Homomorphism of Powers/Integers
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Theorem
Let $\struct {T_1, \odot}$ and $\struct {T_2, \oplus}$ be monoids.
Let $\phi: \struct {T_1, \odot} \to \struct {T_2, \oplus}$ be a (semigroup) homomorphism.
Let $a$ be an invertible element of $T_1$.
Let $n \in \Z$.
Let $\odot^n$ and $\oplus^n$ be as defined as in Index Laws for Monoids.
Then:
- $\forall n \in \Z: \map \phi {\map {\odot^n} a} = \map {\oplus^n} {\map \phi a}$
Proof
By Homomorphism of Powers: Natural Numbers, we need show this only for negative $n$, that is:
- $\forall n \in \N^*: \map \phi {\map {\odot^{-n} } a} = \map {\oplus^{-n} } {\map \phi a}$
But by Homomorphism with Identity Preserves Inverses:
- $\map \phi {a^{-1} } = \paren {\map \phi a}^{-1}$
Hence by Homomorphism of Powers: Natural Numbers:
- $\map {\oplus^{-n} } {\map \phi a} = \map {\oplus^n} {\map \phi {a^{-1} } } = \map \phi {\map {\odot^n} {a^{-1} } } = \map \phi {\map {\odot^{-n} } a}$
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $20$. The Integers: Theorem $20.13$