Abelian group order divides lcm

I want to prove the following:

Let $G$ be an abelian group and let $a_1,...,a_r \in G$ then $|a_1...a_r| $ (the order of $a_1...a_r$ ) divides $ {\rm lcm}(|a_1|,...,|a_r|) $

So far I have shown that $(a_1,...,a_r)^{{\rm lcm}(|a_1|,...,|a_r|) } = 1 $ but I don't think this implies the desired result.

Any help? Thank you in advance

Solution 1:

It does imply the desired result. Note that if $a^n = 1,$ then $|a|$ divides $n.$ Simply write $n = |a|q + r$ for $0 \leq r < |a|,$ and then $$1 = a^n = a^{|a|q} \cdot a^r = 1^q \cdot a^r = a^r,$$ and so since $r < |a|$ and $a^r = 1,$ we have $r= 0.$