semigroup with two elements

Perhaps the simplest non-trivial example of a semigroup which is not a group is a particular semigroup with two elements. The underlying set of thissemigroup is $\{a,b\}$ and the operation is defined as follows:

$a\cdot a$$=$$a$

$a\cdot b$$=$$b$

$b\cdot a$$=$$b$

$b\cdot b$$=$$b$

It is rather easy to check that this operation is associative, as itshould be:

$a\cdot (a\cdot a)=a\cdot a=$$a$$=a\cdot a=(a\cdot a)\cdot a$

$a\cdot (a\cdot b)=a\cdot b=$$b$$=a\cdot b=(a\cdot a)\cdot b$

$a\cdot (b\cdot b)=a\cdot b=$$b$$=b\cdot b=(a\cdot b)\cdot b$

$b\cdot (a\cdot a)=b\cdot a=$$b$$=a\cdot a=(a\cdot a)\cdot a$

$a\cdot (b\cdot b)=a\cdot b=$$b$$=b\cdot b=(a\cdot b)\cdot b$

$b\cdot (a\cdot b)=b\cdot b=$$b$$=b\cdot b=(b\cdot a)\cdot b$

$b\cdot (b\cdot a)=b\cdot b=$$b$$=b\cdot a=(b\cdot b)\cdot a$

$b\cdot (b\cdot b)=b\cdot b=$$b$$=b\cdot b=(b\cdot b)\cdot b$

It is worth noting that this semigroup is commutative and has an identityelement, which is $a$. It is not a group because the element $b$ doesnot have an inverse. In fact, it is not even a cancellative semigroupbecause we cannot cancel the $b$ in the equation $a\cdot b=b\cdot b$.

This semigroup also arises in various contexts. For instance,if we choose $a$ to be the truth value ”true” and $b$ to be the truthvalue ”false” and the operation $\cdot$ to be the logical connective ”and”,we obtain this semigroup in logic. We may also represent it by matriceslike so: