Order of Group Product

Theorem
Let $\left({G, \circ}\right)$ be a group whose identity is $e$.

Then:
 * $\forall a, b \in \left({G, \circ}\right): \left|{a \circ b}\right| = \left|{b \circ a}\right|$

where $\left|{a \circ b}\right|$ is the order of $a \circ b$ in G.

Proof
We have:
 * $a \circ b = \left({a \circ b}\right) \circ \left({a \circ a^{-1}}\right) = a \circ \left({b \circ a}\right) \circ a^{-1}$

We also have:
 * $\left|{b \circ a}\right| = \left|{a \circ \left({b \circ a}\right) \circ a^{-1}}\right|$

from Order of Conjugate.

The result follows.