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Let $G$ be a solvable group, and $S$ a subgroup. Prove that $S$ is solvable.

Let $G$ be solvable. Then there exists a sequence of subgroups $$ \{ e\}\triangleleft H_r\triangleleft H_{r-1}\cdots \triangleleft H_1 \triangleleft H_0=G $$ such that $H_{i+1}\triangleleft H_i$ for $i=0, ...r-1,$ and $H_i/H_{i+1}$ is abelian. If $S=H_i$, where $H_i$ is as above for $i\neq r$, then it is trivial. So we need to assume that $S\neq H_i.$ Let $N_S$ be the normalizer of $S$ in $G$. We know that $S\triangleleft N_S$ and $N_S$ is a subgroup of $G.$ Here I am stuck.

Can somebody propose a solution or give me some hint. Thanks.

user249018
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1 Answers1

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suppose that $G$ is resolvable and that $\{H_i\}_0^n$ is a series of composition of $G$ such that $H_{i+1}/H_i$ is cyclic $0\leq i\leq n-1$. Let $T$ be a subgroup $G$ either. For each $i$ we can define $T_i=T\cap H_i$ so that $T_i$ is a subgroup of $T$. Avoiding the duplication of those $T_i$ that are repeated, that is, those that comply with $T\cap H_i=T\cap H_{i+1}$ we can form a series of $T$ composition. $$\{e\}=T_0’\leq \cdots T_m’=T$$ Then $T_{i+1}’/T_i’=(T\cap H_{i+1})/(T\cap H_i)$ is a cyclic group. Therefore $T$ es resolvable.

Julio Trujillo Gonzalez
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