Let $N$ be an Abelian normal subgroup of $G$. If $G/N$ is perfect, then also $G'$ is perfect.
$begingroup$
I'm reading Hans Kurzweil's The Theory of Infinite Group, where it says
1.5.3 Let $N$ be an Abelian normal subgroup of $G$. If $G/N$ is perfect, then also $G'$ is perfect.
Proof. From 1.5.1, applied to the natural epimorphism, we obtain
$$G/N = (G/N)' = G'N/N$$ and thus $G = G'N$. Since also $G'/N cap G' (cong G/N)$ is perfect, the same argument gives $G' = G''(N cap G')$. It
follows that $G = G''N$ and $G/G'' cong N/N cap G''$. Now 1.5.2
implies $G' = G''$ since $N$ is Abelian. $square$
My first question is the motivation of the theorem.
By motivation, for example, about commutator subgroup $G'$, wiki explains,
The commutator subgroup is important because it is the smallest normal
subgroup such that the quotient group of the original group by this
subgroup is abelian. In other words, $G/N$ is abelian if and only if
$N$ contains the commutator subgroup. So in some sense it provides a
measure of how far the group is from being abelian; the larger the
commutator subgroup is, the "less abelian" the group is.
So if a group/subgroup is perfect, it's quite "non-abelian".
What is the motivation of the theorem here? 1.5.3 says if $G/N$ is perfect, i.e. quite non-Abelian, then also $G'$ is perfect, i.e. quite non-Abelian, what's the idea of linking $G/N$ with $G'$? What is the character of Abelian normal subgroup this theorem is describing? or it's simple a useful tool like some integral formula that would be used later?
Secondly, what is the relationship between $G/N$ and $G'$? $G=G'N$, does it mean $G = G' times N$?
abstract-algebra
asked Jan 4 at 23:33
athosathos
98411340
$endgroup$
add a comment |
$begingroup$
I'm reading Hans Kurzweil's The Theory of Infinite Group, where it says
1.5.3 Let $N$ be an Abelian normal subgroup of $G$. If $G/N$ is perfect, then also $G'$ is perfect.
Proof. From 1.5.1, applied to the natural epimorphism, we obtain
$$G/N = (G/N)' = G'N/N$$ and thus $G = G'N$. Since also $G'/N cap G' (cong G/N)$ is perfect, the same argument gives $G' = G''(N cap G')$. It
follows that $G = G''N$ and $G/G'' cong N/N cap G''$. Now 1.5.2
implies $G' = G''$ since $N$ is Abelian. $square$
My first question is the motivation of the theorem.
By motivation, for example, about commutator subgroup $G'$, wiki explains,
The commutator subgroup is important because it is the smallest normal
subgroup such that the quotient group of the original group by this
subgroup is abelian. In other words, $G/N$ is abelian if and only if
$N$ contains the commutator subgroup. So in some sense it provides a
measure of how far the group is from being abelian; the larger the
commutator subgroup is, the "less abelian" the group is.
So if a group/subgroup is perfect, it's quite "non-abelian".
What is the motivation of the theorem here? 1.5.3 says if $G/N$ is perfect, i.e. quite non-Abelian, then also $G'$ is perfect, i.e. quite non-Abelian, what's the idea of linking $G/N$ with $G'$? What is the character of Abelian normal subgroup this theorem is describing? or it's simple a useful tool like some integral formula that would be used later?
Secondly, what is the relationship between $G/N$ and $G'$? $G=G'N$, does it mean $G = G' times N$?
abstract-algebra
asked Jan 4 at 23:33
athosathos
98411340
$endgroup$
$begingroup$
No, $G’N$ is the set ${gnmid gin G’, nin N}$; it need not be isomorphic to the (or equal to the internal) direct product of $G’$ and $N$, except in the case where $G’cap N={e}$. That need not be the ase in general.
$endgroup$
– Arturo Magidin
Jan 5 at 1:51
$begingroup$
You would want to see where this theorem is used in the book; off the top of my head, I would note that since $N$ is abelian, it is solvable, and hence the solvability of $G$ would depend on that of $G/N$. This tells you that if $G/N$ is perfect, then the derived series of $G$ gets “stuck” after a single application, which gives you a fair amount of information on $G’$ (being perfect is more than simply a measure of non-abelianness, as you will no doubt see further along in the book).
$endgroup$
– Arturo Magidin
Jan 5 at 1:53
add a comment |
$begingroup$
I'm reading Hans Kurzweil's The Theory of Infinite Group, where it says
1.5.3 Let $N$ be an Abelian normal subgroup of $G$. If $G/N$ is perfect, then also $G'$ is perfect.
Proof. From 1.5.1, applied to the natural epimorphism, we obtain
$$G/N = (G/N)' = G'N/N$$ and thus $G = G'N$. Since also $G'/N cap G' (cong G/N)$ is perfect, the same argument gives $G' = G''(N cap G')$. It
follows that $G = G''N$ and $G/G'' cong N/N cap G''$. Now 1.5.2
implies $G' = G''$ since $N$ is Abelian. $square$
My first question is the motivation of the theorem.
By motivation, for example, about commutator subgroup $G'$, wiki explains,
The commutator subgroup is important because it is the smallest normal
subgroup such that the quotient group of the original group by this
subgroup is abelian. In other words, $G/N$ is abelian if and only if
$N$ contains the commutator subgroup. So in some sense it provides a
measure of how far the group is from being abelian; the larger the
commutator subgroup is, the "less abelian" the group is.
So if a group/subgroup is perfect, it's quite "non-abelian".
What is the motivation of the theorem here? 1.5.3 says if $G/N$ is perfect, i.e. quite non-Abelian, then also $G'$ is perfect, i.e. quite non-Abelian, what's the idea of linking $G/N$ with $G'$? What is the character of Abelian normal subgroup this theorem is describing? or it's simple a useful tool like some integral formula that would be used later?
Secondly, what is the relationship between $G/N$ and $G'$? $G=G'N$, does it mean $G = G' times N$?
abstract-algebra
asked Jan 4 at 23:33
athosathos
98411340
$endgroup$
I'm reading Hans Kurzweil's The Theory of Infinite Group, where it says
1.5.3 Let $N$ be an Abelian normal subgroup of $G$. If $G/N$ is perfect, then also $G'$ is perfect.
Proof. From 1.5.1, applied to the natural epimorphism, we obtain
$$G/N = (G/N)' = G'N/N$$ and thus $G = G'N$. Since also $G'/N cap G' (cong G/N)$ is perfect, the same argument gives $G' = G''(N cap G')$. It
follows that $G = G''N$ and $G/G'' cong N/N cap G''$. Now 1.5.2
implies $G' = G''$ since $N$ is Abelian. $square$
My first question is the motivation of the theorem.
By motivation, for example, about commutator subgroup $G'$, wiki explains,
The commutator subgroup is important because it is the smallest normal
subgroup such that the quotient group of the original group by this
subgroup is abelian. In other words, $G/N$ is abelian if and only if
$N$ contains the commutator subgroup. So in some sense it provides a
measure of how far the group is from being abelian; the larger the
commutator subgroup is, the "less abelian" the group is.
So if a group/subgroup is perfect, it's quite "non-abelian".
What is the motivation of the theorem here? 1.5.3 says if $G/N$ is perfect, i.e. quite non-Abelian, then also $G'$ is perfect, i.e. quite non-Abelian, what's the idea of linking $G/N$ with $G'$? What is the character of Abelian normal subgroup this theorem is describing? or it's simple a useful tool like some integral formula that would be used later?
Secondly, what is the relationship between $G/N$ and $G'$? $G=G'N$, does it mean $G = G' times N$?
abstract-algebra
abstract-algebra
asked Jan 4 at 23:33
athosathos
98411340
asked Jan 4 at 23:33
athosathos
98411340
asked Jan 4 at 23:33
athosathos
98411340
asked Jan 4 at 23:33
athosathos
98411340
asked Jan 4 at 23:33
athosathos
98411340
98411340
$begingroup$
No, $G’N$ is the set ${gnmid gin G’, nin N}$; it need not be isomorphic to the (or equal to the internal) direct product of $G’$ and $N$, except in the case where $G’cap N={e}$. That need not be the ase in general.
$endgroup$
– Arturo Magidin
Jan 5 at 1:51
$begingroup$
You would want to see where this theorem is used in the book; off the top of my head, I would note that since $N$ is abelian, it is solvable, and hence the solvability of $G$ would depend on that of $G/N$. This tells you that if $G/N$ is perfect, then the derived series of $G$ gets “stuck” after a single application, which gives you a fair amount of information on $G’$ (being perfect is more than simply a measure of non-abelianness, as you will no doubt see further along in the book).
$endgroup$
– Arturo Magidin
Jan 5 at 1:53
add a comment |
$begingroup$
No, $G’N$ is the set ${gnmid gin G’, nin N}$; it need not be isomorphic to the (or equal to the internal) direct product of $G’$ and $N$, except in the case where $G’cap N={e}$. That need not be the ase in general.
$endgroup$
– Arturo Magidin
Jan 5 at 1:51
$begingroup$
You would want to see where this theorem is used in the book; off the top of my head, I would note that since $N$ is abelian, it is solvable, and hence the solvability of $G$ would depend on that of $G/N$. This tells you that if $G/N$ is perfect, then the derived series of $G$ gets “stuck” after a single application, which gives you a fair amount of information on $G’$ (being perfect is more than simply a measure of non-abelianness, as you will no doubt see further along in the book).
$endgroup$
– Arturo Magidin
Jan 5 at 1:53
$begingroup$
No, $G’N$ is the set ${gnmid gin G’, nin N}$; it need not be isomorphic to the (or equal to the internal) direct product of $G’$ and $N$, except in the case where $G’cap N={e}$. That need not be the ase in general.
$endgroup$
– Arturo Magidin
Jan 5 at 1:51
$begingroup$
No, $G’N$ is the set ${gnmid gin G’, nin N}$; it need not be isomorphic to the (or equal to the internal) direct product of $G’$ and $N$, except in the case where $G’cap N={e}$. That need not be the ase in general.
$endgroup$
– Arturo Magidin
Jan 5 at 1:51
$begingroup$
You would want to see where this theorem is used in the book; off the top of my head, I would note that since $N$ is abelian, it is solvable, and hence the solvability of $G$ would depend on that of $G/N$. This tells you that if $G/N$ is perfect, then the derived series of $G$ gets “stuck” after a single application, which gives you a fair amount of information on $G’$ (being perfect is more than simply a measure of non-abelianness, as you will no doubt see further along in the book).
$endgroup$
– Arturo Magidin
Jan 5 at 1:53
$begingroup$
You would want to see where this theorem is used in the book; off the top of my head, I would note that since $N$ is abelian, it is solvable, and hence the solvability of $G$ would depend on that of $G/N$. This tells you that if $G/N$ is perfect, then the derived series of $G$ gets “stuck” after a single application, which gives you a fair amount of information on $G’$ (being perfect is more than simply a measure of non-abelianness, as you will no doubt see further along in the book).
$endgroup$
– Arturo Magidin
Jan 5 at 1:53
add a comment |
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$begingroup$
No, $G’N$ is the set ${gnmid gin G’, nin N}$; it need not be isomorphic to the (or equal to the internal) direct product of $G’$ and $N$, except in the case where $G’cap N={e}$. That need not be the ase in general.
$endgroup$
– Arturo Magidin
Jan 5 at 1:51
$begingroup$
You would want to see where this theorem is used in the book; off the top of my head, I would note that since $N$ is abelian, it is solvable, and hence the solvability of $G$ would depend on that of $G/N$. This tells you that if $G/N$ is perfect, then the derived series of $G$ gets “stuck” after a single application, which gives you a fair amount of information on $G’$ (being perfect is more than simply a measure of non-abelianness, as you will no doubt see further along in the book).
$endgroup$
– Arturo Magidin
Jan 5 at 1:53