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)|).

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Comments

  • 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|))…

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