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I'm afraid at this ungodly hour I've found myself a bit stumped. I'm attempting to answer the following homework question:

If $p_1,\dots,p_s$ are distinct primes, show that an abelian group of order $p_1p_2\cdots p_s$ must be cyclic.

Cauchy's theorem is the relevant theorem to the chapter that precedes this question...

So far (and quite trivially), I know the element in question has to be the product of the elements with orders $p_1,\dots, p_s$ respectively. I've also successfully shown that the order of this element must divide the product of the $p$'s. However, showing that the order is exactly this product (namely that the product also divides the orders) has proven a bit elusive. Any helpful clues/hints are more than welcome and very much appreciated!

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up vote 3 down vote accepted

Start by proving that, in an abelian group, if $g$ has order $a$, and $h$ has order $b$, and $\gcd(a,b)=1$, then $gh$ has order $ab$. Clearly, $(gh)^{ab}=1$, so $gh$ has order dividing $ab$. Now show that if $(gh)^s=1$ for some $s\lt ab$ then you can find some $r$ such that $(gh)^{rs}$ is either a power of $g$ or of $h$ and not the identity. Details left to the reader.

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