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 this semigroup is $\{a,b\}$ and the operation is defined as follows: \begin{eqnarray*} a \cdot a &=& a \\ a \cdot b &=& b \\ b \cdot a &=& b \\ b \cdot b &=& b \end{eqnarray*}It is rather easy to check that this operation is associative, as it should be: \begin{eqnarray*} 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 \end{eqnarray*} It is worth noting that this semigroup is commutative and has an identity element, which is $a$ . It is not a group because the element $b$ does not have an inverse. In fact, it is not even a cancellative semigroup because 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 truth value "false" and the operation $\cdot$ to be the logical connective "and", we obtain this semigroup in logic. We may also represent it by matrices like so:$$a = \left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right) \qquad b = \left( \begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix} \right)$$