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Let $G$ be a group having a Sylow Tower, as defined here. I want to prove that for every $N\triangleleft G$, $G/N$ has a nontrivial normal Sylow subgroup.

I started like this:

Let first $N=1$. Then $P_1$ is a nontrivial normal Sylow subgroup of $G\cong G/N$.

Here is the tricky part.

Now, let's suppose $N>1$, and let $k\in \{1,\dots,r\}$ such that $P_k\nleq N$. Then if $P$ is a Sylow subgroup of $G$ such that $P_k/P_{k-1}\cong P$, then $PN/N$ is a Sylow subgroup of $G$, and since $P\trianglelefteq G$ and $N\trianglelefteq G$, $PN\trianglelefteq G$ and then $PN/N\trianglelefteq G$.

But there is a wrong step in there. I never had that $P\trianglelefteq G$, so I can't say anything else there. So, is there any way to correct the proof so that it works, or must it be redone?

Another thing: The proposition is in fact an equivalence. I don't need the converse, but, by curiosity, is there any place where I could find it, or any idea to prove it? Or in general, is there any book/place where I could find these kind of results for Sylow Towers? The ones I have searched in don't have much more than the definition (So I'm trying to prove this needed result by myself).

Edit: Another Idea I just thought was, since $P_1$ is a normal Sylow subgroup of $G$, then $P_1N/N$ is a normal Sylow subgroup of $G/N$. But I can't say that $P_1N/N$ is not trivial. Furthermore in my original proof what I needed is that $P\nleq N$, not that $P_k\neq N$.

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    $\begingroup$ Take the smallest $k$ such that $P_k\not\leq N$, then consider $P_kN/N$. $\endgroup$
    – verret
    Sep 7 '17 at 4:14
  • $\begingroup$ It seems it works. Thanks. $\endgroup$ Sep 7 '17 at 5:01
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    $\begingroup$ I'll make it an answer, just so this doesn't remain unanswered. $\endgroup$
    – verret
    Sep 7 '17 at 18:44
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Take the smallest $k$ such that $P_k\not\leq N$, then consider $P_kN/N$.

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