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.

  • 3
    $\begingroup$ 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? $\endgroup$
    – Derek Holt
    Dec 3, 2019 at 11:22
  • $\begingroup$ Also, you should probably explain what you mean by a Sylow subgroup in an infinite group $\endgroup$ Dec 3, 2019 at 11:36
  • 1
    $\begingroup$ 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},$\subseteq$) $\endgroup$
    – user149426
    Dec 3, 2019 at 12:16
  • 1
    $\begingroup$ 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. $\endgroup$ Dec 3, 2019 at 12:46
  • $\begingroup$ 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. $\endgroup$
    – Derek Holt
    Dec 3, 2019 at 13:24

1 Answer 1


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.)


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