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Homesemigroup with two elements

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# 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 this semigroup is $\{a,b\}$ and the operation is defined as follows:

$\displaystyle a\cdot a$ | $\displaystyle=$ | $\displaystyle a$ | ||

$\displaystyle a\cdot b$ | $\displaystyle=$ | $\displaystyle b$ | ||

$\displaystyle b\cdot a$ | $\displaystyle=$ | $\displaystyle b$ | ||

$\displaystyle b\cdot b$ | $\displaystyle=$ | $\displaystyle b$ |

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

$\displaystyle a\cdot(a\cdot a)=a\cdot a=$ | $\displaystyle a$ | $\displaystyle=a\cdot a=(a\cdot a)\cdot a$ | ||

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

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

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

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

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

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

$\displaystyle b\cdot(b\cdot b)=b\cdot b=$ | $\displaystyle b$ | $\displaystyle=b\cdot b=(b\cdot b)\cdot b$ |

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:

$a=\left(\begin{matrix}1&0\\ 0&1\end{matrix}\right)\qquad b=\left(\begin{matrix}1&0\\ 0&0\end{matrix}\right)$ |

## Mathematics Subject Classification

20M99*no label found*

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