Transplanting Theorem

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Theorem

Let $\struct {S, \circ}$ be an algebraic structure.

Let $f: S \to T$ be a bijection.


Then there exists one and only one operation $\oplus$ such that $f: \struct {S, \circ} \to \struct {T, \oplus}$ is an isomorphism.

The operation $\oplus$ is defined by:

$\forall x, y \in T: x \oplus y = \map f {\map {f^{-1} } x \circ \map {f^{-1} } y}$


The operation $\oplus$ is called the transplant of $\circ$ under $f$.


Corollary

Let $\struct {S, \circ}$ be an algebraic structure.

Let $f: S \to S$ be an automorphism on $\struct {S, \circ}$.


Then the transplant of $\circ$ under $f$ is $\circ$ itself.


Proof

Existence

To show that $\oplus$ as defined above exists:

Let $u, v \in S$, and let $x = \map f u, y = \map f v$.

Then as $f$ is a bijection, $u = \map {f^{-1} } x, v = \map {f^{-1} } y$.

Thus:

\(\ds \map f {u \circ v}\) \(=\) \(\ds \map f {\map {f^{-1} } x \circ \map {f^{-1} } y}\) Definition of $x$ and $y$
\(\ds \) \(=\) \(\ds \map f {\map {f^{-1} } x} \oplus \map f {\map {f^{-1} } y}\) Definition of Morphism Property
\(\ds \) \(=\) \(\ds x \oplus y\) Inverse of Inverse of Bijection
\(\ds \) \(=\) \(\ds \map f u \oplus \map f v\) Definition of $x$ and $y$


It is seen that $f$ is an isomorphism as required.

$\Box$


Uniqueness

Let $f$ be the isomorphism whose existence has been proven above.

Thus:

\(\ds \map f {\map {f^{-1} } x \circ \map {f^{-1} } y}\) \(=\) \(\ds \map f {\map {f^{-1} } x} \oplus \map f {\map {f^{-1} } y}\) Definition of Morphism Property
\(\ds \) \(=\) \(\ds \paren {f \circ \map {f^{-1} } x} \oplus \paren {f \circ \map {f^{-1} } y}\) Definition of Composition of Mappings
\(\ds \) \(=\) \(\ds \paren {\map {I_T} x} \oplus \paren {\map {I_T} y}\) Bijection has Left and Right Inverse
\(\ds \) \(=\) \(\ds x \oplus y\) Definition of Identity Mapping


So, if $\oplus$ is an operation on $T$ such that $f$ is an isomorphism from $\struct {S, \circ} \to \struct {T, \oplus}$, then $\oplus$ must be defined as by this theorem, and there can be no other such operations.

$\blacksquare$


Sources

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