Order of Group Product

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

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

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.