By hypothesis, all the elements of $\struct {R, *}$ are cancellable. Thus Inverse Completion of Commutative Semigroup is Abelian Group can be applied. So $\struct {T, *}$ is an abelian group. {{explain|Where is the above used?}} === Existence of Distributive Operation === For each $m \in R$, we define $\lambda_m: R \to T$ as: :$\forall x \in R: \map {\lambda_m} x = m \circ x$ Then: {{begin-eqn}} {{eqn | l = \map {\lambda_m} {x * y}       | r = m \circ \paren {x * y}       | c =  }} {{eqn | r = \paren {m \circ x} * \paren {m \circ y}       | c = as $\circ$ distributes over $*$ }} {{eqn | r = \map {\lambda_m} x * \map {\lambda_m} y       | c =  }} {{end-eqn}} So $\lambda_m$ is a homomorphism from $\struct {R, *}$ into $\struct {T, *}$. Now, by the Extension Theorem for Homomorphisms, every homomorphism from $\struct {R, *}$ into $\struct {T, *}$ is the restriction to $R$ of a unique endomorphism of $\struct {T, *}$. We have just shown that $\lambda_m$ is such a homomorphism. Therefore there exists a unique endomorphism $\lambda'_m: T \to T$ which extends $\lambda_m$. Now: {{begin-eqn}} {{eqn | lo= \forall m, n, z \in R:       | l = \map {\lambda_{m * n} } z       | r = \paren {m * n} \circ z       | c = Definition of $\lambda$ }} {{eqn | r = \paren {m \circ z} * \paren {n \circ z}       | c = Distributivity of $\circ$ over $*$ }} {{eqn | r = \map {\lambda_m} z * \map {\lambda_n} z       | c = Definition of $\lambda$ }} {{eqn | r = \map {\paren {\lambda_m * \lambda_n} } z       | c = Here $*$ is the operation induced on $T^T$ by $*$ }} {{end-eqn}} By Homomorphism on Induced Structure, $\lambda'_m * \lambda'_n$ is an endomorphism of $\struct {T, *}$ that, as we have just seen, coincides on $R$ with $\lambda'_{m * n}$. Hence $\lambda'_{m * n} = \lambda'_m * \lambda'_n$. Similarly, for each $z \in T$, we define $\rho_z: R \to T$ as: :$\forall m \in R: \map {\rho_z} m = \map {\lambda'_m} z$ Then: {{begin-eqn}} {{eqn | l = \map {\rho_z} {m * n}       | r = \map {\lambda'_{m * n} } z       | c = Definition of $\rho_z$ }} {{eqn | r = \map {\lambda'_m} z * \map {\lambda'_n} z       | c = Behaviour of $\lambda'_{m * n}$ }} {{eqn | r = \map {\rho_z} m * \map {\rho_z} n       | c = Definition of $\rho_z$ }} {{end-eqn}} Therefore $\rho_z$ is a homomorphism from $\struct {R, *}$ into $\struct {T, *}$. Consequently there exists a unique endomorphism $\rho'_z: T \to T$ extending $\rho_z$. {{begin-eqn}} {{eqn | lo= \forall y, z \in T, m \in R:       | l = \map {\rho_{y * z} } m       | r = \map {\lambda'_m} {y * z}       | c = Definition of $\rho_{y * z}$ }} {{eqn | r = \map {\lambda'_m} y * \map {\lambda'_m} z       | c = $\lambda'_m$ is a homomorphism }} {{eqn | r = \map {\paren {\rho_y * \rho_z} } m       | c = Definition of $\rho_y$ and $\rho_z$ }} {{end-eqn}} By Homomorphism on Induced Structure, $\rho'_y * \rho'_z$ is an endomorphism on $\struct {T, *}$ that coincides (as we have just seen) with $\rho'_{y * z}$ on $R$. Hence $\rho'_{y * z} = \rho'_y * \rho'_z$. Now we define an operation $\circ'$ on $T$ by: :$\forall x, y \in T: x \circ' y = \map {\rho'_y} x$ Now suppose $x, y \in R$. Then: {{begin-eqn}} {{eqn | l = x \circ' y       | r = \map {\rho_y} x       | c = as $\rho'_y = \rho_y$ on $R$ }} {{eqn | r = \map {\lambda_x} y       | c = Definition of $\rho_x$ }} {{eqn | r = x \circ y       | c = Definition of $\lambda_x$ }} {{end-eqn}} so $\circ'$ is an extension of $\circ$. Next, let $x, y, z \in T$. Then: {{begin-eqn}} {{eqn | l = \paren {x * y} \circ' z       | r = \map {\rho'_z} {x * y}       | c =  }} {{eqn | r = \map {\rho'_z} x * \map {\rho'_z} y       | c =  }} {{eqn | r = x \circ' z * y \circ' z       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = x \circ' \paren {y * z}       | r = \map {\rho'_{y * z} } x       | c =  }} {{eqn | r = \map {\rho'_y} x * \map {\rho'_z} x       | c =  }} {{eqn | r = x \circ' y * x \circ' z       | c =  }} {{end-eqn}} So $\circ'$ is distributive over $*$. {{qed}} === Uniqueness of Distributive Operation === To show that $\circ'$ is unique, let $\circ_1$ be any operation on $T$ distributive over $*$ that induces $\circ$ on $R$. Since $\circ'$ and $\circ_1$ both distribute over $*$, for every $m \in R$, the mappings: {{explain|Should this be $y \in T$? If not, how does it define a $T$-endomorphism?}} {{begin-eqn}} {{eqn | l = y \mapsto m \circ_1 y       | o = ,       | r = y \in R       | c =  }} {{eqn | l = y \mapsto m \circ' y       | o = ,       | r = y \in R       | c =  }} {{end-eqn}} are endomorphisms of $\struct {T, *}$ that coincide on $R$ so must be the same mapping. Therefore $\forall m \in R, y \in T: m \circ_1 y = m \circ' y$. Similarly, for every $y \in T$, the mappings: {{begin-eqn}} {{eqn | l = x \mapsto x \circ_1 y       | o = ,       | r = x \in T       | c =  }} {{eqn | l = x \mapsto x \circ' y       | o = ,       | r = x \in T       | c =  }} {{end-eqn}} are endomorphisms of $\struct {T, *}$ that coincide on $R$ by what we have just proved, so must be the same mapping. Hence $\forall x, y \in T: x \circ_1 y = x \circ' y$. Thus $\circ'$ is the only operation on $T$ which extends $\circ$ and distributes over $*$. {{qed}} === Proof of Associativity === Suppose $\circ$ is associative. As $\circ'$ distributes over $*$, for all $n, p \in R$, the mappings: {{begin-eqn}} {{eqn | l = x \mapsto \paren {x \circ' n} \circ' p       | o = ,       | r = x \in T       | c =  }} {{eqn | l = x \mapsto x \circ' \paren {n \circ' p}       | o = ,       | r = x \in T       | c =  }} {{end-eqn}} are endomorphisms of $\struct {T, *}$ that coincide on $R$ by the associativity of $\circ$ and hence are the same mapping. Therefore: :$\forall x \in T, n, p \in R: \paren {x \circ' n} \circ' p = x \circ' \paren {n \circ' p}$ Similarly, for all $x \in T, p \in R$, the mappings: {{begin-eqn}} {{eqn | l = y \mapsto \paren {x \circ' y} \circ' p       | o = ,       | r = y \in T       | c =  }} {{eqn | l = y \mapsto x \circ' \paren {y \circ' p}       | o = ,       | r = y \in T       | c =  }} {{end-eqn}} are endomorphisms of $\struct {T, *}$ that coincide on $R$ by what we have proved and hence are the same mapping. Therefore: :$\forall x, y \in T, p \in R: \paren {x \circ' y} \circ' p = x \circ' \paren {y \circ' p}$ Finally, for all $x, y \in T$, the mappings: {{begin-eqn}} {{eqn | l = z \mapsto \paren {x \circ' y} \circ' z       | o = ,       | r = z \in T       | c =  }} {{eqn | l = z \mapsto x \circ' \paren {y \circ' z}       | o = ,       | r = z \in T       | c =  }} {{end-eqn}} are endomorphisms of $\struct {T, *}$ that coincide on $R$ by what we have proved and hence are the same mapping. Therefore $\circ'$ is associative. {{qed}} === Proof of Commutativity === Suppose $\circ$ is commutative. As $\circ'$ distributes over $*$, for all $n \in R$, the mappings: {{begin-eqn}} {{eqn | l = x \mapsto x \circ' n       | o = ,       | r = x \in T       | c =  }} {{eqn | l = x \mapsto n \circ' x       | o = ,       | r = x \in T       | c =  }} {{end-eqn}} are endomorphisms of $\struct {T, *}$ that coincide on $R$ by the commutativity of $\circ$ and hence are the same mapping. Therefore $\forall x \in T, n \in R: x \circ' n = n \circ' x$. Finally, for all $y \in T$, the mappings: {{begin-eqn}} {{eqn | l = z \mapsto z \circ' y       | o = ,       | r = z \in T       | c =  }} {{eqn | l = z \mapsto y \circ' z       | o = ,       | r = z \in T       | c =  }} {{end-eqn}} are endomorphisms of $\struct {T, *}$ that coincide on $R$ by what we have proved and hence are the same mapping. Therefore $\circ'$ is commutative. {{qed}} === Proof of Identity === Let $e$ be the identity element of $\struct {R, \circ}$. Then the restrictions to $R$ of the endomorphisms $\lambda_e: x \mapsto e \circ' x$ and $\rho_e: x \mapsto x \circ' e$ of $\struct {T, *}$ are monomorphisms. But then $\lambda_e$ and $\rho_e$ are monomorphisms by the Extension Theorem for Homomorphisms, so $e$ is the identity element of $T$. {{explain|How does the above explain why $e$ is the identity element of $T$?}} {{qed}} === Proof of Cancellability === To prove that every element of $R$ cancellable for $\circ$ is also cancellable for $\circ'$: Let $a$ be an element of $R$ cancellable for $\circ$. Then the restrictions to $R$ of the endomorphisms $\lambda_a: x \mapsto a \circ' x$ and $\rho_a: x \mapsto x \circ' a$ of $\struct {T, *}$ are monomorphisms. But then $\lambda_a$ and $\rho_a$ are monomorphisms by the Extension Theorem for Homomorphisms, so $a$ is cancellable for $\circ'$. {{qed}}