# 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?

• Suppose $a_1*a_2 = a_1*a_3 = b$. What would $a_1^{-1}*b$ be? – Rahul Mar 5 '16 at 5:33
• This follows from uniqueness of inverses. – Math1000 Mar 5 '16 at 5:33
• Good question. I'm presuming the book is considering this as a criterion of definition. It's true for groups. I don't think it is usually considered a property of binary operations. Actually I'm confused by this. The usual definition is simply two ordered elements result in an element of the set. I was under the impression that a constant operation would be a binary operation if the constant was a member of the set. – fleablood Mar 5 '16 at 5:48
• Math 100, that's assuming the table is for a group. A binary operation on a set need not, have an identity, an inverse, need not be associative, need not have any significance at all. However, if the book is "group centric" it can simply consider only group type operations are significant. – fleablood Mar 5 '16 at 5:57
• If the elements in the table weren't elements of the set or if some cells in the tables didn't have values then it wouldn't be a binary operation. The condition that each element appears exactly once in a row isn't a requirement. However if the set and operation defines a group, it would be a condition as a result of group properties. Out of context the statement is odd, but in context it maybe is clear that they are determining the operation to be binary as a precurser to declaring it a group. – fleablood Mar 5 '16 at 6:18

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.

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!