Why in order to be a binary operation on $S$, each element of $S$ has to appear 'once and only once' in each row and column in Cayley Table? I was reading about Composition table or Cayley Table; one of the points my book presents is that 

If all the entries of the table are elements of set $S$ and each element of $S$ appears once and only once in each row and column, then the operation is a binary operation.

The first row of the table contains elements $a_1*a_1,\,a_1*a_2,\,a_1*a_3,\,\cdots$
Now, according to the statement, in order for $*$ to be a binary operation, $a_1*a_1\ne \,a_1*a_2\ne \,a_1*a_3\ne \,\cdots\,$ is it so?
If yes, I'm not getting the reason behind it. What would be the problem if $a_1*a_1= \,a_1*a_2 = \,a_1*a_3\;?$
Can anyone please explain to me what the bold statement means?
 A: The statement isn't false, but it is misleading. A binary operation is a mapping from $S \times S$ to $S$; therefore it is true that if all the entries in the table are elements of $S$, then the operation is a binary operation (regardless of any further restrictions). The part you put in bold has no bearing on the operation being binary.
A: On my third reading, I realize that the statement states IF all elements appear exactly once THEN it is a binary operation.  The converse (if binary than every element appears once) is NOT actually stated and shouldn't be implied.
As stated this is actually a trivial statement.  If every element of the set appears exactly once, then all possible pairs have a result and the only results are members of the set.  So it's binary.  It'd also be binary if every element didn't appear exactly once.
Notice it's an IF, not an IF AND ONLY IF. Nor is it a definition.
All it's saying is if it's a precise and specific type of binary operation, then it is a binary operation.  This is a bit like saying, if someone is a New York male resident between 20 and 30 years old with a social security number, than he is a human being.  ... well, he is!
