Commutator subgroup $G'$ of a solvable group $G$ is not $G$
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Here is the question:
Let $G$ be a solvable group. Prove that $G neq G'$.
Is this correct?
Suppose $G=G'$, then the only abelian quotient of $G$ must be ${1}=G/G'$. So there is no proper normal subgroup $H$ of $G$ that gives abelian quotient $G/H$. So then in the normal serie, it should end with $G triangleright G triangleright {1}$. but $G/{1}$ doesn't give abelian composition factor. I'm just unsure if each $G_i$ is the definition of composition series must be proper normal subgroup of $G_{i-1}$?
Here is definition of solvable group(before lemma 25.1.1):
group-theory
asked Nov 20 at 14:54
mathnoob
1,747322
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show 4 more comments
up vote
0
down vote
favorite
Here is the question:
Let $G$ be a solvable group. Prove that $G neq G'$.
Is this correct?
Suppose $G=G'$, then the only abelian quotient of $G$ must be ${1}=G/G'$. So there is no proper normal subgroup $H$ of $G$ that gives abelian quotient $G/H$. So then in the normal serie, it should end with $G triangleright G triangleright {1}$. but $G/{1}$ doesn't give abelian composition factor. I'm just unsure if each $G_i$ is the definition of composition series must be proper normal subgroup of $G_{i-1}$?
Here is definition of solvable group(before lemma 25.1.1):
group-theory
asked Nov 20 at 14:54
mathnoob
1,747322
3
Not every solvable group has a composition series. For example $({mathbb Z},+)$ has no such series. So you should prove it without using composition series.
– Derek Holt
Nov 20 at 14:56
Ok I guess I meant normal series. That is how my note defined solvable groups
– mathnoob
Nov 20 at 15:01
It is not correct for the trivial group $G=1$, where we have $G=G'$. Otherwise it follows from the derived series.
– Dietrich Burde
Nov 20 at 15:04
OK, so $G$ has a normal series $G =G_0 > G_1 > cdots G_k = 1$, where $G/G_1$ is abelian, so $G' le G_1$ and we are done. Of course, as Dietrich Burde points out, this only works if $|G|>1$.
– Derek Holt
Nov 20 at 15:07
It'd perhaps be helpful, for the question, to write down what your definition of "solvable group" is. Sometimes it requires something on the derived series, some other times it says something about a subnormal series with abelian factors...
– DonAntonio
Nov 20 at 15:07
|
show 4 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Here is the question:
Let $G$ be a solvable group. Prove that $G neq G'$.
Is this correct?
Suppose $G=G'$, then the only abelian quotient of $G$ must be ${1}=G/G'$. So there is no proper normal subgroup $H$ of $G$ that gives abelian quotient $G/H$. So then in the normal serie, it should end with $G triangleright G triangleright {1}$. but $G/{1}$ doesn't give abelian composition factor. I'm just unsure if each $G_i$ is the definition of composition series must be proper normal subgroup of $G_{i-1}$?
Here is definition of solvable group(before lemma 25.1.1):
group-theory
asked Nov 20 at 14:54
mathnoob
1,747322
Here is the question:
Let $G$ be a solvable group. Prove that $G neq G'$.
Is this correct?
Suppose $G=G'$, then the only abelian quotient of $G$ must be ${1}=G/G'$. So there is no proper normal subgroup $H$ of $G$ that gives abelian quotient $G/H$. So then in the normal serie, it should end with $G triangleright G triangleright {1}$. but $G/{1}$ doesn't give abelian composition factor. I'm just unsure if each $G_i$ is the definition of composition series must be proper normal subgroup of $G_{i-1}$?
Here is definition of solvable group(before lemma 25.1.1):
group-theory
group-theory
asked Nov 20 at 14:54
mathnoob
1,747322
asked Nov 20 at 14:54
mathnoob
1,747322
edited Nov 20 at 15:35
asked Nov 20 at 14:54
mathnoob
1,747322
asked Nov 20 at 14:54
mathnoob
1,747322
asked Nov 20 at 14:54
mathnoob
1,747322
1,747322
3
Not every solvable group has a composition series. For example $({mathbb Z},+)$ has no such series. So you should prove it without using composition series.
– Derek Holt
Nov 20 at 14:56
Ok I guess I meant normal series. That is how my note defined solvable groups
– mathnoob
Nov 20 at 15:01
It is not correct for the trivial group $G=1$, where we have $G=G'$. Otherwise it follows from the derived series.
– Dietrich Burde
Nov 20 at 15:04
OK, so $G$ has a normal series $G =G_0 > G_1 > cdots G_k = 1$, where $G/G_1$ is abelian, so $G' le G_1$ and we are done. Of course, as Dietrich Burde points out, this only works if $|G|>1$.
– Derek Holt
Nov 20 at 15:07
It'd perhaps be helpful, for the question, to write down what your definition of "solvable group" is. Sometimes it requires something on the derived series, some other times it says something about a subnormal series with abelian factors...
– DonAntonio
Nov 20 at 15:07
|
show 4 more comments
3
Not every solvable group has a composition series. For example $({mathbb Z},+)$ has no such series. So you should prove it without using composition series.
– Derek Holt
Nov 20 at 14:56
Ok I guess I meant normal series. That is how my note defined solvable groups
– mathnoob
Nov 20 at 15:01
It is not correct for the trivial group $G=1$, where we have $G=G'$. Otherwise it follows from the derived series.
– Dietrich Burde
Nov 20 at 15:04
OK, so $G$ has a normal series $G =G_0 > G_1 > cdots G_k = 1$, where $G/G_1$ is abelian, so $G' le G_1$ and we are done. Of course, as Dietrich Burde points out, this only works if $|G|>1$.
– Derek Holt
Nov 20 at 15:07
It'd perhaps be helpful, for the question, to write down what your definition of "solvable group" is. Sometimes it requires something on the derived series, some other times it says something about a subnormal series with abelian factors...
– DonAntonio
Nov 20 at 15:07
3
3
Not every solvable group has a composition series. For example $({mathbb Z},+)$ has no such series. So you should prove it without using composition series.
– Derek Holt
Nov 20 at 14:56
Not every solvable group has a composition series. For example $({mathbb Z},+)$ has no such series. So you should prove it without using composition series.
– Derek Holt
Nov 20 at 14:56
Ok I guess I meant normal series. That is how my note defined solvable groups
– mathnoob
Nov 20 at 15:01
Ok I guess I meant normal series. That is how my note defined solvable groups
– mathnoob
Nov 20 at 15:01
It is not correct for the trivial group $G=1$, where we have $G=G'$. Otherwise it follows from the derived series.
– Dietrich Burde
Nov 20 at 15:04
It is not correct for the trivial group $G=1$, where we have $G=G'$. Otherwise it follows from the derived series.
– Dietrich Burde
Nov 20 at 15:04
OK, so $G$ has a normal series $G =G_0 > G_1 > cdots G_k = 1$, where $G/G_1$ is abelian, so $G' le G_1$ and we are done. Of course, as Dietrich Burde points out, this only works if $|G|>1$.
– Derek Holt
Nov 20 at 15:07
OK, so $G$ has a normal series $G =G_0 > G_1 > cdots G_k = 1$, where $G/G_1$ is abelian, so $G' le G_1$ and we are done. Of course, as Dietrich Burde points out, this only works if $|G|>1$.
– Derek Holt
Nov 20 at 15:07
It'd perhaps be helpful, for the question, to write down what your definition of "solvable group" is. Sometimes it requires something on the derived series, some other times it says something about a subnormal series with abelian factors...
– DonAntonio
Nov 20 at 15:07
It'd perhaps be helpful, for the question, to write down what your definition of "solvable group" is. Sometimes it requires something on the derived series, some other times it says something about a subnormal series with abelian factors...
– DonAntonio
Nov 20 at 15:07
|
show 4 more comments
1 Answer
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I think that you are requiring that $G$ is not trivial.
We have two cases:
1) If $G$ is an abelian group. Then $G'=0$, but $G=G'$ is not trivial. Contradiction.
2) If $G$ is not abelian, and $G'=G$ fails the fact that $G$ is solvable.
answered Nov 20 at 15:37
José Alejandro Aburto Araneda
41019
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I think that you are requiring that $G$ is not trivial.
We have two cases:
1) If $G$ is an abelian group. Then $G'=0$, but $G=G'$ is not trivial. Contradiction.
2) If $G$ is not abelian, and $G'=G$ fails the fact that $G$ is solvable.
answered Nov 20 at 15:37
José Alejandro Aburto Araneda
41019
add a comment |
up vote
0
down vote
accepted
I think that you are requiring that $G$ is not trivial.
We have two cases:
1) If $G$ is an abelian group. Then $G'=0$, but $G=G'$ is not trivial. Contradiction.
2) If $G$ is not abelian, and $G'=G$ fails the fact that $G$ is solvable.
answered Nov 20 at 15:37
José Alejandro Aburto Araneda
41019
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
I think that you are requiring that $G$ is not trivial.
We have two cases:
1) If $G$ is an abelian group. Then $G'=0$, but $G=G'$ is not trivial. Contradiction.
2) If $G$ is not abelian, and $G'=G$ fails the fact that $G$ is solvable.
answered Nov 20 at 15:37
José Alejandro Aburto Araneda
41019
I think that you are requiring that $G$ is not trivial.
We have two cases:
1) If $G$ is an abelian group. Then $G'=0$, but $G=G'$ is not trivial. Contradiction.
2) If $G$ is not abelian, and $G'=G$ fails the fact that $G$ is solvable.
answered Nov 20 at 15:37
José Alejandro Aburto Araneda
41019
edited Nov 21 at 1:54
answered Nov 20 at 15:37
José Alejandro Aburto Araneda
41019
answered Nov 20 at 15:37
José Alejandro Aburto Araneda
41019
answered Nov 20 at 15:37
José Alejandro Aburto Araneda
41019
41019
add a comment |
add a comment |
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Not every solvable group has a composition series. For example $({mathbb Z},+)$ has no such series. So you should prove it without using composition series.
– Derek Holt
Nov 20 at 14:56
Ok I guess I meant normal series. That is how my note defined solvable groups
– mathnoob
Nov 20 at 15:01
It is not correct for the trivial group $G=1$, where we have $G=G'$. Otherwise it follows from the derived series.
– Dietrich Burde
Nov 20 at 15:04
OK, so $G$ has a normal series $G =G_0 > G_1 > cdots G_k = 1$, where $G/G_1$ is abelian, so $G' le G_1$ and we are done. Of course, as Dietrich Burde points out, this only works if $|G|>1$.
– Derek Holt
Nov 20 at 15:07
It'd perhaps be helpful, for the question, to write down what your definition of "solvable group" is. Sometimes it requires something on the derived series, some other times it says something about a subnormal series with abelian factors...
– DonAntonio
Nov 20 at 15:07