Identity Property in Semigroup

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Theorem

Let $\struct {S, \circ}$ be a semigroup.

Let $s \in S$ be such that:

$\forall a \in S: \exists x, y \in S: s \circ x = a = y \circ s$


Then $\struct {S, \circ}$ has an identity.


Proof

Suppose that:

$\forall a \in S: \exists x, y \in S: s \circ x = a = y \circ s$.


Since $s \in S$, it follows that:

$\exists v, w \in S: s \circ v = s = w \circ s$.


Let $a \in S$.

Then:

$\exists x, y \in S: s \circ x = a = y \circ s$.


Thus:

\(\displaystyle a\) \(=\) \(\displaystyle s \circ x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle w \circ a\) \(=\) \(\displaystyle w \circ s \circ x\)
\(\displaystyle \) \(=\) \(\displaystyle s \circ x\)
\(\displaystyle \) \(=\) \(\displaystyle a\)


\(\displaystyle a\) \(=\) \(\displaystyle y \circ s\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle a \circ v\) \(=\) \(\displaystyle y \circ s \circ v\)
\(\displaystyle \) \(=\) \(\displaystyle y \circ s\)
\(\displaystyle \) \(=\) \(\displaystyle a\)


Hence:

$w \circ a = a$ and $a \circ v = a$

for any $a \in S$.


In particular:

Letting $a = v$ in the first of these gives $w \circ v = v$
Letting $a = w$ in the second gives $w \circ v = w$.


Thus $v = w \circ v = w$ is the identity element in $S$.

$\blacksquare$


Sources