# The Order of $ab$ and $ba$ in a Group are the Same ## Problem 291

Let $G$ be a finite group. Let $a, b$ be elements of $G$.

Prove that the order of $ab$ is equal to the order of $ba$.
(Of course do not assume that $G$ is an abelian group.) Add to solve later

## Proof.

Let $n$ and $m$ be the order of $ab$ and $ba$, respectively. That is,
$(ab)^n=e, (ba)^m=e,$ where $e$ is the identity element of $G$.

We compute
\begin{align*}
e&=(ab)^n=\underbrace{(ab)\cdot (ab) \cdot (ab) \cdots (ab)}_{n\text{ times}}\6pt] &=a\underbrace{\cdot (ba)(ba)\cdot (ba) \cdots (ba)}_{n-1\text{ times}}b\\[6pt] &=a(ba)^{n-1}b. \end{align*} From this, we obtain \[(ba)^{n-1}=a^{-1}b^{-1}=(ba)^{-1}, and thus we have
$(ba)^n=e.$ Therefore the order $m$ of $ba$ divides $n$.

Similarly, we see that $n$ divides $m$, and hence $m=n$.
Thus the orders of $ab$ and $ba$ are the same. Add to solve later

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