Prove that if a and b are elements of a finite group G, then
ord(ab) = ord(ba).
If $\displaystyle ord(ab) = n$ then $\displaystyle (ab)^n = e\quad \Rightarrow \quad \underbrace {(ab)(ab) \cdots (ab)}_n = e$.
So $\displaystyle a\left[ {\underbrace {(ba)(ba) \cdots (ba)}_{n - 1}} \right]b = e\quad \Rightarrow \quad \left[ {\underbrace {(ba)(ba) \cdots (ba)}_{n - 1}} \right] = a^{ - 1} b^{ - 1}$
$\displaystyle (ba)^n = \underbrace {(ba)(ba) \cdots (ba)}_{n - 1}(ba) = \left( {a^{ - 1} b^{ - 1} } \right)(ba) = e$
Look in my first reply, I deliberately leave some point for the questioner to answer.
I showed that $\displaystyle ord(ba) \leqslant n$, correct?
Now suppose that $\displaystyle ord(ba) = m$.
As Opalg pointed out the same argument shows that $\displaystyle n \leqslant m$, so $\displaystyle n = m$.