Structure Induced by Group Operation is Group

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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 $f, g \in G^S$.

As $\left({G, \circ}\right)$ is a group, it is closed by group axiom $G0$.

From Closure of Pointwise Operation on Algebraic Structure it follows that $\left({G^S, \oplus}\right)$ is likewise closed.

$\Box$


G1: Associativity

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

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

$\Box$


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 $\left({G^S, \oplus}\right)$.

$\Box$


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 $\circ$.

$\Box$


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

$\blacksquare$


Sources