associative binary operation and unique table

Question

Suppose that * is an associative binary operation on $S=\{ e,a,b \} $ where $e$ is the identity element and every element of S has an inverse. Show that there's only way/table for this operation.

So this is my result, after completing the table from what the problem ask but i don't know how to prove that this is the unique/only table for this. Do I have to prove every property for this (associative, identity...)? \begin{array}{|c|c|c|c|c|c|} \hline *& e & a & b \\ \hline e& e&a &b\\ \hline a& a&e &b\\ \hline b& b &b &e\\ \hline \end{array}

Answer 1

If you can use some basic group theory: A set with an associative binary operation with inverse and symmetric element is a group. There is only one group with three elements.

Otherwise: Since $e$ is the identity, we have to determine $a*a$, $a*b$, $b*a$ and $b* b$. The inverse of $a$ can be $a$ or $b$.
If it is $a$, then $a*a=e$. Then $a*(a*b)=(a*a)*b=e*b=b$. Therefore $a*b$ is not $a$ or $e$ (because $a*a=e$ and $a*e=a$), so $a*b=b$. But this implies $a=e$, a contradiction. We conclude that $a*b=e$.
Can you finish?

Answer 2

To prove it is the unique solution, you need to prove two things:

1. Your solution is a solution, i.e. you do indeed need to show that your solution is associative, that $e$ is indeed the identity element, and that every element has an inverse.

2. There are no other solutions.

Now, to show that there are no other solutions, you would go through an argument like:

$e$ being the identity element immediately forces 5 of the 9 values in the table:

\begin{array}{|c|c|c|c|c|c|} \hline *& e & a & b \\ \hline e& e&a &b\\ \hline a& a& &\\ \hline b& b & &\\ \hline \end{array}

$a$ has an inverse element, i.e. there is some element $a^{-1}$ such that $a * a^{-1} = e$, so $a^{-1} \not = e$, since $a * e = a$ and $a \not = e$.

Now, if $a^{-1} = a$ is inverse of $a$, then $aa=e$, so $(b * a) * a = b * (a * a) = b * e = b$, and so $b * a = b$, since if $b * a = e$ then $(b * a) * a = e * a = a \not = b$, and if $b * a = a$ then $(b * a) * a = a * a = e \not = b$. But since $b$ has to have an inverse $b^{-1}$ such that $b * b^{-1} = e$, and since $b * e = b \not = e$, and $b * a = b \not = e$, that means that $b^{-1} = b$. But then $(a * b) * b = b * b = e$,while $a * (b * b) = a * e= a$, and so $*$ is not associative. Hence, we have a contradiction if $a^{-1}= a$, and hence the only way things can work is if $a^{-1}= b$, i.e. $a * b = e$

\begin{array}{|c|c|c|c|c|c|} \hline *& e & a & b \\ \hline e& e&a &b\\ \hline a& a& &e\\ \hline b& b & &\\ \hline \end{array}

But this means that $a*a = b$, for we have $(a*a)*b = a * (a * b) = a * e = a$, and if $a*a = e$ then $(a * a) * b = e * b = b \not = a$, and if $a * a = a$ then $(a * a) * b = a * b = e \not = a$.

\begin{array}{|c|c|c|c|c|c|} \hline *& e & a & b \\ \hline e& e&a &b\\ \hline a& a& b&e\\ \hline b& b & &\\ \hline \end{array}

Also, since $a * (b * a) = (a * b) * a = e * a = a$, we get that $b * a = e$, for if $b * a = a$ then $a * (b * a) = a * a = e \not = a$, and if $b * a = b$, then $a * (b * a) = a * b = e \not = a$.

\begin{array}{|c|c|c|c|c|c|} \hline *& e & a & b \\ \hline e& e&a &b\\ \hline a& a& b&e\\ \hline b& b & e&\\ \hline \end{array}

And finally, since $a * (b * b) = (a * b) * b = e * b = b$, it must be that $b * b = a$, for if $b * b = e$, then $a * (b * b) = a * e = a \not = b$, and if $b * b = b$, then $a * (b * b) = a * b = e \not = b$

\begin{array}{|c|c|c|c|c|c|} \hline *& e & a & b \\ \hline e& e&a &b\\ \hline a& a& b&e\\ \hline b& b & e&a\\ \hline \end{array}

OK, so what we have now established is that there cannot be more than 1 solution.

But we have not yet established that this actually is a solution!

That is, if you look back at the proof, you see that we rules out all kinds of values, but we never established that the values that were forced actually did satisfy the requirements!

So, yes, you need to show that $e$ works like an identity element (easy), that ever element has an inverse (easy), and that this $*$ is associative (not hard, but tedious)