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Let $G$ be a finite solvable group and denote by $\pi(G)$ the set of prime divisors of $|G|$.

Suppose there is a $p\in \pi(G)$ for which there is an element of order $pq$ in $G$ for every $q\in \pi(G)\setminus\{p\}$. If we assume that $P\in \operatorname{Syl}_p(G)$ is not a direct factor of $G$, is it true that there exists a subgroup $H\leqslant G$ such that

  1. $\pi(H)=\pi(G)$, and

  2. there exists a $q\in \pi(H)\setminus\{p\}$ such that there is no element of order $pq$ in $H$?

Note that we may assume without loss of generality that every Sylow subgroup of $G$ is elementary abelian (though I'm not sure whether this will help).

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I believe you have to add some extra hypothesis, because it seems to me that in the current formulation $G = C \times S_3$ is a counterexample, with $p=3$ and $H = S_3$. –  Andreas Caranti Feb 4 '13 at 7:29
    
@AndreasCaranti Why is that a counterexample? $S_3$ has no element of order $6$, so that's the $H$, with $p=3$ and $q=2$. –  Samuel Handwich Feb 4 '13 at 7:30
    
Sorry, I misread the question. –  Andreas Caranti Feb 4 '13 at 7:46
    
@AndreasCaranti No problem. Thanks for reading it at all! –  Samuel Handwich Feb 4 '13 at 7:47
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1 Answer

up vote 1 down vote accepted

I think that ${\rm SL}(2,3)$ is a counterexample with $p =3.$ The only proper subgroup of order divisible by $6$ is cyclic of order $6,$ and contains an element of order two in its centre.

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