Homomorphism of Powers

Theorem

Let $\left({T_1, \odot_1}\right)$ and $\left({T_2, \odot_2}\right)$ be semigroups.

Let $\phi: \left({T_1, \odot_1}\right) \to \left({T_2, \odot_2}\right)$ be a (semigroup) homomorphism.

Naturally Ordered Semigroup

Let $\left({S, \circ, \preceq}\right)$ be a naturally ordered semigroup.

Let $\odot_1^n$ and $\odot_2^n$ be as defined as in Recursive Mapping to Semigroup.

Then:

$\forall a \in T_1: \forall n \in \left({S^*, \circ, \preceq}\right): \phi \left({\odot_1^n \left({a}\right)}\right) = \odot_2^n \left({\phi \left({a}\right)}\right)$

Natural Numbers

Let $n \in \N$.

Let $\odot_1^n$ and $\odot_2^n$ be as defined as in Index Laws for Semigroups.

Then:

$\forall a \in T_1: \forall n \in \N: \phi \left({\odot_1^n \left({a}\right)}\right) = \odot_2^n \left({\phi \left({a}\right)}\right)$

Integers

Let $\left({T_1, \odot_1}\right)$ and $\left({T_2, \odot_2}\right)$ be monoids.

Let $\phi: \left({T_1, \odot_1}\right) \to \left({T_2, \odot_2}\right)$ be a (semigroup) homomorphism.

Let $a$ be an invertible element of $T_1$.

Let $n \in \Z$.

Let $\odot_1^n$ and $\odot_2^n$ be as defined as in Index Laws for Monoids.

Then:

$\forall n \in \Z: \phi \left({\odot_1^n \left({a}\right)}\right) = \odot_2^n \left({\phi \left({a}\right)}\right)$