Powers of Commutative Elements in Semigroups
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Theorem
Let $\left ({S, \circ}\right)$ be a semigroup.
Let $a, b \in S$ both be cancellable elements of $S$.
Then the following results hold:
Commutativity of Powers
- $\forall m, n \in \N_{>0}: a \circ b = b \circ a \implies a^m \circ b^n = b^n \circ a^m$
but it is not necessarily the case that:
- $\forall m, n \in \N_{>0}: a^m \circ b^n = b^n \circ a^m \implies a \circ b = b \circ a$
Product of Commutative Elements
- $\forall n \in \N_{>1}: \paren {x \circ y}^n = x^n \circ y^n \iff x \circ y = y \circ x$