# Structure Induced by Group Operation is Group

## Contents

## 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

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