# Showing abelian group order that is the product of primes is cyclic

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?

• Take a look at the methods used here and generalize them to this case. Commented Oct 30, 2015 at 17:59

Lemma: Let $m$ and $n$ be relatively prime, and suppose $a$ has order $m$ and $b$ has order $n$. Then $ab$ has order $mn$.
Proof: Let $k$ be the order of $ab$. It is clear that $k\le mn$. We have $$b^{km}=(a^{km})(b^{km})=(ab)^{km}=e,$$ and therefore $n$ divides $km$. But $m$ and $n$ are relatively prime, so $n$ divides $k$.
Similarly, $m$ divides $k$, and since $m$ and $n$ are relatively prime, it follows that $mn$ divides $k$, and therefore $k=mn$.
Remark: One can use the same argument to show directly that if $g_i$ has order $p_i$, $i=1$ to $t$, then $g_1g_2\cdots g_t$ has order $p_1p_2\cdots p_t$. But I prefer to go through the Lemma, and then, in principle, induction.