Variations on Hall Theorem for solvability in Dummit and Foote In "Abstract Algebra" by Dummit and Foote p.105 there is a theorem:

The finite group $G$ is solvable if and only if for every divisor $n$
of $|G|$ such that $(n, \frac{|G|}{n}) = 1$, $G$ has a subgroup of
order $n$.

It refers to the Hall theorem p.196:

If for every prime $p$ dividing $|G|$ we factor the order of $G$ as
$|G| = p^a m$ where $(p, m) = 1$, and $G$ has a subgroup of order $m$,
then $G$ is solvable (i.e., if for all primes $p$, $G$ has a subgroup
whose index equals the order of a Sylow $p$-subgroup, then $G$ is
solvable - such subgroups are called Sylow $p$-complements).

The two theorems look similar. But, take a group $G$ which order can be decomposed as the product of primes : $|G| = p_1 p_2 p_3$.
In the first theorem the solvability is equivalent to the existence of subgroups of orders $\{p_1, p_2, p_3, p_1 p_2, p_2 p_3, p_1 p_3\}$ while the second theorem only require the existence of subgroups of orders $\{p_1 p_2, p_2 p_3, p_1 p_3\}$.
My question: is the first theorem just a stronger version of the second one, with more requirements, but also providing an equivalence instead of a sufficient condition for solvability ? Or am I missing some other similarities / differences ?
 A: A Hall $\pi$-subgroup of a finite group $G$ is a subgroup $H$ of $G$ such that $H$ is a $\pi$-group and no prime in $\pi$ divides $|G:H|$. Hall's theorem is that a finite group $G$ is soluble if and only if $G$ possesses Hall $p'$-subgroups for all primes $p$, where $p'$ is the set of all primes other than $p$. If $p\nmid |G|$ then a Hall $p'$-subgroup is simply $G$ itself, so this is only interesting for $p\mid |G|$.
Suppose that $G$ is soluble and let $\pi$ be a set of primes. We prove by induction that $G$ has a Hall $\pi$-subgroup. For $p\in \pi$, $G$ possesses a Hall $p'$-subgroup $H$. But $|H|<|G|$ and $H$ is soluble, so $H$ has a Hall $\pi$-subgroup $K$. Notice that $K$ is also a Hall $\pi$-subgroup of $G$, and we are done.
Thus we see that all finite soluble groups $G$ possess Hall $\pi$-subgroups for all $\pi$, not just Hall $p'$-subgroups, from just knowing about Hall $p'$-subgroups. Obviously if $G$ possesses Hall $\pi$-subgroups for all $\pi$ then it possesses Hall $p'$-subgroups.
