Mapping to Power is Endomorphism iff Abelian
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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.
Proof
Necessary Condition
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 $\phi$ | |||||||||||
\(\ds \) | \(=\) | \(\ds a^n \circ b^n\) | Power of Product of Commutative Elements in Group | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi a \circ \map \phi b\) | 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.
$\Box$
Sufficient Condition
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 $\phi$ |
From Power of Product of Commutative Elements in Group it follows that $G$ is an abelian group.
$\blacksquare$
Sources
- 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 $6$