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Can someone help me finish the proof? I am basing off this answer.

Let $G$ be an abelian group of order $pq$ with $\gcd(p,q)=1$. If there exist elements $a$ and $b$ such that $|a|=p$ and $|b|=q$, show that $G$ is cyclic.

After showing that $g^{pq}=e$, what do I do for the cases where $g=e$, $g^{p}=e$, and $g^{q}=e$? Any hints?

Shaun
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    You need more than $p,q$ coprime. After all we can write the order of any finite group as the product of two coprime integers. I suspect you mean to assume $p,q$ are distinct (positive) primes, and from this the conclusion follows. – hardmath Aug 19 '18 at 00:31
  • @hardmath I see! That helps a lot. For some reason, the question in the textbook does not state this assumption so it must be common knowledge. –  Aug 19 '18 at 16:36
  • My guess is that $p,q $ are assumed to be primes, perhaps throughout a chapter or section of the book. It’s easy to lose sight of when looking at an exercise or proposition in isolation. Citing the text would help. – hardmath Aug 19 '18 at 16:55

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Now $g=ab$.
Case 1: $g=e$. Then $a=b^{-1}$. Hence $p=|a|=|b^{-1}|=|b|=q$, a contradiction.
Case 2: $g^p=e$. That is, $$(ab)^p=e\implies a^pb^p=e\implies b^p=e$$ which means that $q$ divides $p$, a contradiction.
Case 3: $g^q=e$. Similar to case 2.

Wang Kah Lun
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