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I have $G$ non-trivial group, $A$ is a non-trivial abelian group with non-trivial automorphism group. I have that every short exact sequence $0\rightarrow A\rightarrow B\rightarrow G\rightarrow 0$ splits. This gives that $G$ is free. So we have non-trivial homomorphism $\phi:G\rightarrow \text{Aut}(A)$ by def of free group. So we look at $0\rightarrow A\overset{f}{\rightarrow}A\rtimes_\phi G\overset{g}{\rightarrow}G\rightarrow 0$ where $f,g$ the obvious maps. This is a short exact sequence. I have a theorem that says there exists section $s$ such that $s(G)\trianglelefteq A\rtimes_\phi G$. The next claim that the author makes in the one that confuses me, he says that we then have $s(G)\cap A=\{e\}$. I am not sure we can make that claim, it seems logical that $s(g)=(e,g)$ but are we guaranteed this? I know that $A\cap G=\{e\}$.

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    $\begingroup$ How are you getting that $G$ is free, exactly? For instance, $G=\mathbb{Z}/5$ and $A=\mathbb{Z}/3$ satisfy the hypotheses you stated initially. $\endgroup$ Commented Mar 8, 2017 at 23:19
  • $\begingroup$ We are actually getting a contradiction out of this. $\endgroup$
    – tmpys
    Commented Mar 8, 2017 at 23:36
  • $\begingroup$ What author? ${}$ $\endgroup$ Commented Mar 8, 2017 at 23:46
  • $\begingroup$ math.leidenuniv.nl/scripties/AponBach.pdf Thm 4.2 $\endgroup$
    – tmpys
    Commented Mar 9, 2017 at 0:04

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Suppose $x\in s(G)\cap A$. Since $x\in s(G)$, we can write $x=s(y)$ for some $y\in G$. We then have $g(x)=g(s(y))=y$. But since $x\in A=\ker(g)$, $g(x)=e$, so $y=e$. Thus $x=s(e)=e$.

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