Powers of Semigroup Element Commute

From ProofWiki
Jump to: navigation, search

Theorem

Let $\struct {S, \odot}$ be a semigroup.

Let $a \in S$.

Let $m, n \in \Z_{>0}$.

Then:

$\forall m, n \in \Z_{>0}: a^n \odot a^m = a^m \odot a^n$


Proof

From Index Laws for Semigroup: Sum of Indices:

$\forall m, n \in \Z_{>0}: a^{n + m} = a^n \odot a^m$

But from Integer Addition is Commutative:

$n + m = m + n$

So:

$a^n \odot a^m = a^{n + m} = a^{m + n} = a^m \odot a^n$

$\blacksquare$


Sources