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