Proof of Cancellation Laws and Basic Inverse Identities in a Group

Throughout, the operation of $G$ is associative (Semigroups (i)) and $e$ is its neutral element (Monoids (i)).

For (i), suppose $ga=gb$. Multiplying on the left by $g^{-1}$ and using associativity gives $(g^{-1}g)a=(g^{-1}g)b$, that is, $ea=eb$, since $g^{-1}g=e$ (Uniqueness of the Identity Element and of Inverses in a Group (ii)). Because $e$ is a neutral element of the operation (Uniqueness of the Identity Element and of Inverses in a Group (i)), this says $a=b$. The right cancellation law is the mirror image: if $ag=bg$, multiplying on the right by $g^{-1}$ gives

$$a=ae=a(gg^{-1})=(ag)g^{-1}=(bg)g^{-1}=b(gg^{-1})=be=b.$$

For (ii), the equation $ee=e$ says that $e$ is an inverse of $e$, and by Uniqueness of the Identity Element and of Inverses in a Group (ii) it is the only one; that is, $e^{-1}=e$.

For (iii), recall that every element of $G$ is a unit of the monoid $(G,\cdot)$ (Groups and Abelian Groups (i)), and that by Uniqueness of the Identity Element and of Inverses in a Group (ii), applied to $g$ and to $g^{-1}$, the group inverses of $g$ and of $g^{-1}$ are their inverses as units in the sense of Units of a Monoid (i). Hence $(g^{-1})^{-1}=g$ by Units of a Monoid: Unique Inverses, Products, and the Group of Units (iv).

For (iv), regard $g$ and $h$ as units of $(G,\cdot)$ in the same way; then Units of a Monoid: Unique Inverses, Products, and the Group of Units (iii) gives $(gh)^{-1}=h^{-1}g^{-1}$.

For (v), the element $x=a^{-1}b$ solves $ax=b$, since by associativity and Uniqueness of the Identity Element and of Inverses in a Group (ii)

$$a(a^{-1}b)=(aa^{-1})b=eb=b.$$

If $x$ and $x'$ both solve $ax=b$, then $ax=ax'$, and $x=x'$ by (i). Symmetrically,

$$(ba^{-1})a=b(a^{-1}a)=be=b,$$

and two solutions of $xa=b$ agree by the right cancellation law of (i).