Schur-Zassenhaus Theorem Im a reading Mac Lane's Book on Homology and now he wants to prove the Schur-Zassenhaus Theorem.

If the integers $m$ and $n$ are relatively prime, any extension of a group of order $m$ by one of order  $n$ splits.

So the takes any  extension $0 \rightarrow G \rightarrow B \rightarrow C \rightarrow 0 $, a prime $p$ dividing $m$ and maximal $p$-subgroup $P$ of $B$. He then goes on to say that "all the conjugates of $P$ must lie in $G$ and are maximal $p$-subgroups there", now why is this true ? I have thinking about this but i cant seem to see why, maybe because $p$ divides $m$ but i cant come up with an argument to convice myself completely so any help is appreciated.Thanks in advance.
 A: So it doesn't stay unanswered: every $p$-element must lie in the kernel of $B\to C$, which is $G$; otherwise, the image would be a nontrivial $p$-element in $C$, but $|C|$ is prime to $p$.

This is Theorem 10.5, pp. 133 of the Springer Classics in Mathematics series.
Added (to an addition now, apparently, deleted): you don't say, but you are looking at the case where $|G|=m$ and $|C|=n$. What's more, you are in the middle of an inductive argument on $|G|$, with the case of an abelian group $G$ already taken care of (regardless of the order).
In addition, $P$ is a maximal $p$-subgroup of $B$ (that is, a Sylow $p$-subgroup of $B$) and $N=N_B(P)$, the normalizer of $P$ in $B$. Note that $P$ is also a Sylow $p$-subgroup of $G$, so any if $K$ is another Sylow $p$-subgroup of $B$, then $K$ is not just conjugate to $P$ in $B$, but in fact it is conjugate to $K$ in $G$.
Now, $N\cap G=N_G(P)$ is the normalizer of $P$ in $G$; note that the index of $N_G(P)$ in $G$ is the number of Sylow $p$-subgroups of $G$, which is the same as the number of Sylow $p$-subgroups of $P$ in $B$, which is the index of $N$ in $B$. That gives the equality $[B:N]=[G:G\cap N]$. 
The claim is that $n=[B:G]=[N:G\cap N]$. Now, it is certainly the case that $n=[B:G]$, because $B/G\cong C$. As to why $[N:G\cap N]=[B:G]$, this follows because 
$$[B:G\cap N] = [B:G][G:G\cap N] = [B:N][N:G\cap N]$$
and we know that $[B:N]=[G:G\cap N]$, hence $[B:G]=[N:G\cap N]$. 
