Identity Property in Semigroup

From ProofWiki
Jump to navigation Jump to search

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:

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


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


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

Retrieved from "https://proofwiki.org/w/index.php?title=Identity_Property_in_Semigroup&oldid=521983"