Notation Let $p'$ denote the set of all primes other than $p$.
Let $G = p^a m$ with $p \not | m$, a Sylow $p$-subgroup $P$ has order $p^a$ whereas a Hall $p'$-subgroup $H$ of $G$ has order $m$. $H \cap P = 1$ and so $|G|=|H||P|$ and $G=HP$.
Lemma If $H,K \le G$ and $|G:H|,|G:K|$ are coprime then $|G:H \cap K| = |G:H||G:K|$ (note, this doesn't assume normality).
proof: Put $a=|G:H|$, $b=|G:K|$, $c=|G:H \cap K|$. Since $|G:H \cap K|=|G:H||H:H\cap K|$ we have $a|c$ and similarly $b|c$ thus $ab|c$. In the other direction we may define a map $$(H \cap K) x \mapsto (Hx,Kx) : \{\text{cosets of }H \cap K\} \to \{\text{cosets of }H\}\times\{\text{cosets of }K\}$$ this is well defined since $(H \cap K) x = (H \cap K) y$ iff $xy^{-1} \in H$ and $K$ iff $Hx=Hy$ and $Kx = Ky$. Since this map is injective $c \le ab$.
Definition Let $G$ be a group whose order factors into powers of primes $p_1, \ldots, p_k$. A Sylow basis for $G$ is a collection of Sylow subgroups $P_1,\ldots,P_k$ such that for all $i$, $P_i$ is a Sylow $p_i$-subgroup and for all $i,j$ $P_i P_j = P_j P_i$.
Any product of a subset of the Sylow basis will give a Hall $\pi$-subgroup and you can get a Hall $\pi$-subgroup for any $\pi$ this way.
Definition Two Sylow bases $P$ and $B$ are said to be conjugate when there exists a single $g$ such that forall $i$, $P_i = B_i^g$.
Theorem (Hall - 1937) If $G$ is solvable then it has a Sylow basis and any two such bases are conjugate.
proof: (Existence) Let $G$ be a solvable group and write $|G| = p_1^{a_1}\cdots p_k^{a_k}$, let $S = \{1,\ldots,k\}$. For each $i \in S$ let $Q_i$ be a Hall $p_i'$-subgroup so $|G:Q_i| = p_i^{a_i}$ by the previous theorem of Hall. Given any $T \subseteq S$ the intersection $\bigcap_{t \in T} Q_t$ is a Hall $\pi$-subgroup (for the appropriate $\pi = \{p_j | j \in S \setminus T \}$). In particular $P_i = \bigcap_{t \not = i} Q_t$ is a Hall $\{p_i\}$-subgroup (A Sylow $p_i$-subgroup) of $G$. To see that this gives a basis take $i,j \in S$ not equal and $P_i \cap P_j = 1$ by coprimality, therefore we have (considered as sets) $|P_i P_j| = |P_i||P_j| = |P_j P_i|$. Write $T = S \setminus \{p_i,p_j\}$ then $\bigcap_{t \in T} Q_t$ is a group that contains $P_i$ and $P_j$ hence $P_i P_j$ and $P_j P_i$ but $|\bigcap_{t \in T} Q_t| = p_i^{a_i} p_j^{a_j}$ as it's a Hall $\pi$-group!
(Uniqueness up to conjugacy) Let $B_1,\ldots,B_k$ be any other Sylow basis for $G$ with (by renumbering) $|B_i|=|P_i|$. For $t in S$ form the Hall $p_t'$-subgroup $C_t = \Pi_{i \not = t}B_i$. Instead of showing each $B_i$ is conjugate to $P_i$, we show that each $Q_i$ is conjugate to $C_i$ then deduce that. Let $d$ be the number of $t$ such that $C_t \not = Q_t$, and we prove by induction on $d$ that there exists some $g$ such that for every $t$, $C_t = Q_t^g$: the base case $d=0$ is trivial. Assume (by renumbering if necessary) that $C_t = Q_t$ for all $t > d$. Write $H = \bigcap_{t > d}Q_t$ by the uniqueness up to conjugacy of Hall $\pi$-subgroups for solvable groups there exists $x$ such that $C_d = Q_d^x$, since $Q_d$ contains each $P_i$ except $P_d$ - which lies in $H$ - $G = Q_d H$ and so $x = gh$ for some $g \in Q_d$, $h \in H$. Now $C_d = Q_d^{gh} = Q_d^h$ and for $t < d$ $C_t = Q_t = Q_t^h$ so there are at most $d-1$ values of $t$ (the $t < d$) such that $C_t \not = Q_t$ therefore we have by induction $z \in G$ such that $\forall t$, $C_t = (Q_t^h)^z = Q_t^{hz}$.
Finally for all $i$, $$B_i = \bigcap_{t \not = i}C_i = \bigcap_{t \not = i}Q_i^g = P_i^g.$$
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