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.