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A group G is called solvable if it has a subnormal series whose factor groups (quotient groups) are all abelian, that is, if there are subgroups {1} = $G_0 < G_1 < ⋅⋅⋅ < G_k = G$ such that $G_{j − 1}$ is normal in $G_j$, and $G_j/G_{j − 1}$ is an abelian group, for $j = 1, 2, …, k$.

Question : How to prove that maximal normal solvable subgroup of $G \le S_n$ is unique ?

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  • $\begingroup$ Well, $S_n$ doesn't have very many normal subgroups... $\endgroup$ – Eric Wofsey Dec 30 '17 at 7:20
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We'll actually show that this result holds for any finite group $G$. I claim that $$M = N_1 \cdots N_k,$$ where $\left\{N_1, \dots, N_k \right\}$ is the set of all solvable normal subgroups of $G$, is the group we want. Note that $\{1\}$ is a solvable normal subgroup of $G$, so this product is not empty.

First, for any two solvable normal subgroups $H, N \leq G$, we know that $HN$ is a normal subgroup of $G$. We want to show that $HN$ is also solvable. For this, note that a group $K$ is solvable if and only if it contains a normal subgroup $L$ for which both $L$ and $K/L$ are solvable; in the case that $K$ is solvable, this property holds for all normal subgroups $L$ of $K$. We'll take $K = HN$ and $L = N$. Then by the second isomorphism theorem, $$HN/N \cong H/\left(H \cap N\right),$$ where $N$ is solvable by assumption and, since $H$ is solvable and $H \cap N$ is normal in $H$, $H/\left( H \cap N\right)$ is also solvable, i.e., $HN/N$ is solvable. This completes the proof that $HN$ is a solvable normal subgroup of $G$.

It follows by induction that any finite product of solvable normal subgroups of $G$ is a solvable normal subgroup of $G$. Since we have taken $G$ to be finite, it can have only finitely many subgroups, so the $M$ we defined above is a solvable normal subgroup of $G$. Indeed, $M$ is the unique maximal such subgroup because any other maximal solvable normal subgroup $M'$ of $G$ is both contained in $M$ (by construction) and contains $M$ (by maximality of $M'$), so $M' = M$.

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