Homomorphism of Powers/Integers

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

Proof
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.