# 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.

• Perhaps you could supply a few more definitions. What is a Sylow $\pi$-subgroup when $\pi$ is not prime? Is it the same as a Hall $\pi$-subgroup? Dec 3, 2019 at 11:22
• Also, you should probably explain what you mean by a Sylow subgroup in an infinite group Dec 3, 2019 at 11:36
• Let a group G and a set of primes $\pi$. 1) An element $g\in G$ is said a $\pi$-element if $|\langle g\rangle|$ is finite and if prime divisors of $|\langle g\rangle|$ are in $\pi$. 2) G is said a $\pi$-group if every element of G is a $\pi$-element. 3) A Sylow $\pi$-subgroup of G is a maximal element in the ordered set ({P$\le$ G: P is a $\pi$-group},$\subseteq$) Dec 3, 2019 at 12:16
• Yes, that is the usual definition, but not everyone would be familiar with it (especially with $\pi$ in place of $p$), and it might have been better include in the body of the question. Dec 3, 2019 at 12:46
• I found your quotation in a paper of Hartley, but he was assuming that the groups were locally finite, in which case 2 follows immediately from 1. I haven't found a reference for 1 yet, although you can easily reduce to the case of a simple group in which all proper subgroups are solvable. Dec 3, 2019 at 13:24

## 1 Answer

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.)