If G solvable and S a subgroup, then S is solvable [duplicate]
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Subgroups of finite solvable groups. Solvable?
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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 $$ { e}triangleleft H_rtriangleleft H_{r-1}cdots triangleleft H_1 triangleleft H_0=G $$ 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 $ineq r$, then it is trivial. So we need to assume that $Sneq H_i.$ Let $N_S$ be the normalizer of $S$ in $G$. We know that $Striangleleft N_S$ and $N_S$ is a subgroup of $G.$ Here I am stuck.
Can somebody propose a solution or give me some hint. Thanks.
abstract-algebra group-theory
asked Dec 27 '18 at 17:22
user249018
393127
marked as duplicate by Jyrki Lahtonen abstract-algebra
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Dec 27 '18 at 17:30
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This question already has an answer here:
Subgroups of finite solvable groups. Solvable?
2 answers
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 $$ { e}triangleleft H_rtriangleleft H_{r-1}cdots triangleleft H_1 triangleleft H_0=G $$ 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 $ineq r$, then it is trivial. So we need to assume that $Sneq H_i.$ Let $N_S$ be the normalizer of $S$ in $G$. We know that $Striangleleft N_S$ and $N_S$ is a subgroup of $G.$ Here I am stuck.
Can somebody propose a solution or give me some hint. Thanks.
abstract-algebra group-theory
asked Dec 27 '18 at 17:22
user249018
393127
marked as duplicate by Jyrki Lahtonen abstract-algebra
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Dec 27 '18 at 17:30
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This question already has an answer here:
Subgroups of finite solvable groups. Solvable?
2 answers
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 $$ { e}triangleleft H_rtriangleleft H_{r-1}cdots triangleleft H_1 triangleleft H_0=G $$ 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 $ineq r$, then it is trivial. So we need to assume that $Sneq H_i.$ Let $N_S$ be the normalizer of $S$ in $G$. We know that $Striangleleft N_S$ and $N_S$ is a subgroup of $G.$ Here I am stuck.
Can somebody propose a solution or give me some hint. Thanks.
abstract-algebra group-theory
asked Dec 27 '18 at 17:22
user249018
393127
This question already has an answer here:
Subgroups of finite solvable groups. Solvable?
2 answers
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 $$ { e}triangleleft H_rtriangleleft H_{r-1}cdots triangleleft H_1 triangleleft H_0=G $$ 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 $ineq r$, then it is trivial. So we need to assume that $Sneq H_i.$ Let $N_S$ be the normalizer of $S$ in $G$. We know that $Striangleleft N_S$ and $N_S$ is a subgroup of $G.$ Here I am stuck.
Can somebody propose a solution or give me some hint. Thanks.
This question already has an answer here:
Subgroups of finite solvable groups. Solvable?
2 answers
abstract-algebra group-theory
abstract-algebra group-theory
asked Dec 27 '18 at 17:22
user249018
393127
asked Dec 27 '18 at 17:22
user249018
393127
edited Dec 27 '18 at 17:28
asked Dec 27 '18 at 17:22
user249018
393127
asked Dec 27 '18 at 17:22
user249018
393127
asked Dec 27 '18 at 17:22
user249018
393127
393127
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Dec 27 '18 at 17:30
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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 $0leq ileq n-1$. Let $T$ be a subgroup $G$ either. For each $i$ we can define $T_i=Tcap H_i$ 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 $Tcap H_i=Tcap H_{i+1}$ we can form a series of $T$ composition.
$${e}=T_0’leq cdots T_m’=T$$
Then $T_{i+1}’/T_i’=(Tcap H_{i+1})/(Tcap H_i)$ is a cyclic group. Therefore $T$ es resolvable.
answered Dec 27 '18 at 17:47
Julio Trujillo Gonzalez
856
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1 Answer
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1 Answer
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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 $0leq ileq n-1$. Let $T$ be a subgroup $G$ either. For each $i$ we can define $T_i=Tcap H_i$ 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 $Tcap H_i=Tcap H_{i+1}$ we can form a series of $T$ composition.
$${e}=T_0’leq cdots T_m’=T$$
Then $T_{i+1}’/T_i’=(Tcap H_{i+1})/(Tcap H_i)$ is a cyclic group. Therefore $T$ es resolvable.
answered Dec 27 '18 at 17:47
Julio Trujillo Gonzalez
856
add a comment |
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 $0leq ileq n-1$. Let $T$ be a subgroup $G$ either. For each $i$ we can define $T_i=Tcap H_i$ 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 $Tcap H_i=Tcap H_{i+1}$ we can form a series of $T$ composition.
$${e}=T_0’leq cdots T_m’=T$$
Then $T_{i+1}’/T_i’=(Tcap H_{i+1})/(Tcap H_i)$ is a cyclic group. Therefore $T$ es resolvable.
answered Dec 27 '18 at 17:47
Julio Trujillo Gonzalez
856
add a comment |
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 $0leq ileq n-1$. Let $T$ be a subgroup $G$ either. For each $i$ we can define $T_i=Tcap H_i$ 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 $Tcap H_i=Tcap H_{i+1}$ we can form a series of $T$ composition.
$${e}=T_0’leq cdots T_m’=T$$
Then $T_{i+1}’/T_i’=(Tcap H_{i+1})/(Tcap H_i)$ is a cyclic group. Therefore $T$ es resolvable.
answered Dec 27 '18 at 17:47
Julio Trujillo Gonzalez
856
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 $0leq ileq n-1$. Let $T$ be a subgroup $G$ either. For each $i$ we can define $T_i=Tcap H_i$ 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 $Tcap H_i=Tcap H_{i+1}$ we can form a series of $T$ composition.
$${e}=T_0’leq cdots T_m’=T$$
Then $T_{i+1}’/T_i’=(Tcap H_{i+1})/(Tcap H_i)$ is a cyclic group. Therefore $T$ es resolvable.
answered Dec 27 '18 at 17:47
Julio Trujillo Gonzalez
856
answered Dec 27 '18 at 17:47
Julio Trujillo Gonzalez
856
answered Dec 27 '18 at 17:47
Julio Trujillo Gonzalez
856
answered Dec 27 '18 at 17:47
Julio Trujillo Gonzalez
856
856
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