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I have an abelian group $G$ and two cyclic subgroups of orders $p,q$, $( p,q)=1$ and I need to show I have a subgroup of order $pq$. Is it enough to say that, from Cauchy's theorem, there is an $a\in G$ of order $pq$ and then build a cyclic subgroup with elements from the previously shown subgroups? What are the elements I should take?

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  • $\begingroup$ No, that is not enough, because that is not what Cauchy's theorem says. $\endgroup$ Mar 15 '19 at 7:53
  • $\begingroup$ Do you know how to form the product of two subgroups? $\endgroup$
    – the_fox
    Mar 15 '19 at 7:54
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Hint: If $x$ has order $p$ and $y$ has order $q$, what is the order of $xy$?

In fact, we do not even need that the given subgroups are cyclic. In most textbooks (e.g. Dummit and Foote), you will find the more general proposition that if $HK = KH$ for subgroups $H,K<G$, then $$ |HK|=\frac{|H||K|}{|H\cap K|}. $$ The conditions on $H$ and $K$ are trivially satisfied since $G$ is abelian. Also, $H\cap K$ is a subgroup of both $H$ and $K$, so its order must divide $|H|$ and $|K|$. But, when $|H|$ and $|K|$ are relatively prime, this implies $|H\cap K |=1$.

(As pointed out by @Tobias in his comment, the identity holds without the condition that $HK = KH$; the condition ensures that $HK$ is a subgroup of $G$.)

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    $\begingroup$ Actually, the condition that $HK = KH$ is only needed for the product to be a subgroup. The size of the product will always be given by that formula. $\endgroup$ Mar 15 '19 at 10:26

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