📂Abstract Algebra

Monoids in Abstract Algebra

Definition ¹

A semigroup $\left< M , \ast\ \right>$ is defined to be a monoid if there exists an element $e$ such that for all elements $a$ of $\left< M , \ast\ \right>$, $a \ast\ e = e \ast\ a = a$ is satisfied. Such an $e$ is called an identity.

Explanation

A monoid is a semigroup with an identity element. Introducing the concept of an identity element considerably broadens the scope of discussion. Let’s look at a typical example that is a semigroup but not a monoid.

The semigroup $\left< \mathbb{N} , +\right>$ is not a monoid.

• Let’s assume that there exists an identity element $ e$ for any natural number $a$ that satisfies $a + e = a$.

Since $e$ is a natural number greater than or equal to $1$, $a + e \ge a + 1$ holds. Meanwhile, since $a + 1 > a$, it follows that $a + e > a$, which contradicts the assumption.

This naturally-occurring counterexample suggests that the existence of an identity element might not be as obvious as it seems.

For the set of all square matrices $\mathbb{R}^{n \times n}$, $\left< \mathbb{R}^{n \times n} , \cdot \right>$ is a monoid.

• By definition of matrix multiplication, it is straightforward to show that $\left< \mathbb{R}^{n \times n} , + \right>$ is a semigroup. Meanwhile, considering the unit matrix $I_{n}$ and any matrix $( a_{ij} )$, $a_{ij} \cdot 1 = a_{ij}$ and $a_{ij} \cdot 0 = 0$ hence $(a_{ij}) I = I (a_{ij}) = (a_{ij})$ follows. Therefore, $I$ is the identity element of $\left< \mathbb{R}^{n \times n} , \cdot \right>$.

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