Proof of Hall's subgroup Theorem So I'm working through Hall's Theorem for Solvable groups and there is one part of it which I cannot seem to prove. I am following through Isaac's book on Finite group Theory for reference. Currently I can prove Hall-E (existence of a Hall-$\pi$-subgroup), Hall-C (any two such are conjugate) however the proof of Hall-D eludes me. For those without the book it is stated as; Let $U \subseteq G$ be a $\pi$ subgroup, where $\pi$ is a set of primes and $G$ is a finite solvable group. Then $U$ is contained in some Hall-$\pi$-subgroup of $G$.

I don't know how the proof in Isaacs' book goes but let's try a standard induction argument.

Since $G$ is soluble, it has a normal subgroup $N$ which is an elementary abelian $p$-group for some prime $p$ and, by induction we may assume that the result is true in $G/N$, so $UN/N$ is contained in a Hall $\pi$-subgroup $H/N$ of $G/N$.

There are two cases $p \in \pi$ and $p \
ot\in \pi$. If $p \in \pi$ then $H$ is a Hall $\pi$-subgroup of $G$ and $U \le H$, so we are done.

If $p \
ot\in \pi$, then by the existence part applied to $H$, there is a Hall $\pi$-subgroup $J$ of $H$ (and hence also of $G$) that complements $N$ in $H$. Then $J(NU) = JN = H$, so $|J \cap NU| =|J||N||U|/|H| = |U|$. Now, by the conjugacy part applied to the complements $J \cap NU$ and $U$ of $N$ in $NU$, there is $h \in H$ with $(J \cap NU)^h = U$, so $U$ is contained in the Hall $\pi$-subgroup $J^h$ of $G$.