Homomorphism of Power of Group Element

Theorem

Let $\struct {G, \circ}$ and $\struct {H, \ast}$ be groups.

Let $\phi: S \to T$ be a group homomorphism.


Then:

$\forall n \in \Z: \forall g \in G: \map \phi {g^n} = \paren {\map \phi g}^n$


Proof

The result for $n \in \N_{>0}$ follows directly from General Morphism Property for Semigroups.

For $n = 0$, we use Homomorphism with Cancellable Codomain Preserves Identity.

For $n < 0$, we use Homomorphism with Identity Preserves Inverses, along with Index Laws for Monoids: Negative Index.

$\blacksquare$


Sources

  • 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 7.2$. Some lemmas on homomorphisms: Example $133$
  • 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 47.5$ Homomorphisms and their elementary properties
  • 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $8$: The Homomorphism Theorem: Exercise $4$
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