# Identity Property in Semigroup

## 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 $S$.

$\blacksquare$