I don't understand well in what way idempotent element is wired to identity element in a magma context.
idempotent: $x \cdot x = x$
identity element: $1 \cdot x = x = x \cdot 1$
For example subtraction has $0$ as a right identity since $x−0=x$, but it doesn’t have a left identity. So is not a unital magma.
An example of a unital magma that is neither a monoid nor a loop is given by this table but I want understand if this is or is not a idempotent semigroup
$\begin{array}{c|rrrr}& 1 & a & b \\\hline {1} & 1 & a & b \\ {a} & a & 1 & a \\ {b} & b & b & a & \end{array}$
I want understand with examples these differences
- an example of unital magma that is not a idempotent semigroup
- an example of unital magma that is not a idempotent magma
- an example of idempotent magma that is not an idempotent semigroup
- if every idempotent magma requires the identity element then is an idempotent element an identity element of itself ?
I'm little confuse between idempotency and identity, I need some examples.
For example, can you provide me a closed operation under some set $S$ but not associative nor commutative but with identity element ?
Best Answer
The example given in your table is a unital magma. It is not an idempotent magma, as $bb = a ≠ b$ and it is not a semigroup as $(aa)b = b ≠ 1 = aa = a(ab)$. Thus this example solves your first two questions.
Here is an example of an idempotent magma which is not a semigroup:
$\begin{array}{c|rrrr} & a & b & c \\ \hline a & a & a & b \\ b & b & b & b \\ c & c & c & c \end{array}$
If you want an idempotent magma with identity, just add an identity
$\begin{array}{c|rrrrr} & 1 & a & b & c \\ \hline 1 & 1 & a & b & c \\ a & a & a & a & b \\ b & b & b & b & b \\ c & c & c & c & c \end{array}$