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.

$\blacksquare$

Sources

• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 41 \delta$