Question

Prove that if elements a,b of group G commutes, then LCM of their orders is multiply of order
of their product.

Answer 1

Question 1. Prove that if elements $a, b$ of group $G$ commute, then LCM of their orders is multiple of order of their product.

Solution. Suppose $a$ and $b$ have finite orders, say $m$ and $n$, respectively (for the infinite orders this statement does not make sense). Set $k = LCM(m,n)$. Since $k$ is divisible by $m$ and $n$, we conclude that

$$a^{k} = (a^{m})^{\frac{k}{m}} = e^{\frac{k}{m}} = e, \quad b^{k} = (b^{n})^{\frac{k}{n}} = e^{\frac{k}{n}} = e,$$

where $e$ denotes the identity of $G$. We are given that $a$ and $b$ commute, therefore

$$(ab)^{k} = a^{k}b^{k} = e \cdot e = e.$$

Thus, the order of $ab$ divides $k$, i.e. $k$ is a multiple of the order of $ab$.