Theorem: Schur-Zassenhaus Theorem

Let $G$ be a finite group and $N$ be a normal subgroup in $G$.
Let $N$ be a Hall subgroup of $G$.
Then there exists $H$, a complement of $N$, such that $G$ is the semidirect product of $N$ and $H$.

Proof:
The proof proceeds by induction.
By definition, $N$ is a Hall subgroup {{iff}} the index and order of $N$ in $G$ are relatively prime numbers.
Let $G$ be a group whose identity is $e$.
We induct on $\order G$, where $\order G$ is the order of $G$.
We may assume that $N \ne \set e$.
Let $p$ be a prime number dividing $\order N$.
Let $\Syl p N$ be the set of Sylow $p$-subgroups of $N$.
By the First Sylow Theorem:
:$\Syl p N \ne \O$
Let:
: $P \in \Syl p N$
: $G_0$ be the normalizer in $G$ of $P$
: $N_0 = N \cap G_0$.
By Frattini's Argument:
:$G = G_0 N$
By the Second Isomorphism Theorem for Groups and thence Lagrange's Theorem (Group Theory), it follows that:
: $N_0$ is a Hall subgroup of $G_0$
: $\index {G_0} {N_0} = \index G H$
Suppose $G_0 < G$.
Then by induction applied to $N_0$ in $G_0$, we find that $G_0$ contains a complement $H \in N_0$.
We have that:
:$\order H = \index {G_0} {N_0}$
and so $H$ is also a complement to $N$ in $G$.
So we may assume that $P$ is normal in $G$ (that is: $G_0 < G$).
{{explain|What is the significance of $G_0 < G$ here? It has already been stated above. What is its purpose at this point in the argument?}}
Let $Z \paren P$ be the center of $P$.
By:
:Center is Characteristic Subgroup
:$P$ is normal in $G$
:Characteristic Subgroup of Normal Subgroup is Normal
$Z \paren P$ is also normal in $G$.
Let $Z \paren P = N$.
Then there exists a long exact sequence of cohomology groups:
:$0 \to H^1 \paren {G / N, P^N} \to H^1 \paren {G, P} \to H^1 \paren {N, P} \to H^2 \paren {G / N, P} \to H^2 \paren {G, P}$
which splits as desired.
{{explain|The link leads to "exact sequence". An explanation is needed as to what a "long exact sequence" is.}}
{{explain|The definition of $P^N$ in this context}}
{{explain|The definition of "splits" in this context}}
Otherwise:
:$Z \paren P \ne N$
In this case $N / Z \paren P$ is a normal (Hall) subgroup of $G / Z \paren P$.
By induction:
:$N / Z \paren P$ has a complement $H / Z \paren P$ in $E // Z \paren P$.
{{explain|The meaning of $E // Z \paren P$ and definition of $E$ in this context.}}
{{Explain|Although it is stated that this proof is by induction, it is unclear what the base case, induction hypothesis and induction step actually are.}}
Let $G_1$ be the preimage of $H // Z \paren P$ in $G$ (under the equivalence relation).
{{explain|Under what equivalence relation?}}
Then:
:$\order {G_1} = \order {K / Z \paren P} \times \order {Z \paren P} = \order {G / N} \times \order {Z \paren P}$
{{explain|The definition of $K$ in $\order {G_1} {{=}} \order {K / Z \paren P} \times \order {Z \paren P} {{=}} \order {G / N}\ \times \order {Z \paren P}$.}}
Therefore, $Z \paren P$ is normal Hall subgroup of $G_1$.
By induction, $Z \paren P$ has a complement in $G_1$ and is also a complement of $N$ in $G$.
{{Explain|Again, although it is stated that this proof is by induction, it is unclear what the base case, induction hypothesis and induction step actually are.}}
{{qed}}