Homomorphism of Powers

Theorem

Let $\struct {T_1, \odot}$ and $\struct {T_2, \oplus}$ be semigroups.

Let $\phi: \struct {T_1, \odot} \to \struct {T_2, \oplus}$ be a (semigroup) homomorphism.

Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.

For a given $a \in T_1$, let $\map {\odot^n} a$ be the $n$th power of $a$ in $T_1$.

For a given $a \in T_2$, let $\map {\oplus^n} a$ be the $n$th power of $a$ in $T_2$.

Then:

$\forall a \in T_1: \forall n \in \struct {S^*, \circ, \preceq}: \map \phi {\map {\odot^n} a} = \map {\oplus^n} {\map \phi a}$

where $S^* = S \setminus \set 0$.

Let $n \in \N$.

Let $\odot^n$ and $\oplus^n$ be the $n$th powers of $\odot$ and $\oplus$, respectively.

Then:

$\forall a \in T_1: \forall n \in \N: \map \phi {\map {\odot^n} a} = \map {\oplus^n} {\map \phi a}$

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:

$\forall n \in \Z: \map \phi {\map {\odot^n} a} = \map {\oplus^n} {\map \phi a}$