Show that $G_i/G_{i+1} twoheadrightarrow (G_i +N)/ (G_{i+1} +N)$.
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From Aluffi's book of Algebra:
in which it refers to Theorem 7.12 :
[Red underlined:] I can't see any connection with Theorem 7.12! Actually I don't know how the $+$ appears. How to show that $G_i/G_{i+1} twoheadrightarrow (G_i +N)/ (G_{i+1} +N)$?
abstract-algebra proof-explanation
asked Nov 18 at 8:43
72D
50916
add a comment |
up vote
1
down vote
favorite
From Aluffi's book of Algebra:
in which it refers to Theorem 7.12 :
[Red underlined:] I can't see any connection with Theorem 7.12! Actually I don't know how the $+$ appears. How to show that $G_i/G_{i+1} twoheadrightarrow (G_i +N)/ (G_{i+1} +N)$?
abstract-algebra proof-explanation
asked Nov 18 at 8:43
72D
50916
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
From Aluffi's book of Algebra:
in which it refers to Theorem 7.12 :
[Red underlined:] I can't see any connection with Theorem 7.12! Actually I don't know how the $+$ appears. How to show that $G_i/G_{i+1} twoheadrightarrow (G_i +N)/ (G_{i+1} +N)$?
abstract-algebra proof-explanation
asked Nov 18 at 8:43
72D
50916
From Aluffi's book of Algebra:
in which it refers to Theorem 7.12 :
[Red underlined:] I can't see any connection with Theorem 7.12! Actually I don't know how the $+$ appears. How to show that $G_i/G_{i+1} twoheadrightarrow (G_i +N)/ (G_{i+1} +N)$?
abstract-algebra proof-explanation
abstract-algebra proof-explanation
asked Nov 18 at 8:43
72D
50916
asked Nov 18 at 8:43
72D
50916
edited Nov 18 at 18:34
asked Nov 18 at 8:43
72D
50916
asked Nov 18 at 8:43
72D
50916
asked Nov 18 at 8:43
72D
50916
50916
add a comment |
add a comment |
1 Answer
1
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up vote
1
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accepted
Since $G$ is understood to be a multiplicative group, I think the $+$ is a typo.
Let $G'=(G_iN)/(G_{i+1}N)$ and
begin{align}
&varphi:G_itwoheadrightarrow G'&
&xmapsto xG_{i+1}N
end{align}
Since $G_{i+1}subseteqoperatorname{Ker}varphi$, theorem 7.12 asserts that there exists one and only one group homomorphism $barvarphi:G_i/G_{i+1}to G'$ making the following diagram commutative:
$require{AMScd}$
begin{CD}
G_i@>varphi>>G'\
@Vpi VV @|\
G_i/G_{i+1}@>>barvarphi>G'
end{CD}
where $pi:G_ito G_i/G_{i+1}$ is the canonical projection onto the factor group.
Since $varphi$ is surjective, $barvarphi$ is surjective as well.
If $G_i/G_{i+1} $ is simple, then $operatorname{Ker}barvarphi={1} $ or $operatorname{Ker}barvarphi=G_i/G_{i+1} $, hence $barvarphi $ is an isomorphism or is trivial.
answered Nov 18 at 20:44
Fabio Lucchini
7,80311326
Why surjectivity of $barvarphi$ implies $G'$ is either trivial or isomorphic to $G_i/G_{i+1}$? I can't figure that out even knowing $G_i/G_{i+1}$ is simple.
– 72D
Nov 18 at 22:14
1
If $G_i/G_{i+1} $ is simple, then $operatorname{Ker}barvarphi={1} $ or $operatorname{Ker}barvarphi=G_i/G_{i+1} $.
– Fabio Lucchini
Nov 18 at 22:32
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Since $G$ is understood to be a multiplicative group, I think the $+$ is a typo.
Let $G'=(G_iN)/(G_{i+1}N)$ and
begin{align}
&varphi:G_itwoheadrightarrow G'&
&xmapsto xG_{i+1}N
end{align}
Since $G_{i+1}subseteqoperatorname{Ker}varphi$, theorem 7.12 asserts that there exists one and only one group homomorphism $barvarphi:G_i/G_{i+1}to G'$ making the following diagram commutative:
$require{AMScd}$
begin{CD}
G_i@>varphi>>G'\
@Vpi VV @|\
G_i/G_{i+1}@>>barvarphi>G'
end{CD}
where $pi:G_ito G_i/G_{i+1}$ is the canonical projection onto the factor group.
Since $varphi$ is surjective, $barvarphi$ is surjective as well.
If $G_i/G_{i+1} $ is simple, then $operatorname{Ker}barvarphi={1} $ or $operatorname{Ker}barvarphi=G_i/G_{i+1} $, hence $barvarphi $ is an isomorphism or is trivial.
answered Nov 18 at 20:44
Fabio Lucchini
7,80311326
Why surjectivity of $barvarphi$ implies $G'$ is either trivial or isomorphic to $G_i/G_{i+1}$? I can't figure that out even knowing $G_i/G_{i+1}$ is simple.
– 72D
Nov 18 at 22:14
1
If $G_i/G_{i+1} $ is simple, then $operatorname{Ker}barvarphi={1} $ or $operatorname{Ker}barvarphi=G_i/G_{i+1} $.
– Fabio Lucchini
Nov 18 at 22:32
add a comment |
up vote
1
down vote
accepted
Since $G$ is understood to be a multiplicative group, I think the $+$ is a typo.
Let $G'=(G_iN)/(G_{i+1}N)$ and
begin{align}
&varphi:G_itwoheadrightarrow G'&
&xmapsto xG_{i+1}N
end{align}
Since $G_{i+1}subseteqoperatorname{Ker}varphi$, theorem 7.12 asserts that there exists one and only one group homomorphism $barvarphi:G_i/G_{i+1}to G'$ making the following diagram commutative:
$require{AMScd}$
begin{CD}
G_i@>varphi>>G'\
@Vpi VV @|\
G_i/G_{i+1}@>>barvarphi>G'
end{CD}
where $pi:G_ito G_i/G_{i+1}$ is the canonical projection onto the factor group.
Since $varphi$ is surjective, $barvarphi$ is surjective as well.
If $G_i/G_{i+1} $ is simple, then $operatorname{Ker}barvarphi={1} $ or $operatorname{Ker}barvarphi=G_i/G_{i+1} $, hence $barvarphi $ is an isomorphism or is trivial.
answered Nov 18 at 20:44
Fabio Lucchini
7,80311326
Why surjectivity of $barvarphi$ implies $G'$ is either trivial or isomorphic to $G_i/G_{i+1}$? I can't figure that out even knowing $G_i/G_{i+1}$ is simple.
– 72D
Nov 18 at 22:14
1
If $G_i/G_{i+1} $ is simple, then $operatorname{Ker}barvarphi={1} $ or $operatorname{Ker}barvarphi=G_i/G_{i+1} $.
– Fabio Lucchini
Nov 18 at 22:32
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Since $G$ is understood to be a multiplicative group, I think the $+$ is a typo.
Let $G'=(G_iN)/(G_{i+1}N)$ and
begin{align}
&varphi:G_itwoheadrightarrow G'&
&xmapsto xG_{i+1}N
end{align}
Since $G_{i+1}subseteqoperatorname{Ker}varphi$, theorem 7.12 asserts that there exists one and only one group homomorphism $barvarphi:G_i/G_{i+1}to G'$ making the following diagram commutative:
$require{AMScd}$
begin{CD}
G_i@>varphi>>G'\
@Vpi VV @|\
G_i/G_{i+1}@>>barvarphi>G'
end{CD}
where $pi:G_ito G_i/G_{i+1}$ is the canonical projection onto the factor group.
Since $varphi$ is surjective, $barvarphi$ is surjective as well.
If $G_i/G_{i+1} $ is simple, then $operatorname{Ker}barvarphi={1} $ or $operatorname{Ker}barvarphi=G_i/G_{i+1} $, hence $barvarphi $ is an isomorphism or is trivial.
answered Nov 18 at 20:44
Fabio Lucchini
7,80311326
Since $G$ is understood to be a multiplicative group, I think the $+$ is a typo.
Let $G'=(G_iN)/(G_{i+1}N)$ and
begin{align}
&varphi:G_itwoheadrightarrow G'&
&xmapsto xG_{i+1}N
end{align}
Since $G_{i+1}subseteqoperatorname{Ker}varphi$, theorem 7.12 asserts that there exists one and only one group homomorphism $barvarphi:G_i/G_{i+1}to G'$ making the following diagram commutative:
$require{AMScd}$
begin{CD}
G_i@>varphi>>G'\
@Vpi VV @|\
G_i/G_{i+1}@>>barvarphi>G'
end{CD}
where $pi:G_ito G_i/G_{i+1}$ is the canonical projection onto the factor group.
Since $varphi$ is surjective, $barvarphi$ is surjective as well.
If $G_i/G_{i+1} $ is simple, then $operatorname{Ker}barvarphi={1} $ or $operatorname{Ker}barvarphi=G_i/G_{i+1} $, hence $barvarphi $ is an isomorphism or is trivial.
answered Nov 18 at 20:44
Fabio Lucchini
7,80311326
edited Nov 18 at 22:36
answered Nov 18 at 20:44
Fabio Lucchini
7,80311326
answered Nov 18 at 20:44
Fabio Lucchini
7,80311326
answered Nov 18 at 20:44
Fabio Lucchini
7,80311326
7,80311326
Why surjectivity of $barvarphi$ implies $G'$ is either trivial or isomorphic to $G_i/G_{i+1}$? I can't figure that out even knowing $G_i/G_{i+1}$ is simple.
– 72D
Nov 18 at 22:14
1
If $G_i/G_{i+1} $ is simple, then $operatorname{Ker}barvarphi={1} $ or $operatorname{Ker}barvarphi=G_i/G_{i+1} $.
– Fabio Lucchini
Nov 18 at 22:32
add a comment |
Why surjectivity of $barvarphi$ implies $G'$ is either trivial or isomorphic to $G_i/G_{i+1}$? I can't figure that out even knowing $G_i/G_{i+1}$ is simple.
– 72D
Nov 18 at 22:14
1
If $G_i/G_{i+1} $ is simple, then $operatorname{Ker}barvarphi={1} $ or $operatorname{Ker}barvarphi=G_i/G_{i+1} $.
– Fabio Lucchini
Nov 18 at 22:32
Why surjectivity of $barvarphi$ implies $G'$ is either trivial or isomorphic to $G_i/G_{i+1}$? I can't figure that out even knowing $G_i/G_{i+1}$ is simple.
– 72D
Nov 18 at 22:14
Why surjectivity of $barvarphi$ implies $G'$ is either trivial or isomorphic to $G_i/G_{i+1}$? I can't figure that out even knowing $G_i/G_{i+1}$ is simple.
– 72D
Nov 18 at 22:14
1
1
If $G_i/G_{i+1} $ is simple, then $operatorname{Ker}barvarphi={1} $ or $operatorname{Ker}barvarphi=G_i/G_{i+1} $.
– Fabio Lucchini
Nov 18 at 22:32
If $G_i/G_{i+1} $ is simple, then $operatorname{Ker}barvarphi={1} $ or $operatorname{Ker}barvarphi=G_i/G_{i+1} $.
– Fabio Lucchini
Nov 18 at 22:32
add a comment |
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