Let $(G, \cdot)$ be a semigroup (i.e. a set with an associative binary operation) and fix some $e\in G$. If

1) $\forall g\in G: e\cdot g=g$ (left identity),

2) $\forall g\in G~ \exists g^{-1}\in G: g\cdot g^{-1}=e$ (right inverse),

3) $\forall g,h\in G: g\cdot h=e\Rightarrow h=g^{-1}, h\cdot g^{-1}=e\Rightarrow h=g$ (unique inverse),

must $(G, \cdot)$ be a group? I know that a left idenitity and right inverse don't necessarily give a group, and that a unique left identity and unique right inverse give a group. Yet this question I have no clue how to attempt.

  • 3
    $\begingroup$ Axiom 2-3 is awkwardly written, since right inverse is not unique. If 2 is existence of a right inverse, then it has no reason to be written $g^{-1}$. If you mean you fix such a function, it should be part of the axiom: you have a function $g\mapsto g^{-1}$ satisfying axioms 2,3. $\endgroup$ – YCor Mar 11 '19 at 17:44

A semigroup with a left identity and unique right inverses is not necessarily a group. Here is an example:

Let $G$ be any set with more than one element and call one of the elements $e$. Define a multiplication on $G$ by $x\cdot y = y$. This is associative, since $x\cdot(y\cdot z) = (x\cdot y)\cdot z = z$. For all $x\in G$ we have $e\cdot x = x$, so $e$ is a left identity (as is every other element of $G$). Also, $x\cdot y = e$ if and only if $y = e$, so each $x$ has a unique right inverse, namely $e$. But $(G,\cdot)$ is not a group, since there is no right identity.

EDIT: Looking closely at your axiom $3$, you seem to be requiring something more than unique right inverses. It seems that you also want an element to be the right inverse of at most one element; equivalently, if an element has a left inverse, it must be unique. In this case, $(G, \cdot)$ must indeed be a group.

To prove this, it suffices to show that a right inverse is also a left inverse. Choose any $x\in G$ and consider the product $x'\cdot x\cdot x'\cdot x''$, where $x'$ denotes a right inverse. On the one hand, $$x'\cdot (x\cdot x')\cdot x'' = x'\cdot e \cdot x'' = x'\cdot x'' = e.$$ On the other, $x'\cdot x\cdot (x'\cdot x'') = x'\cdot x\cdot e$, which must also be $e$. Since the right inverse of $x'$ is unique, $x'' = x\cdot e$. It follows that $$ x''\cdot x' = (x\cdot e)\cdot x'= x\cdot(e\cdot x') = x\cdot x' = e. $$ Hence $x'$ is a right inverse for both $x$ and $x''$, so $x = x''$. Finally, $x'\cdot x = x'\cdot x'' = e$, so the right inverse of $x$ is also its left inverse.


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