|x| = order of x
Assume |x|=n and |y|=m. Suppose that x and y commute: xy = yx. Prove that |xy| divides the least common multiple of m and n.
Can anyone help?
if $\displaystyle \text{lcm}(m,n)=\ell,$ then obviously $\displaystyle x^{\ell}=y^{\ell}=1.$ thus since $\displaystyle xy=yx,$ we have: $\displaystyle (xy)^{\ell}=x^{\ell}y^{\ell}=1.$ hence $\displaystyle |xy| \mid \ell. \ \ \Box$
Remark: here are three related questions: suppose G is an abelian group and $\displaystyle |x|=n, \ |y|=m.$
Q1: show that it's possible to have $\displaystyle |xy| \neq \text{lcm}(m,n).$ (by giving an example!)
Q2: what can we say about $\displaystyle |xy|$ if $\displaystyle \gcd(m,n)=1$?
and a much more interesting question:
Q3: show that G has an element of order $\displaystyle \text{lcm}(m,n).$