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

Naturally Ordered Semigroup
Can be proved by the Principle of Finite Induction.

For all $n \in \left({S^*, \circ, \preceq}\right)$, let $P \left({n}\right)$ be the proposition:
 * $\phi \left({\odot_1^n \left({a}\right)}\right) = \odot_2^n \left({\phi \left({a}\right)}\right)$

Basis for the Induction
$P(1)$ is true, as this just says:

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\phi \left({\odot_1^k \left({a}\right)}\right) = \odot_2^k \left({\phi \left({a}\right)}\right)$

Then we need to show:
 * $\phi \left({\odot_1^{k+1} \left({a}\right)}\right) = \odot_2^{k+1} \left({\phi \left({a}\right)}\right)$

Induction Step
This is our induction step:

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore, by the Principle of Finite Induction:
 * $\forall n \in \left({S^*, \circ, \preceq}\right): \phi \left({\odot_1^n \left({a}\right)}\right) = \odot_2^n \left({\phi \left({a}\right)}\right)$

as we wanted to show.

Natural Numbers
Follows directly from Homomorphism of Powers: Naturally Ordered Semigroup as the Natural Numbers are a Naturally Ordered Semigroup.

Integers
By Homomorphism of Powers: Natural Numbers, we need show this only for negative $n$, that is:


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

But $\phi \left({a^{-1}}\right) = \left({\phi \left({a}\right)}\right)^{-1}$ by Homomorphism with Identity Preserves Inverses.

Hence:


 * $\odot_2^{-n} \left({\phi \left({a}\right)}\right) = \odot_2^n \left({\phi \left({a^{-1}}\right)}\right) = \phi \left({\odot_1^{-n} \left({a^{-1}}\right)}\right) = \phi \left({\odot_1^{-n} \left({a}\right)}\right)$

by Homomorphism of Powers: Natural Numbers.