Homomorphism of Powers/Naturally Ordered Semigroup

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

Proof
The proof proceeds by the Principle of Mathematical Induction for a Naturally Ordered Semigroup.

Let $A := \set {n \in S^*: \forall a \in T_1: \map \phi {\map {\odot^n} a} = \map {\oplus^n} {\map \phi a} }$

That is, $A$ is defined as the set of all $n$ such that:
 * $\forall a \in T_1 \map \phi {\map {\odot^n} a} = \map {\oplus^n} {\map \phi a}$

Basis for the Induction
We have that:

So $1 \in A$.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $k \in A$ where $k \ge 1$, then it logically follows that $k \circ 1 \in A$.

So this is our induction hypothesis:
 * $\forall a \in T_1: \map \phi {\map {\odot^k} a} = \map {\oplus^k} {\map \phi a}$

Then we need to show:
 * $\forall a \in T_1: \map \phi {\map {\odot^{k \circ 1} } a} = \map {\oplus^{k \circ 1} } {\map \phi a}$

Induction Step
This is our induction step:

So $k \in A \implies k \circ 1 \in A$ and the result follows by the Principle of Mathematical Induction:


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