Mapping to Power is Endomorphism iff Abelian

Theorem

Let $\struct {G, \circ}$ be a group.

Let $n \in \Z$ be an integer.

Let $\phi: G \to G$ be defined as:

$\forall g \in G: \map \phi g = g^n$

Then $\struct {G, \circ}$ is abelian if and only if $\phi$ is a (group) endomorphism.

Let $\struct {G, \circ}$ be an abelian group.

Let $a, b \in G$ be arbitrary.

Then:

$\ds \map \phi {a \circ b}$

$=$

$\ds \paren {a \circ b}^n$

Definition of $\phi$

$\ds $

$=$

$\ds a^n \circ b^n$

$\ds $

$=$

$\ds \map \phi a \circ \map \phi b$

Definition of $\phi$

As $a$ and $b$ are arbitrary, the above holds for all $a, b \in G$.

Thus $\phi$ is a group homomorphism from $G$ to $G$.

So by definition, $\phi$ is a group endomorphism.

$\Box$

Let $\phi: G \to G$ as defined above be a group endomorphism.

Then:

$\ds \forall a, b \in G: \, $

$\ds \map \phi {a \circ b}$

$=$

$\ds \map \phi a \circ \map \phi b$

Definition of Group Endomorphism

$\ds \leadsto \ \ $

$\ds \forall a, b \in G: \, $

$\ds \paren {a \circ b}^n$

$=$

$\ds a^n \circ b^n$

Definition of $\phi$

$\blacksquare$

1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $7$: Homomorphisms: Exercise $2$
1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.2$: Groups; the axioms: Exercise $(9)$
1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): $\S 3$: Exercise $6$