Exercise 2.3.16 (Order of a product $xy$)

Question

Assume that $|x| = n$ and $|y| = m$. Suppose that $x$ and $y$ {\bf commute}: $xy = yx$. prove that $|xy|$ divides the least common multiple of $m$ and $n$. Need this be true if $x$ and $y$ do {\bf not} commute? Give an example of commuting elements $x,y$ such that the order of $xy$ is not equal to the least common multiple of $|x|$ and $|y|$.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

Let $n \vee m$ denote the least common multiple of $n$ and $m$.
\begin{proof} 
The l.c.m. $n \vee m$ is a common multiple of $n$ and $m$, so $n\vee m = \lambda n,\ n\vee m = \mu m$ for some integers $n,m$.

Since $x^n = 1$ and $y^m = 1$, then $x^{n \vee m} = (x^n)^\lambda = 1$ and $y^{n \vee m} = (y^m)^\mu = 1$.

Moreover, $x$ and $y$ commute. Therefore $(xy)^k = x^k y^k$ for all integers $k$. Hence
$$(xy)^{n \vee m} = x^{n \vee m} y^{n \vee m} = 1.$$
Therefore $|xy|$ divides $n \vee m$.

\bigskip
As a counterexample if $x$ and $y$ do not commute, take $x = (1\ 2\ 3)$ and $y = (1\ 2 \ 4)$ in the group $S_4$. Then $ |x| = |y| = 3$, so $|x| \vee |y| = 3$. But
$$x y = (1\ 2\ 3)  (1\ 2 \ 4) = (1\ 3) (2\ 4)$$
has order $2$, so $2 = |xy| 
mid |x| \vee |y| = 3$.

\bigskip
Finally, consider the elements $\overline{2}, \overline{3} \in \Z/5\Z$. Then $\overline{2}$ and $\overline{3}$ commute, but $|\overline{2}| = |\overline{3}| = 5$, and $|\overline{2}\cdot \overline{3}| = |\overline{1}| = 1$ is not the least common multiple of $n = 5$ and $m = 5$.
\end{proof}

Exercise 2.3.16 (Order of a product $xy$)

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Exercise 2.3.16 (Order of a product $xy$)

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