Sylow subgroups conjugates in a soluble group Recall that a group G is said:
1) Sylow $\pi$-connected if the Sylow $\pi$-subgroups of G are conjugates in G.
2) Sylow $\pi$-integrated if every subgroup of G is Sylow $\pi$-connected.
3) Completely Sylow integrated if G is Sylow $\pi$-integrated for every set of primes $\pi$.
Now i found in some work (for example in Dixon or Hartley) that:
1) ''For by a well known theorem of P. Hall, a finite completely Sylow
integrated group is soluble...''
2) ''...and so an arbitrary completely Sylow integrated group is locally soluble''. 
I don't found nowhere a proof for 1); for 2) i must consider a finitely generated (and no necessarily finite) subgroup of a completely Sylow integrated group for proof the locally solubility? 
(pheraps G is assumed locally finite?)
Help me, please.
 A: I think Question 1 reduces easily to the genuinely well known result of Philip Hall, that if the finite group $G$ has a $p$-complement for all prime divisors $p$ of $|G|$, then $G$ is solvable.
(Recall that a $p$-complement is a subgroup of order $t$, where $|G|=p^kt$ and $p$ does not divide $t$.)
To see that, let $\pi$ be the set of prime divisors of $|G|$, and let $p \in \pi$.
Now let  $\pi' =\pi \setminus \{p\}$, and let
$Q \in {\rm Syl}_q(G)$ with $q \in \pi'$. Then $Q$ is a $\pi'$-group, so $Q$ is contained in some Sylow $\pi'$-subgroup $R$ of $G$. This applies to all $q \in \pi'$, and by hypothesis all Sylow $\pi'$-subgroups are conjugate. So $|R|$ contains a Sylow $q$-subgroup of $G$ for all $q \in \pi'$, and hence $R$ is a $p$-complement in $G$. Now we can apply Hall's theorem to deduce solvability of $G$.
As I said in my comment, I think Hartley was assuming that the groups in question are locally finite, in which case 2 follows immediately from 1. (In any case without some extra assumption, a Tarski Monster would be a counterexample to 2.)
