Proof of Cancellation Laws and Basic Inverse Identities in a Group

Throughout, associativity refers to condition 1 of Group and Abelian Group, the defining property of $e_G$ to condition 2 there, and the defining property of inverses to $aa^{-1}=e_G=a^{-1}a$, valid by Uniqueness of the Identity Element and of Inverses in a Group. Each application of associativity below is to an explicitly named triple of elements of $G$.

Claim 1. Suppose $ab=ac$. Multiplying on the left by $a^{-1}$ gives $a^{-1}(ab)=a^{-1}(ac)$. By associativity applied to the triples $a^{-1},a,b$ and $a^{-1},a,c$,

that is $e_Gb=e_Gc$, hence $b=c$ by the defining property of $e_G$. The right-hand cancellation is proved in the same way, multiplying on the right by $a^{-1}$ and using associativity for the triples $b,a,a^{-1}$ and $c,a,a^{-1}$.

Claim 2. The element $a$ satisfies $a^{-1}a=e_G$ and $aa^{-1}=e_G$, which are exactly the two equations required of an inverse of $a^{-1}$. By the uniqueness of inverses in Uniqueness of the Identity Element and of Inverses in a Group, $(a^{-1})^{-1}=a$.

Claim 3. Put $x=b^{-1}a^{-1}$. By associativity applied to the triple $b,b^{-1},a^{-1}$,

and hence, by associativity applied to the triple $a,b,x$,

Similarly, by associativity applied to the triple $a^{-1},a,b$,

and hence, by associativity applied to the triple $b^{-1},a^{-1},ab$,

Thus $x$ satisfies both equations required of an inverse of $ab$, so $(ab)^{-1}=x=b^{-1}a^{-1}$ by uniqueness of inverses.

Claim 4. By the defining property of $e_G$ we have $e_Ge_G=e_G$, so $e_G$ satisfies both equations required of an inverse of $e_G$. By uniqueness of inverses, $e_G^{-1}=e_G$.

Claim 5. Put $x=a^{-1}b$. By associativity applied to the triple $a,a^{-1},b$,

so $x$ is a solution. If $x'$ is any solution of $ax'=b$, then $ax'=ax$, and Claim 1 gives $x'=x$. The statement for $ya=b$ is proved in the same way, using associativity for the triple $b,a^{-1},a$ to get $(ba^{-1})a=b(a^{-1}a)=be_G=b$ and then the right-hand cancellation from Claim 1.