We need to prove the following statement:
Let $A$ be an abelian group, $x_1,\dots,x_h \in A$ and let $q_1,\dots,q_h$ be prime numbers. Suppose that, for all $1 \leq i \leq h$, there exists $k_i \in \mathbb{N}$ such that $o(x_i) = q_i^{k_i}$. Then $o(x_1 + \dots + x_n) = \prod_{i = 1}^h q_i^{k_i}$.
First, since we're assuming that $q_1,\dots,q_h$ are prime numbers, we have that $\gcd(q_1^{k_1},\dots,q_h^{k_h}) = 1$. Thus $\text{lcm}(q_1^{k_1},\dots,q_h^{k_h}) = \prod_{i=1}^h q_i^{k_i}$. Obviously, we have that: \begin{equation} \prod_{i = 1}^h q_i^{k_i} (x_1 + \dots + x_h) = 0 \end{equation} This implies that $o(x_1+\dots+x_n)$ divides $\prod_{i=1}^h q_i^{k_i}$. Now I'm stuck. How to complete the proof?
