Structure Induced by Abelian Group Operation is Abelian Group

Theorem
Let $$\left({T, \oplus}\right)$$ be an group whose identity is $$e_T$$, and let $$S$$ be a set.

Let $$\left({T^S, \oplus}\right)$$ be the structure on $$T^S$$ induced by $$\oplus$$.

Then $$\left({T^S, \oplus}\right)$$ is a group, and the inverse of a given mapping $$f$$ is defined as $$f^*$$ in the following way:

$$\forall f \in T^S: \forall x \in S: f^* \left({x}\right) = \left({f \left({x}\right)}\right)^{-1}$$

If $$\left({T, \oplus}\right)$$ is abelian, then so is $$\left({T^S, \oplus}\right)$$.

Proof

 * From Induced Structure Associative and Induced Structure Identity, $$\left({T^S, \oplus}\right)$$ is associative and has an identity.


 * $$\left({T^S, \oplus}\right)$$ is patently closed, because:

$$\forall f, g \in T^S: f \oplus g \in T^S$$


 * We now need to show that the inverse of $$f$$ is $$f^*$$ for all $$f \in T^S$$.

... and similarly for $$\left({f^* \oplus f}\right) \left({x}\right)$$.


 * If $$\left({T, \oplus}\right)$$ is abelian, then $$\oplus$$ is commutative on $$T$$

Then from Induced Structure Commutative, so is the operation it induces on $$T^S$$, and the result follows.