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Let $\mathbb{Z}_p\lhd H\leq\text{AGL}(1,p),\mathbb{Z}_q\lhd K\leq\text{AGL}(1,q)$ with $p,q$ prime. Let $G=H.K$. Can one show that $G$ contains a normal subgroup of order $pq$?

Note: Here $G=H.K$ means $H\lhd G$ and $G/H\cong K$. And $p,q$ are distinct.

The following is some of my attempt: \begin{align} G=&(\mathbb{Z}_p{:}\mathbb{Z}_r).(\mathbb{Z}_q{:}\mathbb{Z}_s)\\ =&((\mathbb{Z}_p{:}\mathbb{Z}_r).\mathbb{Z}_q).\mathbb{Z}_s\\ =&(\mathbb{Z}_p{:}(\mathbb{Z}_r.\mathbb{Z}_q)).\mathbb{Z}_s\\ =&\mathbb{Z}_p.((\mathbb{Z}_r.\mathbb{Z}_q).\mathbb{Z}_s). \end{align} Applying N/C-Theorem w.r.t $\mathbb{Z}_p$, we have $G/\text{C}_G(\mathbb{Z}_p)$ is cyclic. Then $\text{C}_G(\mathbb{Z}_p)$ should somehow contains $\mathbb{Z}_q$. But I am not sure what is the next...

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I guess you are assuming that $s \ne 1$? In that case, ${\mathbb Z}_q = K'$, and so $q$ divides $|G'|$. But $G/C_G({\mathbb Z}_p)$ cyclic implies that $G' \le C_G({\mathbb Z}_p)$. –  Derek Holt Mar 18 '13 at 8:44
    
@DerekHolt, $\mathbb{Z}_p$ and $\mathbb{Z}_q$ are both proper subgroups. Of course we don't want the trivial counterexample. –  Easy Mar 18 '13 at 8:45
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Note also that $G' \le ({\mathbb Z}_p : {\mathbb Z}_r).{\mathbb Z}_q$. But the subgroupof order $r$ does not lie in $C_G({\mathbb Z}_p)$ and hence not in $G'$, so in fact $|G'|=pq$ and $G'$ is your normal subgroup. Sorry, I don't have time now to write this as a proper answer! –  Derek Holt Mar 18 '13 at 8:55
    
@Derek, I am feeling it is close, but still not sure how to write it up. Also, I wonder if $G=H.K$, then is it true that $G'=H'.K'$? If we have this, the rest should be pretty easy. –  Easy Mar 18 '13 at 14:12
    
Yes, $G' = H'.K'$. Clearly $H.K' \le G'$, and clearly $G' \le H.K'$. But since $G' \le C_G({\mathbb Z}_p)$ and $H \cap C_G({\mathbb Z}_p) = H'$, we must have $G' = H'.K'$. –  Derek Holt Mar 18 '13 at 22:57
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