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 $pq$.

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?

• No, that is not enough, because that is not what Cauchy's theorem says. Mar 15 '19 at 7:53
• Do you know how to form the product of two subgroups? Mar 15 '19 at 7:54

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 $$H,K, then $$|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$$.)
• 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. Mar 15 '19 at 10:26