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.