Semigroup with left neutral element where every element is right-invertible not equals group

Statement

Version with left neutral element and right-invertibility

Suppose $S$ is a set with more than one element. It is possible to construct a binary operation $*$ on $S$ and find an element $e\in S$ such that:

1. $*$ is an associative binary operation, so that $(S,*)$ is a semigroup
2. $e$ is a left neutral element for $*$: $e*a=a\ \forall \ a\in S$
3. For all $a\in S$, there exists a right inverse $b\in S$ with respect to $e$, i.e., $a*b=e$.
4. $S$ is not a group under $*$. In particular, $e$ is not a two-sided neutral element.

Version with right neutral element and left-invertibility

Suppose $S$ is a set with more than one element. It is possible to construct a binary operation $*$ on $S$ and find an element $e\in S$ such that:

1. $*$ is an associative binary operation, so that $(S,*)$ is a semigroup
2. $e$ is a right neutral element for $*$: $a*e=a\ \forall \ a\in S$
3. For all $a\in S$, there exists a left inverse $b\in S$ with respect to $e$, i.e., $b*a=e$.
4. $S$ is not a group under $*$. In particular, $e$ is not a two-sided neutral element.

Related facts

Opposite facts

• Semigroup with left neutral element where every element is left-invertible equals group: The key difference is that we are assuming invertibility on the same side as the side of the one-sided neutral element.
• Monoid where every element is left-invertible equals group (note that we could replace left by right and the statement would still be true.

Proof

Proof for left neutral element and right-invertibility

Define $*$ as follows:

$x*y:=y$

In other words, every element is a right nil.

Let $e$ be any element of $S$.

We check all the four conditions:

Condition to check	What it says	Why it's true
associativity	$(a*b)*c=a*(b*c)\ \forall a,b,c\in S$	The left side becomes $b*c=c$. The right side becomes $a*c=c$. Thus, both sides simplify to $c$.
left neutral element	$e*a=a\ \forall \ a\in S$	direct from definition
right-invertibility	for all $a\in S$, there exists $b\in S$ such that $a*b=e$.	Take $b=e$.
not a group	If $S$ has more than one element, we can see that it has no right neutral element. Alternatively, we note that it is not right-cancellative.

Proof for right neutral element and left-invertibility

Define $*$ as follows:

$x*y:=x$

In other words, every element is a left nil.

Let $e$ be any element of $S$.

We check all the four conditions:

Condition to check	What it says	Why it's true
associativity	$(a*b)*c=a*(b*c)\ \forall a,b,c\in S$	The left side becomes $a*c=a$. The right side becomes $a*b=a$. Thus, both sides simplify to $a$.
right neutral element	$a*e=a\ \forall \ a\in S$	direct from definition
right-invertibility	for all $a\in S$, there exists $b\in S$ such that $b*a=e$.	Take $b=e$.
not a group	If $S$ has more than one element, we can see that it has no left neutral element. Alternatively, we note that it is not left-cancellative.