[[Group order]]
# $ab$ and $ba$ have the same order

In any group, $\abs{ab} = \abs{ba}$, 
so if $(ab)^n = e$ then also $(ba)^n = e$. #m/thm/group 

> [!check]- Proof
> If $(ab)^n = e$ for some $n > 0$ then
> $$
> \begin{align}
> a(ba)^{n-1}b &= e \\
> (ba)^{n-1} &= a^{-1}b^{-1} \\
> (ba)^{n} &= e \qquad\square
> \end{align}
> $$

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