## The order of a product of commuting group elements divides the least common multiple of the orders of the elements

Let $G$ be a group with $x,y \in G$. Assume $|x| = n$ and $|y| = m$. Suppose that $x$ and $y$ commute; i.e., that $xy = yx$. Prove that $|xy|$ divides the least common multiple of $m$ and $n$. Need this be true if $x$ and $y$ do not commute? Give an example of commuting elements $x$ and $y$ such that the order of $xy$ is not equal to the least common multiple of $|x|$ and $|y|$.

Supposing that $xy = yx$, we saw in a previous theorem that $(xy)^k = x^ky^k$ for all $k$. In particular, $(xy)^{\mathsf{lcm}(|x|,|y|)} = (x^{|x|})^{|y|/\mathsf{gcd}(|x|,|y|)} (y^{|y|})^{|x|/\mathsf{gcd}(|x|,|y|)} = 1$. Thus by a previous exercise, $|xy|$ divides $\mathsf{lcm}(|x|,|y|)$.

Consider $S_3$. Note that $|(1\ 2)| = |(1\ 3)| = 2$, so that $\mathsf{lcm}(|(1\ 2)|, |(1\ 3)|) = 2$. However $(1\ 2)(1\ 3) = (1\ 3\ 2)$ has order 3, and 3 does not divide 2.

For a trivial example, consider $x \neq 1$ and let $y = x^{-1}$. Less trivially, consider $(x,y),(1,y) \in Z_2 \times Z_4$, where $Z_2 = \langle x \rangle$ and $Z_4 = \langle y \rangle$. Note that $|(x,y)(1,y)| = |(x,y^2)| = 2$, while $|(x,y)| = |(1,y)| = 4$. Thus $|(x,y)(1,y)| \neq \mathsf{lcm}(|(x,y)|,|(1,y)|)$.

• Matt  On September 20, 2010 at 11:12 am

For the example in D_{2(2k+1)}….

The only issue is that s and r do not commute =)

• nbloomf  On September 20, 2010 at 1:39 pm

Fixed- Thanks!

• Bill Baritompa  On July 19, 2011 at 6:10 pm

slight typo in proof … = (x^|x|)^(|y|/gcd(|x|,|y|))…

• nbloomf  On July 20, 2011 at 12:50 pm

Thanks!