Transplanting Theorem

Theorem
Let $\left({S, \circ}\right)$ be an algebraic structure.

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

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

The operation $\oplus$ is defined by:


 * $\forall x, y \in T: x \oplus y = f \left({f^{-1} \left({x}\right) \circ f^{-1} \left({y}\right)}\right)$

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

Proof

 * First we show that $\oplus$ as defined above has the required properties:

Let $u, v \in S$, and let $x = f \left({u}\right), y = f \left({v}\right)$.

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

Thus:

... and we see that $f$ is an isomorphism as required.


 * Then we show that $\oplus$ is the only operation to have these properties.

If $f$ is an isomorphism, then $f \circ f^{-1} = I_T$. Thus:

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

Comment
If $S = T$, that is, if $f$ is an automorphism on $\left({S, \circ}\right)$, then the transplant of $\circ$ under $f$ is $\circ$ itself.