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Let $G$ be a group that contains normal subgroups of prime orders $p$ and $q$, respectively, where $p$ and $q$ are different. Prove that $G$ contains an element of order $pq$.

I tried using Lagrange's theorem but I'm not sure if it applies since it isn't said that $G$ is finite. Can it be shown that this is the case? i.e. Does a group with non-trivial subgroups of finite order necessarily have finite order itself?

If it does hold, using Lagrange's theorem I could say that $pq$ divides $|G|$, but I'm not sure what follows. Can I then somehow show that there exists a subgroup in $G$ of order $pq$, and so there must be some element in $G$ of order $pq$ (since the group operation for the subgroup is the same as that of $G$)?

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    $\begingroup$ If you have a group of order a prime $p$ what can you say about that group? $\endgroup$ – Alex J Best Aug 19 '13 at 15:58
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    $\begingroup$ Hint: Consider the subgroup $HK$, where $H$ and $K$ are the normal subgroups of order $p$ and $q$ respectively. $\endgroup$ – Prahlad Vaidyanathan Aug 19 '13 at 16:03
  • $\begingroup$ No one else has addressed this, so I'll go ahead and make a comment. "I tried using Lagrange's theorem but I'm not sure if it applies since it isn't said that G is finite…" What were you inclined to say if $G$ were finite? That is, assume that you are working with finite groups to begin with. How would you use Lagrange's theorem? $\endgroup$ – the_fox Aug 19 '13 at 20:21
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Hint: let $H=<a>$ where $|a|=p$ and $K=<b>$ where $ |b|=q$. Look at $aba^{-1}b^{-1}$, use normality and the fact that $H\cap K=\{e\}$ (Relatively prime to say something about the commutativity of the element $ab$. What can we say about $<ab>?$

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  • $\begingroup$ Ohhh. So the order of $\langle ab \rangle$ will be $lcm(|a|,|b|)=lcm(p,q)=pq$. Since $a$ and $b$ commute we can write $(ab)^n=abab...abab$ ($n$ times) which we reorder as $aa...aabb...bb$ which will have order divisible by both $p$ and $q$. The commutator has popped up in a few proofs that I've seen lately and it has been very useful, I may have to start trying to use it myself! Thanks. $\endgroup$ – hotstuff69 Aug 19 '13 at 16:43
  • $\begingroup$ Exactly. I am glad that helped. The commutator is an extremely useful tool. $\endgroup$ – LASV Aug 20 '13 at 4:49
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Hints:

$$P,Q\lhd G\;,\;\;|P|=p\;,\;|Q|=q\implies PQ\cong P\times Q\lhd G$$

Now, since $\,PQ\,$ abelian and $\,P=\langle p\rangle\;,\;\;Q=\langle q\rangle\;$ and $\,ord(pq)=ord(p)\cdot ord(q)\;$ (why?) , thus we're done...

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Let $A=\langle a\rangle,B=\langle b\rangle$ be normal subgroups of $G$ of order $p,q$, respectively. Then $(ab)^n\in a^nB=B$ and $(ab)^n\in b^nA$ for all $n$, especially $(ab)^{pq}\in A\cap B=\{1\}$, i.e. the order of $ab$ is one of $1,p,q,pq$. Why can it be neither $1$, nor $p$, nor $q$?

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