Homomorphism of Powers/Integers

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Theorem

Let $\left({T_1, \odot}\right)$ and $\left({T_2, \oplus}\right)$ be monoids.

Let $\phi: \left({T_1, \odot}\right) \to \left({T_2, \oplus}\right)$ 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:

$\forall n \in \Z: \phi \left({\odot^n \left({a}\right)}\right) = \oplus^n \left({\phi \left({a}\right)}\right)$


Proof

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

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

But by Homomorphism with Identity Preserves Inverses:

$\phi \left({a^{-1}}\right) = \left({\phi \left({a}\right)}\right)^{-1}$

Henceby Homomorphism of Powers: Natural Numbers:

$\oplus^{-n} \left({\phi \left({a}\right)}\right) = \oplus^n \left({\phi \left({a^{-1}}\right)}\right) = \phi \left({\odot^n \left({a^{-1}}\right)}\right) = \phi \left({\odot^{-n} \left({a}\right)}\right)$

$\blacksquare$


Sources