This is question 13.2 in Armstrongs "Groups and Symmetry". I've googled quite a bit but I can't find an overview of the solutions of the the book. Variations of this particular question come back in multiple instances and I can't really figure it out.
"If $p_1,p_2,...p_s$ are distinct primes, show that an abelian group of order $p_1p_2...p_s$ must be cyclic."
Now, to show this I want to find an element with order $p_1p_2...p_n$. By Cauchy's theorem, the group contains elements $g_1, g_2$ ..., of order $p_1$, $p_2$, etc. and it is clear that the product of these elements has an order that must be a divisor of $p_1p_2...p_n$. However, I am not sure how to show that the order of this element actually is equal to $p_1p_2...p_n$; i.e., how to make sure there are no cancellations before that?