Structure Induced by Group Operation is Group

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Theorem

Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $S$ be a set.


Let $\struct {G^S, \oplus}$ be the structure on $G^S$ induced by $\circ$.


Then $\struct {G^S, \oplus}$ is a group.


Proof

Taking the group axioms in turn:


$\text G 0$: Closure

Let $f, g \in G^S$.

As $\struct {G, \circ}$ is a group, it is closed by group axiom $G0$.

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

$\Box$


$\text G 1$: Associativity

As $\struct {G, \circ}$ is a group, $\circ$ is associative.

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

$\Box$


$\text G 2$: Identity

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

$\forall x \in S: \map {f_e} x = e$

is the identity for $\struct {G^S, \oplus}$.

$\Box$


$\text G 3$: Inverses

Let $f \in G^S$.

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

$\forall f \in G^S: \forall x \in S: \map {f^*} x = \paren {\map f x}^{-1}$

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

$\Box$


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

$\blacksquare$


Sources