Prove that every finite group $G$ contains a (unique) soluble normal subgroup $N$ such that $G/N$ has no nontrivial abelian normal subgroups.

Thanks a lot.

To show that $G/N$ has no nontrivial abelian normal subgroup, I think we can use the fact that if $H$ is a group such that $N$ is normal in $H$ with the property that $H/N$ and $N$ are soluble then $H$ is soluble. Choose $H$ such that $H/N$ is abelian.

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    $\begingroup$ Can you please share what you've tried, and explain what's giving you trouble? I notice that you've asked quite a few questions without showing any thoughts or efforts on any - please edit your post to include this information. $\endgroup$ – user61527 Dec 11 '13 at 6:07
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    $\begingroup$ This is slightly advanced group theory: you should by now know some stuff as to, at least, give a little more self effort and work. $\endgroup$ – DonAntonio Dec 11 '13 at 6:11

If $M$ and $N$ are normal soluble subgroups of a group $G$, then $MN$ is again a soluble normal subgroup. For $MN/N \cong M/(M \cap N)$, and $M/(M \cap N)$ is soluble as quotient of the soluble $M$, so $MN/N$ is soluble, $N$ is soluble, so (also according to your valid remark) $MN$ must be soluble. This shows that in every group there is a unique maximal soluble normal subgroup (it could be trivial). As you did call this group $N$.

Now assume $K/N$ is a non-trivial abelian normal subgroup of $G/N$. Then $N \lt K \lhd G$, with the commutator subgroup $K' \subseteq N$. Since $N$ is soluble, $K'$ is soluble and hence $K$ must be soluble. By the soluble normal maximality of $N$, $K \subseteq N$, which is a contradiction.

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    $\begingroup$ You are welcome, and I do not see why people downvoted your question. $\endgroup$ – Nicky Hekster Dec 11 '13 at 7:00
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    $\begingroup$ @NickyHekster This was explained in the comments. $\endgroup$ – Did Dec 11 '13 at 7:18
  • $\begingroup$ I am not clear with how the existence of maximal normal subgroup is proven. My idea is : Let $M$ be a normal soluble subgroup. Suppose it is not maximal, then $M<N<G$ for some normal soluble subgroup $N$. Then $MN$ is normal soluble. If $MN$ is not maximal, repeat the procedure to get $M<N\leq MN < \dots$. Since $G$ is finite, the sequence must terminate. If my proof is correct, yet I am not sure how to show that it is unique. $\endgroup$ – Alan Wang Mar 20 '17 at 8:26
  • $\begingroup$ Your proof is correct. The normal soluble subgroup asked for by the OP is the set-theoretic product of all the soluble normal subgroups of $G$. Call this $N$. Then if $N \subseteq M \unlhd G$ for some soluble subgroup $M$, then by definition of $N$, we must have $M \subseteq N$, whence $M=N$. If $N_1$ would be another largest soluble normal subgroup of $G$, then $NN_1$ is also soluble and normal, so $NN_1 \subseteq N$, which implies $N_1 \subseteq N$. And this reasoning works the other way around, hence $N \subseteq N_1$, so $N=N_1$, proving uniqueness. $\endgroup$ – Nicky Hekster Mar 20 '17 at 12:59
  • $\begingroup$ What if N=G? Then you can't use the maximality to reason that N is a subset of N1. I'm stuck at this problem too. Can anybody explain? $\endgroup$ – Minh Khôi Dec 26 '19 at 13:37

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