Powers of Commutative Elements in Semigroups

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

 * $$\forall m, n \in \N^*: a^m \circ b^n = b^n \circ a^m \iff a \circ b = b \circ a$$

Product of Commutative Elements

 * $$\forall n \in \N, n > 1: \left({x \circ y}\right)^n = x^n \circ y^n \iff x \circ y = y \circ x$$

Commutativity of Powers
Let $$a, b \in S: a \circ b = b \circ a$$.

Because $$\left({S, \circ}\right)$$ is a semigroup, $$\circ$$ is associative on $$S$$.

Let $$T$$ be the set of all $$n \in \N^*$$ such that:
 * $$a^n \circ b = b \circ a^n$$

We have:
 * $$a \circ b = b \circ a \implies a^1 \circ b = b \circ a^1$$.

So $$1 \in T$$.

Now suppose $$n \in T$$. Then we have:

$$ $$ $$ $$ $$ $$ $$

So $$n + 1 \in T$$.

Thus by the Principle of Finite Induction, $$T = \N^*$$. Thus:


 * $$\forall m \in \N^*: a^m \circ b = b \circ a^m$$

Thus, from the preceding: $$\forall m, n \in \N^*: a^m$$ and $$b^n$$ also commute with each other.


 * For the above relationships and equalities to hold, it follows that $$a$$ and $$b$$ must commute.

The result follows.

Product of Commutative Elements
Let $$P \left({n}\right)$$ be the proposition that $$\left({x \circ y}\right)^n = x^n \circ y^n$$ iff $$x$$ and $$y$$ commute.

We note in passing that $$P \left({1}\right)$$ does not hold: it says $$x \circ y = x \circ y \iff x \circ y = y \circ x$$ which is just wrong.


 * Next we note that $$P \left({2}\right)$$ holds, as follows:

$$ $$ $$


 * Now suppose $$P \left({n}\right)$$ holds. We need to show that $$P \left({n+1}\right)$$ holds as a result of this.

That is, that $$\left({x \circ y}\right)^{n+1} = x^{n+1} \circ y^{n+1} \iff x \circ y = y \circ x$$.

Suppose $$x \circ y = y \circ x$$ commute. Then:

$$ $$ $$ $$

As $$x \circ y^n = y^n \circ x$$ if and only if $$x$$ and $$y$$ commute, the result follows.