# Structure Induced by Group Operation is Group

## Contents

## 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

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 13$: Theorem $13.6: \ 3^\circ$