the set). But I don't know about associativity; in fact I'm not even sure the if the column and row representing the identity will actually be the same element. So is there a counter-example or if it's true, a proof?
 

Answers and Replies

  • #2
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Never mind, the array need not represent an associative structure. There is an example on the wikipedia article on Latin Squares that shows one of a quasigroup with identity.
 

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