Structure Induced by Group Operation is Group

Theorem
Let $\left({G, \circ}\right)$ be a group whose identity is $e$.

Let $S$ be a set.

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

Then $\left({G^S, \oplus}\right)$ is a group.

Proof
Taking the group axioms in turn:

G0: Closure
Let $f, g \in G^S$.

That is, $f$ and $g$ are mapping from $S$ to $G$.

Then $f \oplus g$ is also a mapping from $S$ to $G$.

Thus $f \oplus g \in G^S$ and so $\left({G^S, \oplus}\right)$ is closed.

G1: Associativity
As $\left({G, \circ}\right)$ is a group, $\circ$ is associative.

So from Induced Structure Associative, $\left({G^S, \oplus}\right)$ is also associative.

G2: Identity
We have from Induced Structure Identity that the constant mapping $f_e: S \to T$ defined as:


 * $\forall x \in S: f_e \left({x}\right) = e$

is the identity for $\left({G^S, \oplus}\right)$.

G3: Inverses
Let $f \in G^S$.

Let $f^* \in G^S$ be defined as follows:


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

From Induced Structure Inverse, $f^*$ is the inverse of $f$ for the operation $\oplus$ induced on $G^S$ by $\circ$.

All the group axioms are thus seen to be fulfilled, and so $\left({G^S, \oplus}\right)$ is a group.