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Let $G$ be an abelian group of order $pq$, where $p$ and $q$ are two distinct prime numbers. Prove that $G$ is cyclic. (Hint: use a theorem of Sylow: if $G$ is a group of order $n$ and $p$ is prime, and if $p$ divides $n$, then $G$ has en element of order $p$.)

Attempt at proof: Since $p$ divides $pq$, Sylow says that $G$ has an element $g_1$ with order $p$. Since $q$ divides $pq$, $G$ has also an element $g_2$ with order $q$. Now consider the element $g = g_1 g_2$. Then this element has order $pq$, since $G$ is abelian we have $$g^{pq} = (g_1 \ast g_2)^{pq} = (g_1^{pq} \ast g_2^{pq}) = (e^q \ast e^p) = (e \ast e) = e. $$ So I have shown that $G$ has an element with the same order as the order of the group. Can I conclude from this that $G$ is cyclic? Or do I still need to show explicitly that $g$ is a generator for $G$?

Kamil
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    You've only shown that $g^{pq} = e$, not that the order of $g$ is $pq$. The order could be anything which divides $pq$, so you still have to rule out the possibilities that $g = e$, $g^p = e$ and $g^q = e$. – Dylan Aug 09 '16 at 13:44
  • Ah, I see, thanks. Once I have shown that, the proof is complete? Is it always the case that if $g$ has the same order as the group $G$, then $g$ is a generator for $G$? – Kamil Aug 09 '16 at 13:55
  • Yes, if $g$ and $G$ both have order $n$, then $e, g, g^2, g^3, \dots, g^{n-1}$ are $n$ distinct elements of the group, and so must be the whole group. – Dylan Aug 09 '16 at 14:04

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