Let $p$ be a prime and let $A$ be an abelian normal $p$-subgroup of a finite group $G$. Hence, for any Sylow $p$-subgroup $H$ of $G$, it holds that $A$ is contained in $H$ and so $A$ is a normal subgroup of any Sylow $p$-subgroup of $G$. So now fix a Sylow $p$-subgroup $P$ of $G$.
How to show that the following two conditions are equivalent?
$1$. There is a split exact sequence $1\to A \to G \to G/A \to 1$.
$2$. There is a split exact sequence $1\to A \to P \to P/A \to 1$.
My try: I think I can prove that $1.$ implies $2.$ Indeed, if $1.$ holds then there is a subgroup $B$ of $G$ such that $B\cong G/A$ and $G=BA$ and $B\cap A=1$. So then $P=P\cap (BA)=(P \cap B)A$, and we moreover have $(P\cap B)\cap A \subseteq B \cap A=1$. Thus $P$ is a semidirect product of $A$ and $P\cap B\cong P/A$. Thus the extension $1\to A \to P \to P/A \to 1$ splits. Is this correct? And I have no idea how to prove that $2.$ implies $1.$
Please help.
What you are asking about is a standard theorem of Gaschütz. You may want to consult the book by Kurzweil and Stellmacher "The Theory of Finite Groups: An Introduction". It is Theorem $3.3.2$ there, pp. $73-76$. Alternatively, you can find it in the original publication
Gaschütz, W.: Zur Erweiterungstheorie der endlichen Gruppen, J. reine angew. Math. 190 (1952), 93–107.