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I’ve read in many places that Hall subgroups are a generalization of Sylow subgroups. However, if I understand correctly, every Hall $\pi$-subgroup of a finite group is also a Sylow $\pi$-subgroup (where $\pi$ is a non-empty subset of $\mathbb{P}$)... here’s my reasoning:

Let $H\in Hall_{\pi}G$, where $G$ is a finite group. Then $H$ is a $\pi$-subgroup such that $|G:H|$ is a $\pi’$-number ($\pi’=\mathbb{P} \setminus \pi$). Let $\pi(n)=\{p\in \mathbb{P} \mid p|n\}$.

Let $T \leq G$ be a $\pi$-subgroup such that $H \leq T \leq G$. Then $$|G:H|=|T:H||G:T|$$ and therefore $|T:H|$ will divide $|G:H|$. But $|T:H|$ divides |T|, and then $$\pi(|T:H|)\subseteq \pi(|T|)\subseteq \pi$$

Finally, $(|H|,|G:H|)=1$ implies $\pi \cap \pi(|G:H|)=1$, then $\pi(|T:H|)=1$, $|T:H|=1$ and finally $T=H$. We conclude that $H\in Syl_{\pi}G$.

However, according to this, every Sylow $p$-subgroup is a Hall $p$-subgroup. This would mean the two concepts are equivalent when $\pi=\{p\}$ and, more generally, Sylow subgroups would be a generalization of Hall subgroups and not the other way around. So I’m clearly confused here.

P.S.:

Here’s the definitions I’m using:

$A\leq G$ is a $\pi$-subgroup of $G$ if every element in $A$ has finite order and $\pi(o(a))\subseteq \pi$ $\forall a \in A$.

$A \leq G$ is a Sylow $\pi$-subgroup of $G$ if

  1. $A$ is a $\pi$-subgroup of $G$,

  2. if $A\leq B \leq G$ such that $B$ is also a $\pi$-subgroup of $G$, then $A=B$.

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    $\begingroup$ I have never heard of a Sylow $\pi$-subgroup, only of a Sylow $p$-subgroup for a prime $p$. Yes, for finite groups, a Hall $\{ p\}$-subgroup is the same thing as a Sylow $p$-subgroup. I would agree with the assertion that the concept of a Hall $\pi$-subgroup is a generalization of Sylow $p$-subgroup to the case when there is more than one prime in $\pi$. $\endgroup$
    – Derek Holt
    Jun 14, 2021 at 12:46
  • $\begingroup$ @DerekHolt I’ve added to the question the definitions I’m using. I’m thinking maybe both Sylow and Hall $\pi$-subgroups are generalizations of Sylow $p$-subgroups in the sense of your comment then. $\endgroup$
    – dahemar
    Jun 14, 2021 at 12:55
  • $\begingroup$ @DerekHolt Could you make the comment into an answer? $\endgroup$
    – Pedro
    Jun 14, 2021 at 12:56

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I am reluctant to answer this question, because I have no particular expertise in the history of these various definitions. I suspect your confusion is resulting from later attempts to extend various definitions to infinite groups.

Let's start with finite groups. For a set $\pi$ of primes, we could define a Hall $\pi$-subgroup $H$ of $G$ to be a subgroup of order $m$ and index $n$ in $G$, where $m$ is divisible only by primes in $\pi$ and $n$ only be primes not in $\pi$. So, for prime $p$, a Hall $\{p\}$-subgroup is the same as a Sylow $p$-subgroup.

The results of Philip Hall extend Sylow's theorems to (finite) solvable groups, and assert that for such groups and for all sets of primes $\pi$ Hall $\pi$-subgroups exist and are conjugate, and are also the maximal $\pi$-subgroups under inclusion. (There are counterexamples showing that none of these properties extend to finite non-solvable groups.)

Given these theorems, there are other possible definitions of Hall $\pi$-subgroups, whicht are equivalent for finite solvable groups. For example, you could define a Hall $\pi$-subgroup to be a maximal $\pi$-subgroup. If you did that, then all finite groups would of course have Hall $\pi$-subgroups, but they would not generally be conjugate, or satify the order and index condition in the original definition. So it's probably a bad idea to do that! But that definition is equivalent to your definition of a Sylow $\pi$-subgroup, which I have not come across before.

There have also been a variety extensions of Sylow $p$-subgroups to infinite groups, which can also be extended to $\pi$-subgroups.

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  • $\begingroup$ Thank you for your answer, it was helpful! The definition I gave of Sylow $\pi$-subgroup was used several times in the lectures notes of the Group Theory course I’ve taken (they’re unfortunately not in pdf), for instance in the proof of Hall’s theorem. I’m guessing this definition is a non-standard generalization of the Hall $\pi$-subgroup one. I think my confusion is cleared up now. $\endgroup$
    – dahemar
    Jun 14, 2021 at 14:37

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