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.