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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 (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 (2) implies (1).

Please help.

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    $\begingroup$ What you said so far was OK. You are trying to prove a known theorem of Gaschutz. I don't know a way to do the other implication without some use of group cohomology. $\endgroup$
    – Geoff Robinson
    Commented Jun 17, 2020 at 10:21
  • $\begingroup$ @Geoff Robinson: Thank you very much for your comment , and I'm okay to use Group Cohomology (as I have also given the tag ...) ... could you please detail out a proof or give any reference ? Thanks again $\endgroup$
    – Louis
    Commented Jun 17, 2020 at 10:29
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    $\begingroup$ This is an old and well-known result, and it also follows easily from basic results on the cohomology of finite groups, which you can find in the classic text by Cartan and Eilenberg on Homological Algebra. The restriction map $H^k(G/A,A) \to H^k(P/A,A)$ is injective for all $k > 0$. So I don't think it is really a suitable question for MO. $\endgroup$
    – Derek Holt
    Commented Jun 17, 2020 at 12:27
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    $\begingroup$ There are proofs that do not use the language of cohomology explicitly. The extension splits if and only if the identity map on $A$ extends to a crossed homomorphism $G \to A$, the kernel of which is a complement. There is a short slick proof in which the existence of such a crossed homomorphism $P \to A$ is used to prove that it exists for $G \to A$ (although it is not true that every such map $P \to A$ extends to $G \to A$). I don't know a reference for that immediately. $\endgroup$
    – Derek Holt
    Commented Jun 17, 2020 at 13:24
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    $\begingroup$ I have found a reference: (10.4) in Aschbacher's book, "Finite Group Theory". $\endgroup$
    – Derek Holt
    Commented Jun 17, 2020 at 13:34

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