Proof of Units of a Monoid: Unique Inverses, Products, and the Group of Units

Throughout, $\cdot$ is associative with neutral element $e$ (Monoids (i), Semigroups (i), Associative and Commutative Binary Operations, and Neutral Elements (i), Associative and Commutative Binary Operations, and Neutral Elements (iii)), so brackets in products are dropped, and units and inverses are as in Units of a Monoid (i).

We begin with (i). Let $v$ and $w$ be inverses of a unit $u$, so that $uv=vu=e$ and $uw=wu=e$. Then

$$v=ve=v(uw)=(vu)w=ew=w,$$

so $u$ has only one inverse.

For (ii), the equation $ee=e$ says that $e$ is a unit and that $e$ is an inverse of itself. Since inverses are unique by (i), $e^{-1}=e$.

For (iii), let $u$ and $v$ be units. Then

$$(uv)(v^{-1}u^{-1})=u(vv^{-1})u^{-1}=ueu^{-1}=uu^{-1}=e,$$

and in the same way $(v^{-1}u^{-1})(uv)=v^{-1}(u^{-1}u)v=v^{-1}v=e$. Thus $v^{-1}u^{-1}$ is an inverse of $uv$, so $uv$ is a unit, and $(uv)^{-1}=v^{-1}u^{-1}$ by (i).

For (iv), let $u$ be a unit. The equations $uu^{-1}=u^{-1}u=e$, read the other way round, say that $u$ is an inverse of $u^{-1}$. Hence $u^{-1}$ is a unit, and $(u^{-1})^{-1}=u$ by (i).

Finally, for (v), let $\cdot'$ be the restriction of $\cdot$ to $M^{\times}\times M^{\times}$, where $M^{\times}$ is the set of units of $M$ (Units of a Monoid (ii)); it is a function with domain $M^{\times}\times M^{\times}$ and $u\cdot' v=uv$ for $u,v\in M^{\times}$ (Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (ix)). By (iii) a product of two units is a unit, so $\cdot'$ takes its values in $M^{\times}$ and is a binary operation on $M^{\times}$ (Binary Operations on a Set (i)). It is associative because $\cdot$ is, so $(M^{\times},\cdot')$ is a semigroup (Semigroups (i)). By (ii), $e$ lies in $M^{\times}$, and it is neutral for $\cdot'$ because it is neutral for $\cdot$. Hence $(M^{\times},\cdot')$ is a monoid (Monoids (i)), and its identity is $e$, since a binary operation has at most one neutral element (A Binary Operation Has at Most One Neutral Element (i)). Now let $u\in M^{\times}$. By (iv), $u^{-1}$ lies in $M^{\times}$, and

$$u\cdot' u^{-1}=u^{-1}\cdot' u=e,$$

so $u$ is a unit of the monoid $(M^{\times},\cdot')$ (Units of a Monoid (i)). Every element of $M^{\times}$ is therefore a unit of $(M^{\times},\cdot')$, which is thus a group (Groups and Abelian Groups (i)).